在XDK中单击按钮登录后如何调用下一页

How to call the next page after button login clicked in XDK

本文关键字：调用 一页 何调用 XDK 单击 按钮 登录      更新时间：2023-09-26

在我的登录脚本中，当用户名和密码都正确时，它必须转到英特尔XDK的下一页(主/菜单页)。我的问题是如何或什么代码，我可以使用调用下一页每当用户名和密码是正确的(登录成功)?

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!!!");
}

明白了。

it will be…像这样……

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