0

由于某种原因，单击此页面中的第 2 步后，我的 Jquery 将无法工作。去看看。我添加了：

<script type="text/javascript">
    $('li').click(function () {
        $(this).slideUp();
    });
</script>

它对第二页没有任何影响，它只影响第一页。这是我到目前为止的代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Jquery Wizard</title>
<link href="css/default.css" rel="stylesheet" type="text/css" />
<script src="js/jquery.js" type="text/javascript"></script>
<script src="js/validate.min.js" type="text/javascript"></script>
<script src="js/init.js" type="text/javascript"></script>
<link href="style.css" rel="stylesheet" type="text/css" />
<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<script type="text/javascript">
function loadnext(divout,divin){

$("." + divout).hide();
$("." + divin).fadeIn("fast");
}



</script>
</head>
<body>
<div id="wizardwrapper">

  <div class="1">

    <h3>Contact Info</h3>
    <div id="wizardcontent">
<div class="wrapper">
Please Make Sure Our Records Are Correct:
        <form id="form" class="blocks" action="#" method="post">
            <p>
                <label>Name:</label>
                <input type="text" class="text" name="name" />
            </p>
            <p>
                <label>Company:</label>
                <input type="text" class="text" name="company" />
            </p>
            <p>
                <label>Your e-mail:</label>
                <input type="text" class="text" name="email" />
            </p>
            <p>
                <label>Contact number:</label>
                <input type="text" class="text" name="phone" />
            </p>

            <p>
                <label>&nbsp;</label>

            </p>
        </form>
    </div>

    </div>
    <div class="buttons">
      <button type="submit" class="previous"  disabled="disabled"> <img src="images/arrow_left.png" alt=""/> Back </button>
      <button type="submit" class="next" onclick="loadnext(1,2);"> Next <img src="images/arrow_right.png" alt="" /> </button>

    </div>
    <ul id="mainNav" class="fiveStep">
      <li class="current"><a title=""><em>Step 1: User Info</em><span>Enter or Correct Contact Information</span></a></li>
      <li><a title=""><em>Step 2: Subscription Management</em><span>Select your Subscription Settings</span></a></li>

    </ul>
    <div style="clear:both"></div>
  </div>

  <div id="wizardpanel" class="2">
    <h3>Step 2</h3>
    <div id="wizardcontent">  
    <script type="text/javascript">
      $('li').click(function () {
      $(this).slideUp();
    });
        </script> 

    <div id="selection">
    <ul>
    <li>asdf</li>
    <li>asdf</li>
    <li>asdf</li>
    </ul>

    </div>


    </div>

    <div class="buttons">
      <button type="submit" class="previous" onclick="loadnext(2,1);"> <img src="images/arrow_left.png" alt="" /> Back </button>

      <button type="submit" class="next" onclick="submit()"> Finish <img src="images/arrow_right.png" alt="" /> </button>
    </div>
    <ul id="mainNav" class="fiveStep">
      <li class="lastDone"><a href="/" title=""><em>Step 1: User Info</em><span>Enter or Correct Contact Information</span></a></li>

      <li class="current"><a title=""><em>Step 2: Subscription Management</em><span>Select your Subscription Settings</span></a></li>

  </div>


</div>
</body>
</html>
4

1 回答 1

0

您必须将函数包装在

$(document).ready(function(){


});

让 jQuery 在初始加载时工作（没有函数（））。

所以你的功能需要是：

$(document).ready(function(){
$('li').click(function () {
$(this).slideUp();
});
});
于 2013-07-30T20:16:19.203 回答