大家好，谢谢阅读。
我有一个问题，我不知道如何保持加载了ajax的页面的jquery代码的正常运行。问题是，在原始页面上执行代码，因此jquery框架已经加载，当ajax调用来加载另一个页面的数据时，我得到的是不起作用的jquery代码。从ajax页面加载的jquery代码不起作用。
如果我不清楚，请澄清，更多详细信息，代码如下：
我加载jquery页面的页面：
<?php
ob_start();
include 'includes/chk.php';
?>
<script type="text/javascript">
$(document).ready(function() {
$('#account_link').click(function(){
$.ajax({
type: "POST",
async:false,
url: "ajax/settings_account.php",
success: function(msg){
$('#div_container_sett_acc').replaceWith('<div id="div_container_sett_acc" style="text-align: center;">'+msg+'</div>');
}
});
});
$('#settings_link').click(function(){
$.ajax({
type: "POST",
async:false,
url: "ajax/settings_account.php",
success: function(msg){
$('#div_container_sett_acc').replaceWith('<div id="div_container_sett_acc" style="text-align: center;">'+msg+'</div>');
}
});
});
});
</script>
<div class="ui-grid-a">
<div class="ui-block-a">
<ul data-role="listview" data-divider-theme="d" data-inset="true">
<li data-theme="c">
<a href="#settings" >
Settings
</a>
</li>
<li data-theme="c">
<a href="#account" id="account_link" >
Account
</a>
</li>
</ul>
</div>
<div id="div_container_sett_acc">greg</div>
</div>
<?php
$pagemaincontent = ob_get_contents();
ob_end_clean();
$pageHead='';
$home_page='ui-btn-active';
$pageTitle ='';
include("master.php");
?>
在正在加载的ajax页面上：
<?php
include_once ("../includes/mysql.php");
include_once ("../includes/chk.php");
?>
<div class="ui-block-b">
<div id="settings_link" data-theme="a">
<div data-role="content" data-theme="a" >
<h2>Two</h2>
<p>I have an id of "two" on my page container. I'm the second page container in this multi-page template.</p>
<p>Notice that the theme is different for this page because we've added a few <code>data-theme</code> swatch assigments here to show off how flexible it is. You can add any content or widget to these pages, but we're keeping these simple.</p>
<p><a href="#one" data-direction="reverse" data-role="button" data-theme="b">Back to page "one"</a></p>
</div><!-- /content -->
</div><!-- /page two -->
<div id="account" data-theme="a">
<div data-role="content" data-theme="a">
<form url="">
<div data-role="fieldcontain">
<fieldset data-role="controlgroup">
<label for="textinput1">
Boh
</label>
<input id="textinput1" placeholder="" value="" type="text" />
</fieldset>
</div>
<div data-role="fieldcontain">
<label for="selectmenu1">
Choose:
</label>
<select name="selectmenu1" id="selectmenu1">
<option value="option1">
Option 1
</option>
</select>
</div>
<div data-role="fieldcontain">
<fieldset data-role="controlgroup" data-type="vertical">
<legend>
Choose:
</legend>
<input name="radiobuttons1" id="radio1" value="" type="radio" />
<label for="radio1">
Option
</label>
</fieldset>
</div>
<div data-role="fieldcontain">
<fieldset data-role="controlgroup" data-type="vertical">
<legend>
Choose:
</legend>
<input name="checkbox1" id="checkbox1" type="checkbox" />
<label for="checkbox1">
Checkbox
</label>
</fieldset>
</div>
<input type="submit" value="Submit" />
</form>
</div><!-- /content -->
</div><!-- /page two -->
</div>
如果有人能帮我，我将是最棒的完整
您应该使用on而不是bind:
$(document).ready(function() {
$('#account_link').on('click', function() {...})
$('#settings_link').on('click', function() {...})
});
您需要在加载的内容中有一个jquery脚本代码。因此，基本页面中必须有相同的内容，并且$.post 调用的代码中也必须包含相同的内容
$.默认$(document).ready(function() 不提供已发布的内容