Google OAuth 2.0 不调用回调函数

Question

我在本地测试过这个passport google策略，效果很好，回调函数调用成功，可以使用google账号登录。但是，当我将此代码推送到生产服务器时，我看到回调函数没有被调用。从用户的角度来看，当他们单击链接以使用 google 策略时，他们会看到 google 的“选择帐户”屏幕，然后加载一段时间，然后再将它们转回到他们之前所在的页面上。我看不出问题出在哪里。提前感谢您的帮助！

passport.use('google', new GoogleStrategy({
    clientID: "redacted",
    clientSecret: "redacted",
    callbackURL: "/google/callback"
  },
    function(accessToken, refreshToken, profile, done) {
          const userQueryString = `
            SELECT firstName, lastName, AgentReference, agent_number, website_password, account_type, status, state, BIN(Flags) as flags 
            FROM compass.agent 
            WHERE googleID = ?`;
          console.log("google profile: \n", profile);
          //pull the user data for the agent with the same googleID as the selected accoutn at login.
          db.query(userQueryString,[profile.id],(err, response, fields)=>{
            if(err) {
              done(err);
            } else if(response.length === 0) {
              done(null, false);
            } else if(response[0].status === 'Terminated') {
              console.log('account is terminated.');
              done(null, false);
            } else {
              let user = {
                id: response[0].AgentReference
                , accountType: response[0].account_type
                , agent_number:response[0].agent_number
                , name:response[0].firstName + ' ' + response[0].lastName
                , state: response[0].state
                , flags: response[0].flags
              };
              let flagRegex = RegExp('^[0-1]*1[0-1]{9}$');
              if(flagRegex.test(user.flags)){
                console.log(JSON.stringify(user));
                return done(null, user);
              } else {
                return done(null, false, {message: "You do not have access to this feature. Please speak with your manager for more information."})
              }
            }
      });
    }
  )
);

Answer 1

将 proxy:true 添加到 google 策略为我解决了这个问题。因此，回调 URL 以 http 而不是 https 的形式发送到 google。