如何在XDK中单击按钮登录后调用下一页

Question

在我的登录脚本中，当用户名和密码都正确时，必须转到intel XDK的下一页（主/菜单页面）。 我的问题是，只要用户名和密码正确（登录成功），我可以用什么代码来调用下一页？

function validateForm() {
    var formUsername = document.forms.login.username.value;
    var formPassword = document.forms.login.password.value;
    var MINLENGTH = 5;

    // Validate username and password
    if (formUsername === null || formUsername === "") {
        alert("Username must be filled out");
    }
    else if (formPassword === null || formPassword === "") {
        alert("Password must be filled out");
    }
    else if (formUsername.length < MINLENGTH || formPassword.length < MINLENGTH) {
        alert("The minimum length of username and password at least " + MINLENGTH);
    }
    else if(formUsername == 'admin' && formPassword == 'admin'){
        alert('welcome');
        //this is where should i put the code to go at the next page of the XDK API. 
        return;
    }        
    alert("Login failed!!!");
}

Answer 1

得到了那些人..

它会......像这样......

else if(formUsername == 'admin' && formPassword == 'admin'){
    alert('welcome');
    activated_page("#menu");
    return;
}