weixin_33691598 2018-08-19 05:58 采纳率: 0%
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在Ajax中找不到页面(404)

I want to make a WEB page move screen transition in ajax . I wrote in one.html

<html>
<head>
    <meta charset="UTF-8">
    <title>Top</title>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
    <script src="jquery.pjax.js"></script>
</head>

<script type="text/javascript">
   $(function() {
    $.pjax({
        area: '#main'
    });
    $(document).bind('pjax:render', function() {
        $('#main').attr({
            'class': 'fadeInUp'
        });
    });
    });
</script>

<body>
<p>
name
</p>
<input type="text" name="name" size="10">

<p class="button"><a href="two.html">Next</a></p>
</body>
</html>

in two.html

<p>
age
</p>
<input type="text" size="10">

<p class="button"><a href="three.html">Next</a></p>

But when I put Next button in one.html,Page not found (404) Request URL:http://127.0.0.1:8000/app/two.html error happens. I cannot understand why such an error happens.What is wrong in my codes?How should I fix this?

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  • weixin_33711647 2018-08-19 06:03
    关注

    http://127.0.0.1:8000/app/two.html

    so the link you can set blow:

    <p class="button"><a href="app/two.html">Next</a></p>
    
    评论

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