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怎么用$.getJSON的返回值赋值给变量?

WBOY
WBOYOriginal
2016-06-20 12:34:001828browse

各位大神,我的原程序是这样的:
 $("#item>a[rel=copylink]").zclip({
path: "/zmgjw/js/ZeroClipboard.swf",
copy: function(){
var data = "aid="+arcId;
 $.getJSON("/plus/ajax_copy.php",data, function(re_data){
alert(re_data);
 })}
});

 $.getJSON后,我用alert能返回值,但我想把返回的值re_data再赋给变量VAR result,以便让ZCLIP程序调用。我该怎么写?谢谢!!


回复讨论(解决方案)

$.ajaxSetup({async:false}); //指定 ajax 以同步方式工作var result;$("#item>a[rel=copylink]").zclip({  path: "/zmgjw/js/ZeroClipboard.swf",  copy: function(){  var data = "aid="+arcId;  $.getJSON("/plus/ajax_copy.php",data, function(re_data){    result = re_data;  })}});

非常感谢!已经搞定!谢谢!

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