I made an ajax request to get the name of each button I click... Then I want to put everything in that url into "#Container".(url define a page that has some codes) It works for me at the first time...but for the other times I have to reload the page to show me the details of each button and it doesn't show me details of other buttons that clicked after first desired button..
$(function () {
$('button').click(function () {
var data= $(this).val();
$.ajax({
url: '/myurl',
type: 'post',
dataType: 'json',
data: {
data: data
},
success: function (data) {
$('#container').html(data);
}
});
});
});
What should I do? Is there something preventing of running the ajax for next times ?
Try with $(document).on('click', instead of $('button').click like below
$(function () {
$(document).on('click','button',function () {
var data= $(this).val();
alert(data);
$.ajax({
url: '/myurl',
type: 'post',
dataType: 'json',
data: {data: data},
success: function (data) {
$('#container').html(data);}
});
});
});
You will need to re-bind ajax event because of its re-rendering DOM after an ajax request completes. All those elements under #container
are not virtual part of DOM and your click event only works for pre-existed elements.
Try this
Wrap ajax event in a separate function
function ajaxEvent(data){
$.ajax({
url: '/myurl',
type: 'post',
dataType: 'json',
data: {data: data},
success: function (data) {
$('#container').html(data);
clickEvent(); //note this function attachment I added below
}
});
}
You can wrap your click event into a function
function clickEvent(){
$('button').off('click');
$('button').on('click',function(){
ajaxEvent($(this).val());
});
}
Now js looks like
<script>
$(function(){
clickEvent();
});
function ajaxEvent(data){
$.ajax({
url: '/myurl',
type: 'post',
dataType: 'json',
data: {data: data},
success: function (data) {
$('#container').html(data);
}
});
}
function clickEvent(){
$('button').off('click');
$('button').on('click',function(){
ajaxEvent($(this).val());
});
}
</script>
Also I see you are passing ajax data as $('button').val()
which is not correct. Use a input type text if you want to send a data to server.