Question

我正在尝试使用他们的Javascript API来使用Google OAuth 2.0流程。我复制了他们网站上的示例代码（API Client，Code）。

一切正常，但退出不起作用。如果我将他们的代码修改为：

      function handleSignoutClick(event) {
         var oauth = gapi.auth2.getAuthInstance();
         console.log("Launch the sign-out....");
         oauth.signOut().then(function() {
            console.log("Sign-in state: "+oauth.isSignedIn.get());
         });
      }

我从oauth.isSignedIn.get（）获得一个“true”值（我希望在执行oauth.sigOut（）之后得到“False”）并且我仍然登录。

有关如何正确退出的任何建议？

<!--
  Copyright (c) 2011 Google Inc.
  Licensed under the Apache License, Version 2.0 (the "License"); you may not
  use this file except in compliance with the License. You may obtain a copy of
  the License at
  http://www.apache.org/licenses/LICENSE-2.0
  Unless required by applicable law or agreed to in writing, software
  distributed under the License is distributed on an "AS IS" BASIS, WITHOUT
  WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the
  License for the specific language governing permissions and limitations under
  the License.
  To run this sample, set apiKey to your application's API key and clientId to
  your application's OAuth 2.0 client ID. They can be generated at:
    https://console.developers.google.com/apis/credentials?project=_
  Then, add a JavaScript origin to the client that corresponds to the domain
  where you will be running the script. Finally, activate the People API at:
    https://console.developers.google.com/apis/library?project=_
-->
<!DOCTYPE html>
<html>
  <head>
    <title>Say hello using the People API</title>
    <meta charset='utf-8' />
  </head>
  <body>
    <p>Say hello using the People API.</p>

    <!--Add buttons to initiate auth sequence and sign out-->
    <button id="authorize-button" style="display: none;">Authorize</button>
    <button id="signout-button" style="display: none;">Sign Out</button>

    <div id="content"></div>

    <script type="text/javascript">
      // Enter an API key from the Google API Console:
      //   https://console.developers.google.com/apis/credentials
      var apiKey = 'YOUR_API_KEY';
      // Enter the API Discovery Docs that describes the APIs you want to
      // access. In this example, we are accessing the People API, so we load
      // Discovery Doc found here: https://developers.google.com/people/api/rest/
      var discoveryDocs = ["https://people.googleapis.com/$discovery/rest?version=v1"];
      // Enter a client ID for a web application from the Google API Console:
      //   https://console.developers.google.com/apis/credentials?project=_
      // In your API Console project, add a JavaScript origin that corresponds
      //   to the domain where you will be running the script.
      var clientId = 'YOUR_WEB_CLIENT_ID.apps.googleusercontent.com';
      // Enter one or more authorization scopes. Refer to the documentation for
      // the API or https://developers.google.com/people/v1/how-tos/authorizing
      // for details.
      var scopes = 'profile';
      var authorizeButton = document.getElementById('authorize-button');
      var signoutButton = document.getElementById('signout-button');
      function handleClientLoad() {
        // Load the API client and auth2 library
        gapi.load('client:auth2', initClient);
      }
      function initClient() {
        gapi.client.init({
            apiKey: apiKey,
            discoveryDocs: discoveryDocs,
            clientId: clientId,
            scope: scopes
        }).then(function () {
          // Listen for sign-in state changes.
          gapi.auth2.getAuthInstance().isSignedIn.listen(updateSigninStatus);
          // Handle the initial sign-in state.
          updateSigninStatus(gapi.auth2.getAuthInstance().isSignedIn.get());
          authorizeButton.onclick = handleAuthClick;
          signoutButton.onclick = handleSignoutClick;
        });
      }
      function updateSigninStatus(isSignedIn) {
        if (isSignedIn) {
          authorizeButton.style.display = 'none';
          signoutButton.style.display = 'block';
          makeApiCall();
        } else {
          authorizeButton.style.display = 'block';
          signoutButton.style.display = 'none';
        }
      }
      function handleAuthClick(event) {
        gapi.auth2.getAuthInstance().signIn();
      }
      function handleSignoutClick(event) {
        gapi.auth2.getAuthInstance().signOut();
      }
      // Load the API and make an API call.  Display the results on the screen.
      function makeApiCall() {
        gapi.client.people.people.get({
          'resourceName': 'people/me',
          'requestMask.includeField': 'person.names'
        }).then(function(resp) {
          var p = document.createElement('p');
          var name = resp.result.names[0].givenName;
          p.appendChild(document.createTextNode('Hello, '+name+'!'));
          document.getElementById('content').appendChild(p);
        });
      }
    </script>
    <script async defer src="https://apis.google.com/js/api.js" 
      onload="this.onload=function(){};handleClientLoad()" 
      onreadystatechange="if (this.readyState === 'complete') this.onload()">
    </script>
  </body>
</html>

Answer 1

好的 - 所以我更多地研究了这个问题。我最初的工作都在'localhost'上。为此，我将代码移到了生产服务器上，并且工作得很好。

那就是说，我仍然不知道为什么这不适用于'localhost'网址。

希望其他人可以更好地了解这个问题。

Google OAuth 2.0 Javascript未退出

1 个答案: