jquery+php怎么实现跳转？

 $.ajax({
        type:"POST",
        url:"login.php",
        dataType:'text',
        data:"&user="+user+"&pwd="+pwd,
        success:function(msg){
            if(msg == "success"){
                document.forms[0].submit();
            }else{
            alert(msg);//如果失败，看看返回什么
                $('#usermsg').text("用户名或密码错误").css("color","red");
            }
        }
    });

if(msg == "success"){
// document.forms[0].submit();
window.location.href='success.php';
}

把后台login.php中的接收post请求改为get请求
在浏览器中手动请求一次,查看返回值是不是success