sql queries on ajax php page
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-1
down vote
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i am trying to get content inside a div using ajax, on the php page that am accessing using ajax, i want to fill dropdown options using another query but it just closes the select tag before the options.
<select class="form-control data" data-t="integer" name='role' required/>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>
But here is what i get inside the div
<select class="form-control data" data-t="integer" name="role"
required=""></select>
<option value='1' >option1</option>
<option value='2' >option2</option>
<option value='3' >option3</option>
php ajax
asked Nov 11 at 9:08
akshay sharma
306
add a comment |
up vote
-1
down vote
favorite
i am trying to get content inside a div using ajax, on the php page that am accessing using ajax, i want to fill dropdown options using another query but it just closes the select tag before the options.
<select class="form-control data" data-t="integer" name='role' required/>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>
But here is what i get inside the div
<select class="form-control data" data-t="integer" name="role"
required=""></select>
<option value='1' >option1</option>
<option value='2' >option2</option>
<option value='3' >option3</option>
php ajax
asked Nov 11 at 9:08
akshay sharma
306
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
i am trying to get content inside a div using ajax, on the php page that am accessing using ajax, i want to fill dropdown options using another query but it just closes the select tag before the options.
<select class="form-control data" data-t="integer" name='role' required/>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>
But here is what i get inside the div
<select class="form-control data" data-t="integer" name="role"
required=""></select>
<option value='1' >option1</option>
<option value='2' >option2</option>
<option value='3' >option3</option>
php ajax
asked Nov 11 at 9:08
akshay sharma
306
i am trying to get content inside a div using ajax, on the php page that am accessing using ajax, i want to fill dropdown options using another query but it just closes the select tag before the options.
<select class="form-control data" data-t="integer" name='role' required/>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>
But here is what i get inside the div
<select class="form-control data" data-t="integer" name="role"
required=""></select>
<option value='1' >option1</option>
<option value='2' >option2</option>
<option value='3' >option3</option>
php ajax
php ajax
asked Nov 11 at 9:08
akshay sharma
306
asked Nov 11 at 9:08
akshay sharma
306
asked Nov 11 at 9:08
akshay sharma
306
asked Nov 11 at 9:08
akshay sharma
306
asked Nov 11 at 9:08
akshay sharma
306
306
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
please remove forward / at the end of opening select tag as below:
<select class="form-control data" data-t="integer" name='role' required>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>answered Nov 11 at 9:22
Khalifa Nikzad
3088
oh snap! XD what the... XD sorry that was such an embarrassing mistake :D Btw... why did it work in the first place, i mean i used the same select (closed in the opening tag itself) on another page (not accessed by an ajax call). I just copied this dropdown from there.... it was working as required on that page, even with this error...
– akshay sharma
Nov 11 at 9:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
please remove forward / at the end of opening select tag as below:
<select class="form-control data" data-t="integer" name='role' required>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>answered Nov 11 at 9:22
Khalifa Nikzad
3088
oh snap! XD what the... XD sorry that was such an embarrassing mistake :D Btw... why did it work in the first place, i mean i used the same select (closed in the opening tag itself) on another page (not accessed by an ajax call). I just copied this dropdown from there.... it was working as required on that page, even with this error...
– akshay sharma
Nov 11 at 9:32
add a comment |
up vote
6
down vote
accepted
please remove forward / at the end of opening select tag as below:
<select class="form-control data" data-t="integer" name='role' required>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>answered Nov 11 at 9:22
Khalifa Nikzad
3088
oh snap! XD what the... XD sorry that was such an embarrassing mistake :D Btw... why did it work in the first place, i mean i used the same select (closed in the opening tag itself) on another page (not accessed by an ajax call). I just copied this dropdown from there.... it was working as required on that page, even with this error...
– akshay sharma
Nov 11 at 9:32
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
please remove forward / at the end of opening select tag as below:
<select class="form-control data" data-t="integer" name='role' required>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>answered Nov 11 at 9:22
Khalifa Nikzad
3088
please remove forward / at the end of opening select tag as below:
<select class="form-control data" data-t="integer" name='role' required>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select><select class="form-control data" data-t="integer" name='role' required>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select><select class="form-control data" data-t="integer" name='role' required>
<option value="">--Select--</option>
<?php
foreach($con -> query("select id,name,permissions from role") as $role)
?>
<option value="<?php echo $role['id']; ?>"><?php echo $role['name']; ?></option>
<?php
?>
</select>answered Nov 11 at 9:22
Khalifa Nikzad
3088
answered Nov 11 at 9:22
Khalifa Nikzad
3088
answered Nov 11 at 9:22
Khalifa Nikzad
3088
answered Nov 11 at 9:22
Khalifa Nikzad
3088
3088
oh snap! XD what the... XD sorry that was such an embarrassing mistake :D Btw... why did it work in the first place, i mean i used the same select (closed in the opening tag itself) on another page (not accessed by an ajax call). I just copied this dropdown from there.... it was working as required on that page, even with this error...
– akshay sharma
Nov 11 at 9:32
add a comment |
oh snap! XD what the... XD sorry that was such an embarrassing mistake :D Btw... why did it work in the first place, i mean i used the same select (closed in the opening tag itself) on another page (not accessed by an ajax call). I just copied this dropdown from there.... it was working as required on that page, even with this error...
– akshay sharma
Nov 11 at 9:32
oh snap! XD what the... XD sorry that was such an embarrassing mistake :D Btw... why did it work in the first place, i mean i used the same select (closed in the opening tag itself) on another page (not accessed by an ajax call). I just copied this dropdown from there.... it was working as required on that page, even with this error...
– akshay sharma
Nov 11 at 9:32
oh snap! XD what the... XD sorry that was such an embarrassing mistake :D Btw... why did it work in the first place, i mean i used the same select (closed in the opening tag itself) on another page (not accessed by an ajax call). I just copied this dropdown from there.... it was working as required on that page, even with this error...
– akshay sharma
Nov 11 at 9:32
add a comment |
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