click function called multiple times
0
The thing is there are various products on my homepage , on clicking on it there is a popup opening after it. There is a like button on the modal popup. On clicking on it i fire ajax and populate the database and close it , but when i open another popup model and click on the like button the ajax fires two times similarly after closing and opening another popup the ajax call increases.
I tried to make modal html to blank on clicking the close button on button. but, it is not working.
On the home page
<div class="container">
<div class="prod_detail">
<div class="modal fade" id="prod_viewd" role="dialog">
</div>
</div>
</div>
this is the container where i am populating the data .
This is the like icon html
<a href="javascript:void(0);" class="vf-item-fullview-icon change">
<span class="ProductFullView_like lstCng">
<img src="<?php echo base_url(); ?>assets/blog/feed_image/likeBlack.png" />
</span>
</a>
and this is the code for click
$(document.body).on('click','.change', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
jquery
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
|
show 5 more comments
0
The thing is there are various products on my homepage , on clicking on it there is a popup opening after it. There is a like button on the modal popup. On clicking on it i fire ajax and populate the database and close it , but when i open another popup model and click on the like button the ajax fires two times similarly after closing and opening another popup the ajax call increases.
I tried to make modal html to blank on clicking the close button on button. but, it is not working.
On the home page
<div class="container">
<div class="prod_detail">
<div class="modal fade" id="prod_viewd" role="dialog">
</div>
</div>
</div>
this is the container where i am populating the data .
This is the like icon html
<a href="javascript:void(0);" class="vf-item-fullview-icon change">
<span class="ProductFullView_like lstCng">
<img src="<?php echo base_url(); ?>assets/blog/feed_image/likeBlack.png" />
</span>
</a>
and this is the code for click
$(document.body).on('click','.change', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
jquery
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
I think the problem comes from the way you name the element. Maybe you have multiple elements which have the same class name. That's how: I do this, but why that?
– Foo
Nov 19 '18 at 9:54
@TânNguyễn i have already checked that , i think when i click to open the modal the number of instances of modal = number of ajax called
– Sanjit Bhardwaj
Nov 19 '18 at 9:56
you sample code is not enough to understand, maybe it's something related to how you add the event listener...
– Plastic
Nov 19 '18 at 9:57
can you show your html or better to make a fiddle for it
– Negi Rox
Nov 19 '18 at 10:00
1
i would suggest to bind a click function on page load
– Negi Rox
Nov 19 '18 at 10:11
|
show 5 more comments
0
0
0
The thing is there are various products on my homepage , on clicking on it there is a popup opening after it. There is a like button on the modal popup. On clicking on it i fire ajax and populate the database and close it , but when i open another popup model and click on the like button the ajax fires two times similarly after closing and opening another popup the ajax call increases.
I tried to make modal html to blank on clicking the close button on button. but, it is not working.
On the home page
<div class="container">
<div class="prod_detail">
<div class="modal fade" id="prod_viewd" role="dialog">
</div>
</div>
</div>
this is the container where i am populating the data .
This is the like icon html
<a href="javascript:void(0);" class="vf-item-fullview-icon change">
<span class="ProductFullView_like lstCng">
<img src="<?php echo base_url(); ?>assets/blog/feed_image/likeBlack.png" />
</span>
</a>
and this is the code for click
$(document.body).on('click','.change', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
jquery
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
The thing is there are various products on my homepage , on clicking on it there is a popup opening after it. There is a like button on the modal popup. On clicking on it i fire ajax and populate the database and close it , but when i open another popup model and click on the like button the ajax fires two times similarly after closing and opening another popup the ajax call increases.
I tried to make modal html to blank on clicking the close button on button. but, it is not working.
On the home page
<div class="container">
<div class="prod_detail">
<div class="modal fade" id="prod_viewd" role="dialog">
</div>
</div>
</div>
this is the container where i am populating the data .
This is the like icon html
<a href="javascript:void(0);" class="vf-item-fullview-icon change">
<span class="ProductFullView_like lstCng">
<img src="<?php echo base_url(); ?>assets/blog/feed_image/likeBlack.png" />
</span>
</a>
and this is the code for click
$(document.body).on('click','.change', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
jquery
jquery
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
edited Nov 19 '18 at 10:07
Sanjit Bhardwaj
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
asked Nov 19 '18 at 9:50
Sanjit BhardwajSanjit Bhardwaj
399111
399111
I think the problem comes from the way you name the element. Maybe you have multiple elements which have the same class name. That's how: I do this, but why that?
– Foo
Nov 19 '18 at 9:54
@TânNguyễn i have already checked that , i think when i click to open the modal the number of instances of modal = number of ajax called
– Sanjit Bhardwaj
Nov 19 '18 at 9:56
you sample code is not enough to understand, maybe it's something related to how you add the event listener...
– Plastic
Nov 19 '18 at 9:57
can you show your html or better to make a fiddle for it
– Negi Rox
Nov 19 '18 at 10:00
1
i would suggest to bind a click function on page load
– Negi Rox
Nov 19 '18 at 10:11
|
show 5 more comments
I think the problem comes from the way you name the element. Maybe you have multiple elements which have the same class name. That's how: I do this, but why that?
– Foo
Nov 19 '18 at 9:54
@TânNguyễn i have already checked that , i think when i click to open the modal the number of instances of modal = number of ajax called
– Sanjit Bhardwaj
Nov 19 '18 at 9:56
you sample code is not enough to understand, maybe it's something related to how you add the event listener...
– Plastic
Nov 19 '18 at 9:57
can you show your html or better to make a fiddle for it
– Negi Rox
Nov 19 '18 at 10:00
1
i would suggest to bind a click function on page load
– Negi Rox
Nov 19 '18 at 10:11
I think the problem comes from the way you name the element. Maybe you have multiple elements which have the same class name. That's how: I do this, but why that?
– Foo
Nov 19 '18 at 9:54
I think the problem comes from the way you name the element. Maybe you have multiple elements which have the same class name. That's how: I do this, but why that?
– Foo
Nov 19 '18 at 9:54
@TânNguyễn i have already checked that , i think when i click to open the modal the number of instances of modal = number of ajax called
– Sanjit Bhardwaj
Nov 19 '18 at 9:56
@TânNguyễn i have already checked that , i think when i click to open the modal the number of instances of modal = number of ajax called
– Sanjit Bhardwaj
Nov 19 '18 at 9:56
you sample code is not enough to understand, maybe it's something related to how you add the event listener...
– Plastic
Nov 19 '18 at 9:57
you sample code is not enough to understand, maybe it's something related to how you add the event listener...
– Plastic
Nov 19 '18 at 9:57
can you show your html or better to make a fiddle for it
– Negi Rox
Nov 19 '18 at 10:00
can you show your html or better to make a fiddle for it
– Negi Rox
Nov 19 '18 at 10:00
1
1
i would suggest to bind a click function on page load
– Negi Rox
Nov 19 '18 at 10:11
i would suggest to bind a click function on page load
– Negi Rox
Nov 19 '18 at 10:11
|
show 5 more comments
4 Answers
4
active
oldest
votes
1
Try to unbind the click event whenever page loads.You can use unbind method as follows.
$(document).unbind('click');
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 9:59
it will not work.
– Negi Rox
Nov 19 '18 at 10:00
add a comment |
0
Try to use off to unbind the events:
$('.commonClose').off('click').click(function(e) {
$('.prod_vbody').html('');
})
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
Try like below. It seems that $(document.body).on event is getting bind each time you open popup.
$(document.body).off('click','.change');
$(document.body).on('click','.change', function(e) {
// Your code
});
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
please add your function on document.ready function something like this.
$(document).ready(function(){
$('.change').on('click', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
})
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Try to unbind the click event whenever page loads.You can use unbind method as follows.
$(document).unbind('click');
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 9:59
it will not work.
– Negi Rox
Nov 19 '18 at 10:00
add a comment |
1
Try to unbind the click event whenever page loads.You can use unbind method as follows.
$(document).unbind('click');
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 9:59
it will not work.
– Negi Rox
Nov 19 '18 at 10:00
add a comment |
1
1
1
Try to unbind the click event whenever page loads.You can use unbind method as follows.
$(document).unbind('click');
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
Try to unbind the click event whenever page loads.You can use unbind method as follows.
$(document).unbind('click');
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
answered Nov 19 '18 at 9:55
Anupam BhattAnupam Bhatt
574
574
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 9:59
it will not work.
– Negi Rox
Nov 19 '18 at 10:00
add a comment |
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 9:59
it will not work.
– Negi Rox
Nov 19 '18 at 10:00
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 9:59
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 9:59
it will not work.
– Negi Rox
Nov 19 '18 at 10:00
it will not work.
– Negi Rox
Nov 19 '18 at 10:00
add a comment |
0
Try to use off to unbind the events:
$('.commonClose').off('click').click(function(e) {
$('.prod_vbody').html('');
})
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
Try to use off to unbind the events:
$('.commonClose').off('click').click(function(e) {
$('.prod_vbody').html('');
})
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
0
0
Try to use off to unbind the events:
$('.commonClose').off('click').click(function(e) {
$('.prod_vbody').html('');
})
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
Try to use off to unbind the events:
$('.commonClose').off('click').click(function(e) {
$('.prod_vbody').html('');
})
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
answered Nov 19 '18 at 10:08
Jose G.Jose G.
3216
3216
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
Try like below. It seems that $(document.body).on event is getting bind each time you open popup.
$(document.body).off('click','.change');
$(document.body).on('click','.change', function(e) {
// Your code
});
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
Try like below. It seems that $(document.body).on event is getting bind each time you open popup.
$(document.body).off('click','.change');
$(document.body).on('click','.change', function(e) {
// Your code
});
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
0
0
Try like below. It seems that $(document.body).on event is getting bind each time you open popup.
$(document.body).off('click','.change');
$(document.body).on('click','.change', function(e) {
// Your code
});
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
Try like below. It seems that $(document.body).on event is getting bind each time you open popup.
$(document.body).off('click','.change');
$(document.body).on('click','.change', function(e) {
// Your code
});
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
edited Nov 19 '18 at 10:16
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
answered Nov 19 '18 at 10:07
KaranKaran
3,0902324
3,0902324
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
it is not working
– Sanjit Bhardwaj
Nov 19 '18 at 10:10
add a comment |
0
please add your function on document.ready function something like this.
$(document).ready(function(){
$('.change').on('click', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
})
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
add a comment |
0
please add your function on document.ready function something like this.
$(document).ready(function(){
$('.change').on('click', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
})
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
add a comment |
0
0
0
please add your function on document.ready function something like this.
$(document).ready(function(){
$('.change').on('click', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
})
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
please add your function on document.ready function something like this.
$(document).ready(function(){
$('.change').on('click', function(e) {
alert('clicked');
$.ajax({
url: base_url + 'Like',
type: 'POST',
data: "product_id=" + $('#product_id').val() + "&from=product",
dataType: "json",
success: function (response)
{
if (response.exists == "1")
{
//$('#success_wish').html(response.message);
//$('#success_wish').show("slow");
// $('.hello').attr('src', swap).attr("data",current);
$(".product_like_li").html(response.likeText);
window.setTimeout(function () {
$('#success_wish').hide("slow")
}, 3000);
}
if (response.exists == "2")
{
//$('#success_error').html(response.message);
//$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
window.setTimeout(function () {
window.location.href = base_url + 'Login'
}, 3000);
}
if (response.exists == "0")
{
$('#success_error').html(response.message);
$('#success_error').show("slow");
window.setTimeout(function () {
$('#success_error').hide("slow")
}, 3000);
}
}
});
});
})
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
edited Nov 19 '18 at 10:18
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
answered Nov 19 '18 at 10:12
Negi RoxNegi Rox
1,7051511
1,7051511
add a comment |
add a comment |
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I think the problem comes from the way you name the element. Maybe you have multiple elements which have the same class name. That's how: I do this, but why that?
– Foo
Nov 19 '18 at 9:54
@TânNguyễn i have already checked that , i think when i click to open the modal the number of instances of modal = number of ajax called
– Sanjit Bhardwaj
Nov 19 '18 at 9:56
you sample code is not enough to understand, maybe it's something related to how you add the event listener...
– Plastic
Nov 19 '18 at 9:57
can you show your html or better to make a fiddle for it
– Negi Rox
Nov 19 '18 at 10:00
1
i would suggest to bind a click function on page load
– Negi Rox
Nov 19 '18 at 10:11