2009-09-24 60 views

回答

2

做AJAX調用,並定義一個回調函數。

在回調函數調用此方法:

function tb_show(caption, url, imageGroup) 

這通常應顯示tickbox

發出Ajax調用:

$.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); 
    } 
}); 
+0

我想你明白我的想法。但請您詳細解釋解決方案。 thanx again .. – assaqqaf 2009-09-24 11:57:21

+0

我按照你的指示它出現只是加載圖像..圖像(由url提供)is'n出現.. – assaqqaf 2009-09-24 12:52:38

+0

首先嚐試顯示警報,以檢查您的ajax是否工作正常工作。 – Natrium 2009-09-24 13:17:04