jquery click doesn't work on ajax generated content

耗尽温柔 提交于 2019-11-26 12:04:45
gdoron

Should be done this way.

$('body').on('click', '.button', function (){
        alert('click!');
    });

If you have a container that doesn't change during the ajax request, this is more performant:

$('.container').on('click', '.button', function (){
        alert('click!');
    });

Always bind the delegate event to the closest static element that will contain the dynamic elements.

Ok i solved my problem by using the .on() function correctly since i was missing one parameter.

instead of

$(".button").on("click", function() { } );

i used

$(".container").on("click", ".button", function() { } );
Macc

Instead of:

$(".button").on("click", function() { } );

I used:

$(".container").on("click", ".button", function() { } );

I have used this and it worked.

Is this what you're trying to do? Note, I'm putting the $.on() on the parent, but selecting the .button for the action.

.on( events [, selector] [, data], handler(eventObject) )

selector A selector string to filter the descendants of the selected elements that trigger the event. If the selector is null or omitted, the event is always triggered when it reaches the selected element.

http://api.jquery.com/on/

<div id="stuff">
    <button class="button">Click me!</button>
    <p>Stuff</p>
</div>

var $stuff = $('#stuff'),
    ajaxContent = $stuff.html();

$stuff.on('click', '.button', function(){
    $.get('/echo/html/', function(){
        $stuff.empty();
        console.log($stuff.html());
        alert($stuff.html()); // Look behind, #stuff is empty.
        $stuff.html(ajaxContent);
        console.log($stuff.html());
    });
});

http://jsfiddle.net/62uSU/1

Another demonstration:

var $stuff = $('#stuff'),
    ajaxContent = $stuff.html(),
    $ajaxContent,
    colors = ['blue','green','red'],
    color = 0;

$stuff.on('click', '.button', function(){
    $.get('/echo/html/', function(){
        color++;
        if (color == colors.length) color = 0;
        console.log($stuff.html());
        alert($stuff.html());
        $ajaxContent = $(ajaxContent);
        $stuff.append($ajaxContent).css('color', colors[color]);
        console.log($stuff.html());
    });
});

http://jsfiddle.net/62uSU/2/

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