这不会打印出功能吗？ [重复]

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• How do I return the response from an asynchronous call? 31 answers

如何从异步调用返回响应? 31个答案

function getUsername() {
    $.get("http://www.roblox.com/mobileapi/userinfo", function(data) {
        return data.UserName;
    });
}

console.log(getUsername());

It says "undefined" but whenever I visit the link manually it shows

它说“未定义”,但每当我手动访问该链接时,它会显示

{"UserID":74798521,"UserName":"AbstractMadness","RobuxBalance":7024,"TicketsBalance":21530,"ThumbnailUrl":"http://t2.rbxcdn.com/c189198f6c0689ce004d9438c70eb1bb","IsAnyBuildersClubMember":true}

It will also log the name if I do console.log(data.UserName) inside of the function.

如果我在函数内部执行console.log(data.UserName),它也会记录名称。

2 个解决方案

#1

It's not synchronous. You are printing the return of that function, which doesn't return anything (so it's undefined). Even if you were to return the $.get call you wouldn't get what you're looking for. You need to log inside of the callback function or use a promise.

这不是同步的。您正在打印该函数的返回,该函数不返回任何内容(因此未定义)。即使你要退回$ .get电话,你也不会得到你想要的东西。您需要登录回调函数或使用promise。

function getUsername() {
    return $.get("http://www.roblox.com/mobileapi/userinfo", function(data) {
        console.log(data.UserName);
        return data.UserName;
    });
}

getUsername().done(function(username) { console.log(username); });

#2

Function getUsername executes a $.getrequest, but it returns nothing and it ends here.

函数getUsername执行$ .getrequest,但它不返回任何内容,它在此处结束。

You have to log your data in the function of your asynchronous $.get request. This function (function(data)) is a callback function. It will be called, if the response of your $.get request arrive.

您必须在异步$ .get请求的函数中记录您的数据。该函数(函数(数据))是一个回调函数。如果您的$ .get请求的响应到达,它将被调用。