What do querySelectorAll, getElementsByClassName and other getElementsBy* methods return?
up vote
94
down vote
favorite
Do getElementsByClassName
(and similar functions like getElementsByTagName
and querySelectorAll
) work the same as getElementById
or do they return an array of elements?
The reason I ask is because I am trying to change the style of all elements using getElementsByClassName
. See below.
//doesn't work
document.getElementsByClassName('myElement').style.size = '100px';
//works
document.getElementById('myIdElement').style.size = '100px';
javascript getelementsbyclassname dom-traversal
add a comment |
up vote
94
down vote
favorite
Do getElementsByClassName
(and similar functions like getElementsByTagName
and querySelectorAll
) work the same as getElementById
or do they return an array of elements?
The reason I ask is because I am trying to change the style of all elements using getElementsByClassName
. See below.
//doesn't work
document.getElementsByClassName('myElement').style.size = '100px';
//works
document.getElementById('myIdElement').style.size = '100px';
javascript getelementsbyclassname dom-traversal
30
The clue is, very much, in the name:getElementsByClassName()
implies a plural, whereasgetElementById()
implies a singular element item.
– David Thomas
May 21 '12 at 23:20
1
I get that, it just didn't make sense to me that you can't change all the elements with that class name using the code above instead of having to loop through an array. jquery way is much better, i was just curious about the js way
– dmo
May 22 '12 at 4:34
1
Might be useful too: stackoverflow.com/questions/3871547/…
– kapa
Jul 31 '14 at 9:19
add a comment |
up vote
94
down vote
favorite
up vote
94
down vote
favorite
Do getElementsByClassName
(and similar functions like getElementsByTagName
and querySelectorAll
) work the same as getElementById
or do they return an array of elements?
The reason I ask is because I am trying to change the style of all elements using getElementsByClassName
. See below.
//doesn't work
document.getElementsByClassName('myElement').style.size = '100px';
//works
document.getElementById('myIdElement').style.size = '100px';
javascript getelementsbyclassname dom-traversal
Do getElementsByClassName
(and similar functions like getElementsByTagName
and querySelectorAll
) work the same as getElementById
or do they return an array of elements?
The reason I ask is because I am trying to change the style of all elements using getElementsByClassName
. See below.
//doesn't work
document.getElementsByClassName('myElement').style.size = '100px';
//works
document.getElementById('myIdElement').style.size = '100px';
javascript getelementsbyclassname dom-traversal
javascript getelementsbyclassname dom-traversal
edited Mar 22 '17 at 16:26
Taryn♦
188k45286351
188k45286351
asked May 21 '12 at 23:17
dmo
1,98421320
1,98421320
30
The clue is, very much, in the name:getElementsByClassName()
implies a plural, whereasgetElementById()
implies a singular element item.
– David Thomas
May 21 '12 at 23:20
1
I get that, it just didn't make sense to me that you can't change all the elements with that class name using the code above instead of having to loop through an array. jquery way is much better, i was just curious about the js way
– dmo
May 22 '12 at 4:34
1
Might be useful too: stackoverflow.com/questions/3871547/…
– kapa
Jul 31 '14 at 9:19
add a comment |
30
The clue is, very much, in the name:getElementsByClassName()
implies a plural, whereasgetElementById()
implies a singular element item.
– David Thomas
May 21 '12 at 23:20
1
I get that, it just didn't make sense to me that you can't change all the elements with that class name using the code above instead of having to loop through an array. jquery way is much better, i was just curious about the js way
– dmo
May 22 '12 at 4:34
1
Might be useful too: stackoverflow.com/questions/3871547/…
– kapa
Jul 31 '14 at 9:19
30
30
The clue is, very much, in the name:
getElementsByClassName()
implies a plural, whereas getElementById()
implies a singular element item.– David Thomas
May 21 '12 at 23:20
The clue is, very much, in the name:
getElementsByClassName()
implies a plural, whereas getElementById()
implies a singular element item.– David Thomas
May 21 '12 at 23:20
1
1
I get that, it just didn't make sense to me that you can't change all the elements with that class name using the code above instead of having to loop through an array. jquery way is much better, i was just curious about the js way
– dmo
May 22 '12 at 4:34
I get that, it just didn't make sense to me that you can't change all the elements with that class name using the code above instead of having to loop through an array. jquery way is much better, i was just curious about the js way
– dmo
May 22 '12 at 4:34
1
1
Might be useful too: stackoverflow.com/questions/3871547/…
– kapa
Jul 31 '14 at 9:19
Might be useful too: stackoverflow.com/questions/3871547/…
– kapa
Jul 31 '14 at 9:19
add a comment |
9 Answers
9
active
oldest
votes
up vote
116
down vote
accepted
Your getElementById()
code works since IDs have to be unique and thus the function always returns exactly one element (or null
if none was found).
However, getElementsByClassName()
, querySelectorAll()
, and other getElementsBy*
methods return an array-like collection of elements. Iterate over it like you would with a real array:
var elems = document.getElementsByClassName('myElement');
for(var i = 0; i < elems.length; i++)
elems[i].style.size = '100px';
If you prefer something shorter, consider using jQuery:
$('.myElement').css('size', '100px');
1
Does that also apply to<iframe>
which is also part of your domain
– JMASTER B
Sep 1 '16 at 1:25
It's 2018... Just create a wrapper function forquerySelectorAll()
and you can have nice short code without a large, old-school dependency.qSA(".myElement").forEach(el => el.style.size = "100px")
Maybe have the wrapper receive a callback.qSA(".myElement", el => el.style.size = "100px")
– user2437417
Apr 26 at 20:40
add a comment |
up vote
8
down vote
You are using a array as an object, the difference between getElementbyId
and getElementsByClassName
is that:
getElementbyId
will return you an object.getElementsByClassName
will return you an array.
getElementsByClassName
The
getElementsByClassName(classNames)
method takes a string that
contains an unordered set of unique space-separated tokens
representing classes. When called, the method must return a live
NodeList
object containing all the elements in the document that
have all the classes specified in that argument, having obtained the
classes by splitting a string on spaces. If there are no tokens
specified in the argument, then the method must return an empty
NodeList.
https://www.w3.org/TR/2008/WD-html5-20080610/dom.html#getelementsbyclassname
getElementById
The getElementById() method accesses the first element with the specified id.
http://www.w3schools.com/jsref/met_doc_getelementbyid.asp
in your code the lines:
1- document.getElementsByClassName('myElement').style.size = '100px';
will NOT work as expected, cos the getElementByClassName
will return an array, and the array will NOT have the style
property, you gonna access each element
by iterating them.
That's why the function getElementById
was working for you, this function will return you the direct object , and so you will be able to access the style
property.
add a comment |
up vote
7
down vote
The following description is taken from this page:
The getElementsByClassName() method returns a collection of all elements in the document with the specified class name, as a NodeList object.
The NodeList object represents a collection of nodes. The nodes can be
accessed by index numbers. The index starts at 0.
Tip: You can use the length property of the NodeList object to determine the number of elements with a specified class name, then you can loop through all elements and extract the info you want.
So, as a parameter getElementsByClassName
would accept a class name.
If this is your HTML body:
<div id="first" class="menuItem"></div>
<div id="second" class="menuItem"></div>
<div id="third" class="menuItem"></div>
<div id="footer"></div>
then var menuItems = document.getElementsByClassName('menuItem')
would return a collection (not an array) of the 3 upper <div>
s, as they match the given class name.
You can then iterate over this nodes (<div>
s in this case) collection with:
for (var menuItemIndex = 0 ; menuItems.length ; menuItemIndex ++)
var currentMenuItem = menuItems[menuItemIndex];
// do stuff with currentMenuItem as a node.
Please refer to this post for more on differences between elements and nodes.
add a comment |
up vote
4
down vote
ES6 provides Array.from()
method, which creates a new Array instance from an array-like or iterable object.
let boxes = document.getElementsByClassName('box');
Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
As you can see inside the code snippet, after using Array.from()
function you are then able to manipulate over each element.
The same solution using jQuery
.
$('.box').css('background':'green');
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
add a comment |
up vote
3
down vote
In Other Words
document.querySelector()
selects only the first one element of the specified selector. So it doesn't spit out an array, it's a single value. Similar todocument.getElementById()
which fetches ID-elements only, since IDs have to be unique.document.querySelectorAll()
selects all elements with the specified selector and returns them in an array. Similar todocument.getElementsByClassName()
for classes anddocument.getElementsByTagName()
tags only.
Why use querySelector?
It's used merely for the sole purpose of ease and brevity.
Why use getElement/sBy?*
Faster performance.
Why this performance difference?
Both ways of selection has the purpose of creating a NodeList for further use.
querySelectors generates a static NodeList with the selectors thus it must be first created from scratch.
getElement/sBy* immediately adapts the existing live NodeList of the current DOM.
So, when to use which method it's up to you/your project/your device.
Infos
Demo of all methods
NodeList Documentation
Performance Test
add a comment |
up vote
2
down vote
It returns Array-like list.
You make that an Array as example
var el = getElementsByClassName("elem");
el = Array.prototype.slice.call(el); //this line
el[0].appendChild(otherElem);
add a comment |
up vote
2
down vote
You could get a single element by running
document.querySelector('.myElement').style.size = '100px';
but it's going to work for the first element with class .myElement.
If you would like apply this for all elements with the class I suggest you to use
document.querySelectorAll('.myElement').forEach(function(element)
element.style.size = '100px';
);
add a comment |
up vote
2
down vote
/*
* To hide all elements with the same class,
* use looping to reach each element with that class.
* In this case, looping is done recursively
*/
const hideAll = (className, i=0) =>
if(!document.getElementsByClassName(className)[i]) //exits the loop when element of that id does not exist
return;
document.getElementsByClassName(className)[i].style.visibility = 'hidden'; //hide element
return hideAll(className, i+1) //loop for the next element
hideAll('appBanner') //the function call requires the class name
add a comment |
up vote
-1
down vote
With ES5+ (any browsed nowadays - 2017) you should be able to do
.forEach.call(document.getElementsByClassName('answer'), function(el)
el.style.color= 'red';
);
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
116
down vote
accepted
Your getElementById()
code works since IDs have to be unique and thus the function always returns exactly one element (or null
if none was found).
However, getElementsByClassName()
, querySelectorAll()
, and other getElementsBy*
methods return an array-like collection of elements. Iterate over it like you would with a real array:
var elems = document.getElementsByClassName('myElement');
for(var i = 0; i < elems.length; i++)
elems[i].style.size = '100px';
If you prefer something shorter, consider using jQuery:
$('.myElement').css('size', '100px');
1
Does that also apply to<iframe>
which is also part of your domain
– JMASTER B
Sep 1 '16 at 1:25
It's 2018... Just create a wrapper function forquerySelectorAll()
and you can have nice short code without a large, old-school dependency.qSA(".myElement").forEach(el => el.style.size = "100px")
Maybe have the wrapper receive a callback.qSA(".myElement", el => el.style.size = "100px")
– user2437417
Apr 26 at 20:40
add a comment |
up vote
116
down vote
accepted
Your getElementById()
code works since IDs have to be unique and thus the function always returns exactly one element (or null
if none was found).
However, getElementsByClassName()
, querySelectorAll()
, and other getElementsBy*
methods return an array-like collection of elements. Iterate over it like you would with a real array:
var elems = document.getElementsByClassName('myElement');
for(var i = 0; i < elems.length; i++)
elems[i].style.size = '100px';
If you prefer something shorter, consider using jQuery:
$('.myElement').css('size', '100px');
1
Does that also apply to<iframe>
which is also part of your domain
– JMASTER B
Sep 1 '16 at 1:25
It's 2018... Just create a wrapper function forquerySelectorAll()
and you can have nice short code without a large, old-school dependency.qSA(".myElement").forEach(el => el.style.size = "100px")
Maybe have the wrapper receive a callback.qSA(".myElement", el => el.style.size = "100px")
– user2437417
Apr 26 at 20:40
add a comment |
up vote
116
down vote
accepted
up vote
116
down vote
accepted
Your getElementById()
code works since IDs have to be unique and thus the function always returns exactly one element (or null
if none was found).
However, getElementsByClassName()
, querySelectorAll()
, and other getElementsBy*
methods return an array-like collection of elements. Iterate over it like you would with a real array:
var elems = document.getElementsByClassName('myElement');
for(var i = 0; i < elems.length; i++)
elems[i].style.size = '100px';
If you prefer something shorter, consider using jQuery:
$('.myElement').css('size', '100px');
Your getElementById()
code works since IDs have to be unique and thus the function always returns exactly one element (or null
if none was found).
However, getElementsByClassName()
, querySelectorAll()
, and other getElementsBy*
methods return an array-like collection of elements. Iterate over it like you would with a real array:
var elems = document.getElementsByClassName('myElement');
for(var i = 0; i < elems.length; i++)
elems[i].style.size = '100px';
If you prefer something shorter, consider using jQuery:
$('.myElement').css('size', '100px');
edited Aug 5 '17 at 8:59
Makyen
20.2k83767
20.2k83767
answered May 21 '12 at 23:18
ThiefMaster♦
236k61461555
236k61461555
1
Does that also apply to<iframe>
which is also part of your domain
– JMASTER B
Sep 1 '16 at 1:25
It's 2018... Just create a wrapper function forquerySelectorAll()
and you can have nice short code without a large, old-school dependency.qSA(".myElement").forEach(el => el.style.size = "100px")
Maybe have the wrapper receive a callback.qSA(".myElement", el => el.style.size = "100px")
– user2437417
Apr 26 at 20:40
add a comment |
1
Does that also apply to<iframe>
which is also part of your domain
– JMASTER B
Sep 1 '16 at 1:25
It's 2018... Just create a wrapper function forquerySelectorAll()
and you can have nice short code without a large, old-school dependency.qSA(".myElement").forEach(el => el.style.size = "100px")
Maybe have the wrapper receive a callback.qSA(".myElement", el => el.style.size = "100px")
– user2437417
Apr 26 at 20:40
1
1
Does that also apply to
<iframe>
which is also part of your domain– JMASTER B
Sep 1 '16 at 1:25
Does that also apply to
<iframe>
which is also part of your domain– JMASTER B
Sep 1 '16 at 1:25
It's 2018... Just create a wrapper function for
querySelectorAll()
and you can have nice short code without a large, old-school dependency. qSA(".myElement").forEach(el => el.style.size = "100px")
Maybe have the wrapper receive a callback. qSA(".myElement", el => el.style.size = "100px")
– user2437417
Apr 26 at 20:40
It's 2018... Just create a wrapper function for
querySelectorAll()
and you can have nice short code without a large, old-school dependency. qSA(".myElement").forEach(el => el.style.size = "100px")
Maybe have the wrapper receive a callback. qSA(".myElement", el => el.style.size = "100px")
– user2437417
Apr 26 at 20:40
add a comment |
up vote
8
down vote
You are using a array as an object, the difference between getElementbyId
and getElementsByClassName
is that:
getElementbyId
will return you an object.getElementsByClassName
will return you an array.
getElementsByClassName
The
getElementsByClassName(classNames)
method takes a string that
contains an unordered set of unique space-separated tokens
representing classes. When called, the method must return a live
NodeList
object containing all the elements in the document that
have all the classes specified in that argument, having obtained the
classes by splitting a string on spaces. If there are no tokens
specified in the argument, then the method must return an empty
NodeList.
https://www.w3.org/TR/2008/WD-html5-20080610/dom.html#getelementsbyclassname
getElementById
The getElementById() method accesses the first element with the specified id.
http://www.w3schools.com/jsref/met_doc_getelementbyid.asp
in your code the lines:
1- document.getElementsByClassName('myElement').style.size = '100px';
will NOT work as expected, cos the getElementByClassName
will return an array, and the array will NOT have the style
property, you gonna access each element
by iterating them.
That's why the function getElementById
was working for you, this function will return you the direct object , and so you will be able to access the style
property.
add a comment |
up vote
8
down vote
You are using a array as an object, the difference between getElementbyId
and getElementsByClassName
is that:
getElementbyId
will return you an object.getElementsByClassName
will return you an array.
getElementsByClassName
The
getElementsByClassName(classNames)
method takes a string that
contains an unordered set of unique space-separated tokens
representing classes. When called, the method must return a live
NodeList
object containing all the elements in the document that
have all the classes specified in that argument, having obtained the
classes by splitting a string on spaces. If there are no tokens
specified in the argument, then the method must return an empty
NodeList.
https://www.w3.org/TR/2008/WD-html5-20080610/dom.html#getelementsbyclassname
getElementById
The getElementById() method accesses the first element with the specified id.
http://www.w3schools.com/jsref/met_doc_getelementbyid.asp
in your code the lines:
1- document.getElementsByClassName('myElement').style.size = '100px';
will NOT work as expected, cos the getElementByClassName
will return an array, and the array will NOT have the style
property, you gonna access each element
by iterating them.
That's why the function getElementById
was working for you, this function will return you the direct object , and so you will be able to access the style
property.
add a comment |
up vote
8
down vote
up vote
8
down vote
You are using a array as an object, the difference between getElementbyId
and getElementsByClassName
is that:
getElementbyId
will return you an object.getElementsByClassName
will return you an array.
getElementsByClassName
The
getElementsByClassName(classNames)
method takes a string that
contains an unordered set of unique space-separated tokens
representing classes. When called, the method must return a live
NodeList
object containing all the elements in the document that
have all the classes specified in that argument, having obtained the
classes by splitting a string on spaces. If there are no tokens
specified in the argument, then the method must return an empty
NodeList.
https://www.w3.org/TR/2008/WD-html5-20080610/dom.html#getelementsbyclassname
getElementById
The getElementById() method accesses the first element with the specified id.
http://www.w3schools.com/jsref/met_doc_getelementbyid.asp
in your code the lines:
1- document.getElementsByClassName('myElement').style.size = '100px';
will NOT work as expected, cos the getElementByClassName
will return an array, and the array will NOT have the style
property, you gonna access each element
by iterating them.
That's why the function getElementById
was working for you, this function will return you the direct object , and so you will be able to access the style
property.
You are using a array as an object, the difference between getElementbyId
and getElementsByClassName
is that:
getElementbyId
will return you an object.getElementsByClassName
will return you an array.
getElementsByClassName
The
getElementsByClassName(classNames)
method takes a string that
contains an unordered set of unique space-separated tokens
representing classes. When called, the method must return a live
NodeList
object containing all the elements in the document that
have all the classes specified in that argument, having obtained the
classes by splitting a string on spaces. If there are no tokens
specified in the argument, then the method must return an empty
NodeList.
https://www.w3.org/TR/2008/WD-html5-20080610/dom.html#getelementsbyclassname
getElementById
The getElementById() method accesses the first element with the specified id.
http://www.w3schools.com/jsref/met_doc_getelementbyid.asp
in your code the lines:
1- document.getElementsByClassName('myElement').style.size = '100px';
will NOT work as expected, cos the getElementByClassName
will return an array, and the array will NOT have the style
property, you gonna access each element
by iterating them.
That's why the function getElementById
was working for you, this function will return you the direct object , and so you will be able to access the style
property.
answered Feb 15 '16 at 2:53
Alvaro Joao
4,69452351
4,69452351
add a comment |
add a comment |
up vote
7
down vote
The following description is taken from this page:
The getElementsByClassName() method returns a collection of all elements in the document with the specified class name, as a NodeList object.
The NodeList object represents a collection of nodes. The nodes can be
accessed by index numbers. The index starts at 0.
Tip: You can use the length property of the NodeList object to determine the number of elements with a specified class name, then you can loop through all elements and extract the info you want.
So, as a parameter getElementsByClassName
would accept a class name.
If this is your HTML body:
<div id="first" class="menuItem"></div>
<div id="second" class="menuItem"></div>
<div id="third" class="menuItem"></div>
<div id="footer"></div>
then var menuItems = document.getElementsByClassName('menuItem')
would return a collection (not an array) of the 3 upper <div>
s, as they match the given class name.
You can then iterate over this nodes (<div>
s in this case) collection with:
for (var menuItemIndex = 0 ; menuItems.length ; menuItemIndex ++)
var currentMenuItem = menuItems[menuItemIndex];
// do stuff with currentMenuItem as a node.
Please refer to this post for more on differences between elements and nodes.
add a comment |
up vote
7
down vote
The following description is taken from this page:
The getElementsByClassName() method returns a collection of all elements in the document with the specified class name, as a NodeList object.
The NodeList object represents a collection of nodes. The nodes can be
accessed by index numbers. The index starts at 0.
Tip: You can use the length property of the NodeList object to determine the number of elements with a specified class name, then you can loop through all elements and extract the info you want.
So, as a parameter getElementsByClassName
would accept a class name.
If this is your HTML body:
<div id="first" class="menuItem"></div>
<div id="second" class="menuItem"></div>
<div id="third" class="menuItem"></div>
<div id="footer"></div>
then var menuItems = document.getElementsByClassName('menuItem')
would return a collection (not an array) of the 3 upper <div>
s, as they match the given class name.
You can then iterate over this nodes (<div>
s in this case) collection with:
for (var menuItemIndex = 0 ; menuItems.length ; menuItemIndex ++)
var currentMenuItem = menuItems[menuItemIndex];
// do stuff with currentMenuItem as a node.
Please refer to this post for more on differences between elements and nodes.
add a comment |
up vote
7
down vote
up vote
7
down vote
The following description is taken from this page:
The getElementsByClassName() method returns a collection of all elements in the document with the specified class name, as a NodeList object.
The NodeList object represents a collection of nodes. The nodes can be
accessed by index numbers. The index starts at 0.
Tip: You can use the length property of the NodeList object to determine the number of elements with a specified class name, then you can loop through all elements and extract the info you want.
So, as a parameter getElementsByClassName
would accept a class name.
If this is your HTML body:
<div id="first" class="menuItem"></div>
<div id="second" class="menuItem"></div>
<div id="third" class="menuItem"></div>
<div id="footer"></div>
then var menuItems = document.getElementsByClassName('menuItem')
would return a collection (not an array) of the 3 upper <div>
s, as they match the given class name.
You can then iterate over this nodes (<div>
s in this case) collection with:
for (var menuItemIndex = 0 ; menuItems.length ; menuItemIndex ++)
var currentMenuItem = menuItems[menuItemIndex];
// do stuff with currentMenuItem as a node.
Please refer to this post for more on differences between elements and nodes.
The following description is taken from this page:
The getElementsByClassName() method returns a collection of all elements in the document with the specified class name, as a NodeList object.
The NodeList object represents a collection of nodes. The nodes can be
accessed by index numbers. The index starts at 0.
Tip: You can use the length property of the NodeList object to determine the number of elements with a specified class name, then you can loop through all elements and extract the info you want.
So, as a parameter getElementsByClassName
would accept a class name.
If this is your HTML body:
<div id="first" class="menuItem"></div>
<div id="second" class="menuItem"></div>
<div id="third" class="menuItem"></div>
<div id="footer"></div>
then var menuItems = document.getElementsByClassName('menuItem')
would return a collection (not an array) of the 3 upper <div>
s, as they match the given class name.
You can then iterate over this nodes (<div>
s in this case) collection with:
for (var menuItemIndex = 0 ; menuItems.length ; menuItemIndex ++)
var currentMenuItem = menuItems[menuItemIndex];
// do stuff with currentMenuItem as a node.
Please refer to this post for more on differences between elements and nodes.
edited May 23 '17 at 11:47
Community♦
11
11
answered Jan 7 '16 at 9:14
remdevtec
2,28811020
2,28811020
add a comment |
add a comment |
up vote
4
down vote
ES6 provides Array.from()
method, which creates a new Array instance from an array-like or iterable object.
let boxes = document.getElementsByClassName('box');
Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
As you can see inside the code snippet, after using Array.from()
function you are then able to manipulate over each element.
The same solution using jQuery
.
$('.box').css('background':'green');
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
add a comment |
up vote
4
down vote
ES6 provides Array.from()
method, which creates a new Array instance from an array-like or iterable object.
let boxes = document.getElementsByClassName('box');
Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
As you can see inside the code snippet, after using Array.from()
function you are then able to manipulate over each element.
The same solution using jQuery
.
$('.box').css('background':'green');
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
add a comment |
up vote
4
down vote
up vote
4
down vote
ES6 provides Array.from()
method, which creates a new Array instance from an array-like or iterable object.
let boxes = document.getElementsByClassName('box');
Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
As you can see inside the code snippet, after using Array.from()
function you are then able to manipulate over each element.
The same solution using jQuery
.
$('.box').css('background':'green');
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
ES6 provides Array.from()
method, which creates a new Array instance from an array-like or iterable object.
let boxes = document.getElementsByClassName('box');
Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
As you can see inside the code snippet, after using Array.from()
function you are then able to manipulate over each element.
The same solution using jQuery
.
$('.box').css('background':'green');
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
let boxes = document.getElementsByClassName('box');
Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
let boxes = document.getElementsByClassName('box');
Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
$('.box').css('background':'green');
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
$('.box').css('background':'green');
.box
width: 50px;
height: 50px;
margin: 5px;
background: blue;
display: inline-block;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
edited Apr 4 '17 at 9:21
answered Feb 21 '17 at 23:07
kind user
20.1k51942
20.1k51942
add a comment |
add a comment |
up vote
3
down vote
In Other Words
document.querySelector()
selects only the first one element of the specified selector. So it doesn't spit out an array, it's a single value. Similar todocument.getElementById()
which fetches ID-elements only, since IDs have to be unique.document.querySelectorAll()
selects all elements with the specified selector and returns them in an array. Similar todocument.getElementsByClassName()
for classes anddocument.getElementsByTagName()
tags only.
Why use querySelector?
It's used merely for the sole purpose of ease and brevity.
Why use getElement/sBy?*
Faster performance.
Why this performance difference?
Both ways of selection has the purpose of creating a NodeList for further use.
querySelectors generates a static NodeList with the selectors thus it must be first created from scratch.
getElement/sBy* immediately adapts the existing live NodeList of the current DOM.
So, when to use which method it's up to you/your project/your device.
Infos
Demo of all methods
NodeList Documentation
Performance Test
add a comment |
up vote
3
down vote
In Other Words
document.querySelector()
selects only the first one element of the specified selector. So it doesn't spit out an array, it's a single value. Similar todocument.getElementById()
which fetches ID-elements only, since IDs have to be unique.document.querySelectorAll()
selects all elements with the specified selector and returns them in an array. Similar todocument.getElementsByClassName()
for classes anddocument.getElementsByTagName()
tags only.
Why use querySelector?
It's used merely for the sole purpose of ease and brevity.
Why use getElement/sBy?*
Faster performance.
Why this performance difference?
Both ways of selection has the purpose of creating a NodeList for further use.
querySelectors generates a static NodeList with the selectors thus it must be first created from scratch.
getElement/sBy* immediately adapts the existing live NodeList of the current DOM.
So, when to use which method it's up to you/your project/your device.
Infos
Demo of all methods
NodeList Documentation
Performance Test
add a comment |
up vote
3
down vote
up vote
3
down vote
In Other Words
document.querySelector()
selects only the first one element of the specified selector. So it doesn't spit out an array, it's a single value. Similar todocument.getElementById()
which fetches ID-elements only, since IDs have to be unique.document.querySelectorAll()
selects all elements with the specified selector and returns them in an array. Similar todocument.getElementsByClassName()
for classes anddocument.getElementsByTagName()
tags only.
Why use querySelector?
It's used merely for the sole purpose of ease and brevity.
Why use getElement/sBy?*
Faster performance.
Why this performance difference?
Both ways of selection has the purpose of creating a NodeList for further use.
querySelectors generates a static NodeList with the selectors thus it must be first created from scratch.
getElement/sBy* immediately adapts the existing live NodeList of the current DOM.
So, when to use which method it's up to you/your project/your device.
Infos
Demo of all methods
NodeList Documentation
Performance Test
In Other Words
document.querySelector()
selects only the first one element of the specified selector. So it doesn't spit out an array, it's a single value. Similar todocument.getElementById()
which fetches ID-elements only, since IDs have to be unique.document.querySelectorAll()
selects all elements with the specified selector and returns them in an array. Similar todocument.getElementsByClassName()
for classes anddocument.getElementsByTagName()
tags only.
Why use querySelector?
It's used merely for the sole purpose of ease and brevity.
Why use getElement/sBy?*
Faster performance.
Why this performance difference?
Both ways of selection has the purpose of creating a NodeList for further use.
querySelectors generates a static NodeList with the selectors thus it must be first created from scratch.
getElement/sBy* immediately adapts the existing live NodeList of the current DOM.
So, when to use which method it's up to you/your project/your device.
Infos
Demo of all methods
NodeList Documentation
Performance Test
edited Mar 19 at 15:01
answered Oct 23 '17 at 11:41
Thielicious
1,9431119
1,9431119
add a comment |
add a comment |
up vote
2
down vote
It returns Array-like list.
You make that an Array as example
var el = getElementsByClassName("elem");
el = Array.prototype.slice.call(el); //this line
el[0].appendChild(otherElem);
add a comment |
up vote
2
down vote
It returns Array-like list.
You make that an Array as example
var el = getElementsByClassName("elem");
el = Array.prototype.slice.call(el); //this line
el[0].appendChild(otherElem);
add a comment |
up vote
2
down vote
up vote
2
down vote
It returns Array-like list.
You make that an Array as example
var el = getElementsByClassName("elem");
el = Array.prototype.slice.call(el); //this line
el[0].appendChild(otherElem);
It returns Array-like list.
You make that an Array as example
var el = getElementsByClassName("elem");
el = Array.prototype.slice.call(el); //this line
el[0].appendChild(otherElem);
answered Jun 17 '16 at 3:21
atilkan
2,1331830
2,1331830
add a comment |
add a comment |
up vote
2
down vote
You could get a single element by running
document.querySelector('.myElement').style.size = '100px';
but it's going to work for the first element with class .myElement.
If you would like apply this for all elements with the class I suggest you to use
document.querySelectorAll('.myElement').forEach(function(element)
element.style.size = '100px';
);
add a comment |
up vote
2
down vote
You could get a single element by running
document.querySelector('.myElement').style.size = '100px';
but it's going to work for the first element with class .myElement.
If you would like apply this for all elements with the class I suggest you to use
document.querySelectorAll('.myElement').forEach(function(element)
element.style.size = '100px';
);
add a comment |
up vote
2
down vote
up vote
2
down vote
You could get a single element by running
document.querySelector('.myElement').style.size = '100px';
but it's going to work for the first element with class .myElement.
If you would like apply this for all elements with the class I suggest you to use
document.querySelectorAll('.myElement').forEach(function(element)
element.style.size = '100px';
);
You could get a single element by running
document.querySelector('.myElement').style.size = '100px';
but it's going to work for the first element with class .myElement.
If you would like apply this for all elements with the class I suggest you to use
document.querySelectorAll('.myElement').forEach(function(element)
element.style.size = '100px';
);
edited Jul 2 '17 at 21:06
JJJ
29k147591
29k147591
answered Jul 2 '17 at 19:29
Sergey
984
984
add a comment |
add a comment |
up vote
2
down vote
/*
* To hide all elements with the same class,
* use looping to reach each element with that class.
* In this case, looping is done recursively
*/
const hideAll = (className, i=0) =>
if(!document.getElementsByClassName(className)[i]) //exits the loop when element of that id does not exist
return;
document.getElementsByClassName(className)[i].style.visibility = 'hidden'; //hide element
return hideAll(className, i+1) //loop for the next element
hideAll('appBanner') //the function call requires the class name
add a comment |
up vote
2
down vote
/*
* To hide all elements with the same class,
* use looping to reach each element with that class.
* In this case, looping is done recursively
*/
const hideAll = (className, i=0) =>
if(!document.getElementsByClassName(className)[i]) //exits the loop when element of that id does not exist
return;
document.getElementsByClassName(className)[i].style.visibility = 'hidden'; //hide element
return hideAll(className, i+1) //loop for the next element
hideAll('appBanner') //the function call requires the class name
add a comment |
up vote
2
down vote
up vote
2
down vote
/*
* To hide all elements with the same class,
* use looping to reach each element with that class.
* In this case, looping is done recursively
*/
const hideAll = (className, i=0) =>
if(!document.getElementsByClassName(className)[i]) //exits the loop when element of that id does not exist
return;
document.getElementsByClassName(className)[i].style.visibility = 'hidden'; //hide element
return hideAll(className, i+1) //loop for the next element
hideAll('appBanner') //the function call requires the class name
/*
* To hide all elements with the same class,
* use looping to reach each element with that class.
* In this case, looping is done recursively
*/
const hideAll = (className, i=0) =>
if(!document.getElementsByClassName(className)[i]) //exits the loop when element of that id does not exist
return;
document.getElementsByClassName(className)[i].style.visibility = 'hidden'; //hide element
return hideAll(className, i+1) //loop for the next element
hideAll('appBanner') //the function call requires the class name
answered Nov 25 at 4:06
Irina Mityugova
211
211
add a comment |
add a comment |
up vote
-1
down vote
With ES5+ (any browsed nowadays - 2017) you should be able to do
.forEach.call(document.getElementsByClassName('answer'), function(el)
el.style.color= 'red';
);
add a comment |
up vote
-1
down vote
With ES5+ (any browsed nowadays - 2017) you should be able to do
.forEach.call(document.getElementsByClassName('answer'), function(el)
el.style.color= 'red';
);
add a comment |
up vote
-1
down vote
up vote
-1
down vote
With ES5+ (any browsed nowadays - 2017) you should be able to do
.forEach.call(document.getElementsByClassName('answer'), function(el)
el.style.color= 'red';
);
With ES5+ (any browsed nowadays - 2017) you should be able to do
.forEach.call(document.getElementsByClassName('answer'), function(el)
el.style.color= 'red';
);
.forEach.call(document.getElementsByClassName('answer'), function(el)
el.style.color= 'red';
);
.forEach.call(document.getElementsByClassName('answer'), function(el)
el.style.color= 'red';
);
edited Dec 2 '16 at 8:39
answered Apr 20 '15 at 16:03
Matas Vaitkevicius
32.3k15161171
32.3k15161171
add a comment |
add a comment |
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30
The clue is, very much, in the name:
getElementsByClassName()
implies a plural, whereasgetElementById()
implies a singular element item.– David Thomas
May 21 '12 at 23:20
1
I get that, it just didn't make sense to me that you can't change all the elements with that class name using the code above instead of having to loop through an array. jquery way is much better, i was just curious about the js way
– dmo
May 22 '12 at 4:34
1
Might be useful too: stackoverflow.com/questions/3871547/…
– kapa
Jul 31 '14 at 9:19