jQuery .live（'click'）仅绑定到页面上的第一个元素

Question

I wrote a jQuery plugin that binds to 8 elements on a page, and I wanted to use .live() to bind the click action to a link in each element. When you click the link it should post a form using ajax. The problem is that all the links for all 8 elements submit the form for the first element on the page. When I use .click() everything works correctly. I would prefer to use .live() since I will have adding more elements dynamically.

Here is some code similar to what I'm doing:

var $container = $(this);
var $form      = $container.find('form.some_form');
var $button    = $container.find('a.some_link');

This will only submit the form for the first element:

$button
.live('click', function() {
  // some code that submits $form via ajax
});

However, this always submits the correct form:

$button
.click( function() {
  // identical code that submits $form via ajax
});

Is there something about .live() that I should know? Stumped.

Answer 1

From the jQuery documentation :

Live events currently only work when used against a selector. For example, this would work: $("li a").live(...) but this would not: $("a", someElement).live(...) and neither would this: $("a").parent().live(...).

In your case, you're calling live() on a variable that's the result of a find() call. That's not a selector. You need to figure out a selector that identifies the elements you want.

---

Edited to add: for anyone finding this later, the preferred approach now is to use the function on() for this. The on() function does not have the same restriction -- since it operates on a jQuery object (rather than implicitly binding to the document) it can be set on a set of elements arrived at by chaining as in the original question.

Answer 2

Try something like this. You can't use "live" except on selectors.

$('a.some_link').live('click', function() {
    // some code that submits $form via ajax
});

Hope it helps!