2、 通过传入一个函数解决这个问题。 复制代码 代码如下: function getAjaxReturn(success_function,fail_function){ var bol=false; $.ajax({ type:"POST", data:"username="+vusername.value, success:function(msg){ if(msg=="ok"){ showtipex(vusername.id,"<img src="images/ok.gif"/><b><font color="#ffff00">该用户名可以使用</font></b>",false) success_function(msg); } else showtipex(vusername.id,"<img src="images/cancel.gif"/><b><font color="#ffff00">该用户已被注册</font></b>",false); vusername.className="bigwrong"; fail_function(msg); //return false; } }); function success_function(info) //do what you want do alert(info); funciont fail_function(info) //do what you want do alert(info); }