React function and “this”
up vote
0
down vote
favorite
I know about "this" binding and all other stuff here, except one thing. I can't understand how "this" is not undefined in the first call, but second?
P.S. I know about function reference and that in the first case it's executing function but in the second case returning reference.
import React from "react";
import ReactDOM from "react-dom";
import "./styles.css";
class App extends React.Component
constructor()
super();
this.name = "MyComponent";
// this.handleClick = this.handleClick.bind(this);
handleClick()
console.log(this);
console.log(this.name);
render()
return (
<div>
<button onClick=this.handleClick()>click 1</button>
<button onClick=this.handleClick>click 2</button>
</div>
);
const rootElement = document.getElementById("root");
ReactDOM.render(<App />, rootElement);
javascript reactjs this
add a comment |
up vote
0
down vote
favorite
I know about "this" binding and all other stuff here, except one thing. I can't understand how "this" is not undefined in the first call, but second?
P.S. I know about function reference and that in the first case it's executing function but in the second case returning reference.
import React from "react";
import ReactDOM from "react-dom";
import "./styles.css";
class App extends React.Component
constructor()
super();
this.name = "MyComponent";
// this.handleClick = this.handleClick.bind(this);
handleClick()
console.log(this);
console.log(this.name);
render()
return (
<div>
<button onClick=this.handleClick()>click 1</button>
<button onClick=this.handleClick>click 2</button>
</div>
);
const rootElement = document.getElementById("root");
ReactDOM.render(<App />, rootElement);
javascript reactjs this
4
By writingonClick=this.handleClick()
you are invoking thethis.handleClick
function directly on render, and it evaluates toonClick=undefined
. That's why clicking on theclick 1
button doesn't do anything. This is a great read on whythis
behaves like it does in theclick 2
button case.
– Tholle
Nov 12 at 0:00
Please include the relevant code in the question text itself. Questions should make sense even if the link breaks in the future.
– Håken Lid
Nov 12 at 0:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know about "this" binding and all other stuff here, except one thing. I can't understand how "this" is not undefined in the first call, but second?
P.S. I know about function reference and that in the first case it's executing function but in the second case returning reference.
import React from "react";
import ReactDOM from "react-dom";
import "./styles.css";
class App extends React.Component
constructor()
super();
this.name = "MyComponent";
// this.handleClick = this.handleClick.bind(this);
handleClick()
console.log(this);
console.log(this.name);
render()
return (
<div>
<button onClick=this.handleClick()>click 1</button>
<button onClick=this.handleClick>click 2</button>
</div>
);
const rootElement = document.getElementById("root");
ReactDOM.render(<App />, rootElement);
javascript reactjs this
I know about "this" binding and all other stuff here, except one thing. I can't understand how "this" is not undefined in the first call, but second?
P.S. I know about function reference and that in the first case it's executing function but in the second case returning reference.
import React from "react";
import ReactDOM from "react-dom";
import "./styles.css";
class App extends React.Component
constructor()
super();
this.name = "MyComponent";
// this.handleClick = this.handleClick.bind(this);
handleClick()
console.log(this);
console.log(this.name);
render()
return (
<div>
<button onClick=this.handleClick()>click 1</button>
<button onClick=this.handleClick>click 2</button>
</div>
);
const rootElement = document.getElementById("root");
ReactDOM.render(<App />, rootElement);
javascript reactjs this
javascript reactjs this
edited Nov 12 at 5:01
Dinesh Pandiyan
2,173824
2,173824
asked Nov 11 at 23:54
Hayk
20435
20435
4
By writingonClick=this.handleClick()
you are invoking thethis.handleClick
function directly on render, and it evaluates toonClick=undefined
. That's why clicking on theclick 1
button doesn't do anything. This is a great read on whythis
behaves like it does in theclick 2
button case.
– Tholle
Nov 12 at 0:00
Please include the relevant code in the question text itself. Questions should make sense even if the link breaks in the future.
– Håken Lid
Nov 12 at 0:00
add a comment |
4
By writingonClick=this.handleClick()
you are invoking thethis.handleClick
function directly on render, and it evaluates toonClick=undefined
. That's why clicking on theclick 1
button doesn't do anything. This is a great read on whythis
behaves like it does in theclick 2
button case.
– Tholle
Nov 12 at 0:00
Please include the relevant code in the question text itself. Questions should make sense even if the link breaks in the future.
– Håken Lid
Nov 12 at 0:00
4
4
By writing
onClick=this.handleClick()
you are invoking the this.handleClick
function directly on render, and it evaluates to onClick=undefined
. That's why clicking on the click 1
button doesn't do anything. This is a great read on why this
behaves like it does in the click 2
button case.– Tholle
Nov 12 at 0:00
By writing
onClick=this.handleClick()
you are invoking the this.handleClick
function directly on render, and it evaluates to onClick=undefined
. That's why clicking on the click 1
button doesn't do anything. This is a great read on why this
behaves like it does in the click 2
button case.– Tholle
Nov 12 at 0:00
Please include the relevant code in the question text itself. Questions should make sense even if the link breaks in the future.
– Håken Lid
Nov 12 at 0:00
Please include the relevant code in the question text itself. Questions should make sense even if the link breaks in the future.
– Håken Lid
Nov 12 at 0:00
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
In the first line
<button onClick=this.handleClick()>click 1</button>
this.handleClick()
will be executed in the render
function of App
component (which is a class
by itself) while the component is being rendered in virtual DOM. So by the time of execution, handleClick
function will be defined in the execution context which is App
class.
In the second line
<button onClick=this.handleClick>click 1</button>
this.handleClick
is attached to the DOM and will execute from the DOM's context when click event happens and the execution will look for handleClick in the DOM's context and will be undefined
.
There are two ways to circumvent this
bind
the method to the class as you did.Pass in an anonymous function which will execute without DOM's context and by default be bound to invoker's context.
like this
<button onClick=(e) => this.handleClick(e)>click 1</button>
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
In the first line
<button onClick=this.handleClick()>click 1</button>
this.handleClick()
will be executed in the render
function of App
component (which is a class
by itself) while the component is being rendered in virtual DOM. So by the time of execution, handleClick
function will be defined in the execution context which is App
class.
In the second line
<button onClick=this.handleClick>click 1</button>
this.handleClick
is attached to the DOM and will execute from the DOM's context when click event happens and the execution will look for handleClick in the DOM's context and will be undefined
.
There are two ways to circumvent this
bind
the method to the class as you did.Pass in an anonymous function which will execute without DOM's context and by default be bound to invoker's context.
like this
<button onClick=(e) => this.handleClick(e)>click 1</button>
add a comment |
up vote
4
down vote
accepted
In the first line
<button onClick=this.handleClick()>click 1</button>
this.handleClick()
will be executed in the render
function of App
component (which is a class
by itself) while the component is being rendered in virtual DOM. So by the time of execution, handleClick
function will be defined in the execution context which is App
class.
In the second line
<button onClick=this.handleClick>click 1</button>
this.handleClick
is attached to the DOM and will execute from the DOM's context when click event happens and the execution will look for handleClick in the DOM's context and will be undefined
.
There are two ways to circumvent this
bind
the method to the class as you did.Pass in an anonymous function which will execute without DOM's context and by default be bound to invoker's context.
like this
<button onClick=(e) => this.handleClick(e)>click 1</button>
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
In the first line
<button onClick=this.handleClick()>click 1</button>
this.handleClick()
will be executed in the render
function of App
component (which is a class
by itself) while the component is being rendered in virtual DOM. So by the time of execution, handleClick
function will be defined in the execution context which is App
class.
In the second line
<button onClick=this.handleClick>click 1</button>
this.handleClick
is attached to the DOM and will execute from the DOM's context when click event happens and the execution will look for handleClick in the DOM's context and will be undefined
.
There are two ways to circumvent this
bind
the method to the class as you did.Pass in an anonymous function which will execute without DOM's context and by default be bound to invoker's context.
like this
<button onClick=(e) => this.handleClick(e)>click 1</button>
In the first line
<button onClick=this.handleClick()>click 1</button>
this.handleClick()
will be executed in the render
function of App
component (which is a class
by itself) while the component is being rendered in virtual DOM. So by the time of execution, handleClick
function will be defined in the execution context which is App
class.
In the second line
<button onClick=this.handleClick>click 1</button>
this.handleClick
is attached to the DOM and will execute from the DOM's context when click event happens and the execution will look for handleClick in the DOM's context and will be undefined
.
There are two ways to circumvent this
bind
the method to the class as you did.Pass in an anonymous function which will execute without DOM's context and by default be bound to invoker's context.
like this
<button onClick=(e) => this.handleClick(e)>click 1</button>
answered Nov 12 at 3:01
Dinesh Pandiyan
2,173824
2,173824
add a comment |
add a comment |
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4
By writing
onClick=this.handleClick()
you are invoking thethis.handleClick
function directly on render, and it evaluates toonClick=undefined
. That's why clicking on theclick 1
button doesn't do anything. This is a great read on whythis
behaves like it does in theclick 2
button case.– Tholle
Nov 12 at 0:00
Please include the relevant code in the question text itself. Questions should make sense even if the link breaks in the future.
– Håken Lid
Nov 12 at 0:00