dongtuan1980 2014-10-05 16:15
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jQuery load()方法不起作用

Hi I am printing the ajax html response to div element like-div but the ajax html response is not working.

index.php

  <?php
$res_code = '6754567435';
?>
<html>
<head>
    <style type="text/css">
ul.social-icon > li {
    float: left;
    list-style: none;
    padding: 0 15px;
}
    </style>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$('ul.social-icon li a.like').click(function()
    { 
        var track = <?php echo $res_code; ?> ;
        $(".like-div").load("remove.php?keyword=delete&trackid="+track);
    });
});
</script>
</head>
<body>

<ul class="social-icon">
              <li><a class="like" href="#123">Like</a></li>
</ul>

<div class="like-div"></div>
</body>
</html> 

I want to diaply the ajax response in div like-div on clicking the hyperlink Like but this is not showing anything in the div trait.

remove.php

   <?php
    $keyword = $_REQUEST['keyword'];
    $trackid = $_REQUEST['trackid'];

    if($trackid != '')
{
    echo "hello";
}
    ?>

remove.php is the page from where response will come on sending the request. what could be the reason that it is not showing the response in div?

Please help me to fix this. Thanks

  • 写回答

1条回答 默认 最新

  • douchuilai2355 2014-10-05 16:23
    关注

    You need to cancel the click

    $('ul.social-icon li a.like').click(function(e) {
      e.preventDefault(); 
    

    Also I would do

    $(function(){
      $('ul.social-icon li a.like').click(function(e) { 
        e.preventDefault(); 
        $(".like-div").load("remove.php?keyword=delete&trackid=<?php echo $res_code; ?>");
      });
    });
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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