jquery ajax 同步异步的执行 return值不能取得的解决方案

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);
}