2009-09-24 61 views

回答

2

做AJAX调用,并定义一个回调函数。

在回调函数调用此方法:

function tb_show(caption, url, imageGroup) 

这通常应显示tickbox

发出Ajax调用:

$.ajax({ 
    url: "yoururl.php", 
    cache: false, 
    success: function(){ // --> this defines what function needs to be called when the ajaxcall was successful. 
    // note that you'll need to fill the variables caption, url and imagegroup here. I don't know what they should be in your case. 
    function tb_show(caption, url, imageGroup); 
    } 
}); 
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我想你明白我的想法。但请您详细解释解决方案。 thanx again .. – assaqqaf 2009-09-24 11:57:21

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我按照你的指示它出现只是加载图像..图像(由url提供)is'n出现.. – assaqqaf 2009-09-24 12:52:38

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首先尝试显示警报,以检查您的ajax是否工作正常工作。 – Natrium 2009-09-24 13:17:04