谷歌注册/ SignIn与颤振+ Back4App

Question

下面的是用过的代码：

import 'package:flutter/material.dart';
import 'package:google_sign_in/google_sign_in.dart';
import 'package:parse_server_sdk_flutter/parse_server_sdk.dart';

class LoginScreen extends StatefulWidget {
  @override
  _LoginScreenState createState() => _LoginScreenState();
}

class _LoginScreenState extends State<LoginScreen> {
  Future<void> _signUpGoogle() async {
    // loading = true;

    final GoogleSignIn _googleSignIn = GoogleSignIn(
        scopes: ['email', 'https://www.googleapis.com/auth/contacts.readonly']);

    sigInGoogle() async {
      GoogleSignInAccount account = await _googleSignIn.signIn();
      GoogleSignInAuthentication authentication = await account.authentication;
      final ParseResponse response = await ParseUser.loginWith(
          'google',
          google(_googleSignIn.currentUser.id,
              authentication.accessToken.toString(), authentication.idToken));

      // print('response.error' + response.error.toString());
      // print('response.results' + response.results.toString());
      // print('response.results' + response.statusCode.toString());
      print(response.toString());
      if (response.success) {
        print('parse google signin successs');
        print(response.success);
      } else {
        print('parse google SignIn Failed');
        print('response.error:  ' + response.error.toString());
        // print(google(_googleSignIn.currentUser.id,
        //     authentication.accessToken.toString(), authentication.idToken));
      }
    }

    sigInGoogle();
  }

  @override
  Widget build(BuildContext context) {
    return Center(
      child: IconButton(
        icon: Icon(Icons.login),
        onPressed: _signUpGoogle,
      ),
    );
  }
}

下面的是日志：

I/颤振(23392)：“ParseResponse”的实例

I/颤振(23392)：解析google SignIn失败

I/颤振(23392)：response.error：

I/颤振(23392)：

I/颤振(23392)：ParseException (类型: ObjectNotFound)：

I/颤振(23392)：代码:101个

I/颤振(23392)：消息: Google auth对此用户无效。

请帮助解决这个问题.

Answer 1

看起来你的参数在传递到'google‘数据类型时是错误的。参数如下；

Map<String, dynamic> google(String token, String id, String idToken)

使用你的代码但按这个顺序传递，

google(authentication.accessToken.toString(), _googleSignIn.currentUser!.id, authentication.idToken!));

对我来说这是个诀窍