javascript - Google SignOut 过程中的错误未捕获错误 : nb JS

Question

为了重构我的代码，我编写了如下的 google SignOut 函数

function googleSignOut() {
    if (typeof module_google_login == 'undefined') {
        return false;
    }
    gapi.load('auth2', function () {
        var gApiAuth = gapi.auth2;
        gApiAuth.init().then(doGoogleLogout(gApiAuth));
    });
};

function doGoogleLogout(gApiAuth) {
    console.log(gApiAuth);
    var googleAuth = gApiAuth.getAuthInstance();
    googleAuth.signOut().then(function () {
        $.ajax({
            type: 'POST',
            url: '/account/logout/',
            success: function () {
                auth2.disconnect();
                window.location = "/account";
            }
        });
    });
}

但是它在控制台中给出这样的错误

Uncaught Error: nb
    at tE (cb=gapi.loaded_0:201)
    at jF.<anonymous> (cb=gapi.loaded_0:248)
    at new _.C (cb=gapi.loaded_0:123)
    at jF.BT (cb=gapi.loaded_0:248)
    at Ay.Qv.a.<computed> [as signOut] (cb=gapi.loaded_0:227)
    at doGoogleLogout (main.js:48)
    at main.js:41
    at platform.js:18
    at Sa (platform.js:10)
    at Y (platform.js:18)

但是我是这样实现的

function googleSignOut() {
    if (typeof module_google_login == 'undefined') {
        return false;
    }
    gapi.load('auth2', function () {
        gapi.auth2.init().then(function () {
            var auth2 = gapi.auth2.getAuthInstance();
            auth2.signOut().then(function () {
                $.ajax({
                    type: 'POST',
                    url: '/account/logout/',
                    success: function () {

                        window.location = "/account";
                    }
                });
            });
        });
    });
};

虽然看起来不太好，但它工作正常。如果有人能告诉我我以前的实现有什么问题以及什么不起作用。

Answer 1

问题出在这一行gApiAuth.init().then(doGoogleLogout(gApiAuth));.

对于 promise 的成功处理程序，您需要传递函数引用而不是直接调用函数。这里 doGoogleLogout(gApiAuth) 将在 promise 被解决之前被调用。将其更改为 gApiAuth.init().then(()=>doGoogleLogout(gApiAuth))

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