druzuz321103 2010-01-25 08:37 采纳率: 100%
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jQuery,PHP表单echo不起作用

Could you guys let me know what is wrong below, for some reason, when I click on the delete image, which is supposed to return the echo from dela.php file, but does not.

<script language="javascript" type="text/javascript">
    $(document).ready(function() {
        $('#del_form').ajaxForm({
            target: '#del',
            success: function() {
                $('#del').fadeIn(40000);
            }
        });
    });
</script>

<div>
    <form action="dela.php" id="del_form" method="post">
        <input type="image" src="del.gif" id="al_del" value="clicked" />
        click the image on the left
    </form>
</div>
<div id="del" style="background-color:#FFFF99; width:200px; height:100px;"></div>

// dela.php
<?
    if ($_POST['al_del']) {
        echo 'variable pass success';
    }
?>
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3条回答

  • duanpo2813 2010-01-25 08:42
    关注

    POST variables are based on input names, not ID's, afaik.

    Also I'd usually go

    if(isset($_POST['al_del']))
    

    But that's a side bar.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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