javascript jQuery - 使用 AJAX 加载内容时，document.ready 不会触发
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jQuery - document.ready not firing when content loaded with AJAX
提问by Eduard Luca
I have a simple custom tabbing module, that loads tabs with an AJAX request (via $(elem).load()). On each page that is loaded with AJAX I have some JavaScript. The first time the page loads (via direct input of URL, not AJAX), the javascript fires up perfectly. When I navigate away from the page via the AJAX tabs, the javascripts from the pages aren't loading anymore.
我有一个简单的自定义选项卡模块，它使用 AJAX 请求（通过$(elem).load()）加载选项卡。在每个加载了 AJAX 的页面上，我都有一些 JavaScript。页面第一次加载时（通过直接输入 URL，而不是 AJAX），javascript 完美启动。当我通过 AJAX 选项卡离开页面时，页面中的 javascript 不再加载。
Is there any way I can force them to execute?
有什么办法可以强迫他们执行吗？
(The javascript that is not firing is placed in a $(document).ready()function if that helps)
（$(document).ready()如果有帮助，未触发的 javascript 被放置在一个函数中）
回答by Samich
You need to use callback of load()function:
您需要使用load()函数回调：
$(elem).load('source.html', function() {
// here you need to perofrm something what you need
ActionOnDocumentReady();
});
You can put all your actions in $(document).readyinto some function (ex ActionOnDocumentReady()) and call it on load()callback.
您可以将所有操作$(document).ready放入某个函数（ex ActionOnDocumentReady()）中并在load()回调中调用它。
回答by oezi
the domeready event fires only when is initial dom is ready (like the name and the jquery-apisuggest).
仅当初始 dom 准备好（如名称和jquery-api建议）时，domeready 事件才会触发。
if you want to use jquerys load() and fire a function if that loading has been done, you'll have to use a callback-function (according to the api).
如果要使用 jquerys load() 并在加载完成后触发函数，则必须使用回调函数（根据 api）。
if you want to do the same thing on domready and load-events, the best way would be to define a new function for that:
如果你想对 domready 和 load-events 做同样的事情，最好的方法是为此定义一个新函数：
function dostuff(){
// do some stuff here
}
and fire this function in both cases:
并在两种情况下触发此功能：
$(function(){ // domready | shortcut for "$(document).ready()"
dostuff();
})
$('#something').load('script.php', function() { // callback on load()
dostuff();
});
回答by wasimbhalli
Put your $(document).ready()code in a new method. Lets call it methodA. Now call this methodAfrom $(document).ready(). Secondly, call the same method after ajax request is successful. That should solve your problem.
将您的$(document).ready()代码放在一个新方法中。让我们称之为methodA。现在methodA从$(document).ready(). 其次，ajax请求成功后调用同样的方法。那应该可以解决您的问题。
