jquery ajax中实现页面的跳转

[Specialaspnet]

jquery ajax中实现页面的跳转

function User_Regist() {
    var username = $('#regist_name').val();
    var useremail = $('#regist_email').val();
    var userpwd = $('#regist_password').val();
    var uservalidate = $('#regist_result').val();
    $.ajax({
        type: "post",
        contentType: "application/json; charset=utf-8",
        url: "../WebService/User.asmx/UserRegist",
        data: "{'u_name':'" + username + "','u_email':'" + useremail + "','u_pwd':'" + userpwd + "'}",
        success: function(result) {
            var json = eval('(' + result + ')');
            if(json.d==1)
{
这里实现页面跳转，怎么写代码
}
        }
    });
}

谢谢大家了

[shine_fly]

window.location = url

[alan_219_2008]

[Quote=引用 1 楼 shine_fly 的回复:]

window.location = url
[/Quote]
.