Question

我有一个简单的游戏，有一个排行榜按钮，它显示了玩家的最高分，在发布此问题之前，我确定了以下内容：

SHA1指纹。
启用了API。
正确包装。
更正应用程序ID。
导入并添加了Google Play服务库。
导入配置文件。
Admob，Google Analytics和崩溃报告工作正常。
GET ACCOUNTS已添加，仍然相同。

我的问题是，没有显示帐号选择器/选择器，甚至没有登录，我收到以下错误：

ConnectionResult{statusCode=SIGN_IN_REQUIRED, resolution=PendingIntent{fbd982b: android.os.BinderProxy@b3d396d}, message=null

代码：

public class MainActivity extends AppCompatActivity implements
    GoogleApiClient.ConnectionCallbacks, GoogleApiClient.OnConnectionFailedListener {

ImageButton mLeaderBoard;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    setContentView(R.layout.activity_welcome);
    mLeaderBoard = (ImageButton) findViewById(R.id.leaderboard);
    mLeaderBoard.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            ShowBoard();
        }
    });

   mGoogleApiClient = new GoogleApiClient.Builder(this)
            .addConnectionCallbacks(this)
            .addOnConnectionFailedListener(this)
            .addApi(Games.API)
            .addScope(Games.SCOPE_GAMES)
            .addScope(SCOPE_PLUS_PROFILE)
            .build();
}

@Override
protected void onStart() {
    super.onStart();
    mGoogleApiClient.connect();

}

@Override
protected void onStop() {
    super.onStop();
    if (mGoogleApiClient.isConnected()) {
        mGoogleApiClient.disconnect();
    }
}

private void ShowBoard() {
    if(!isSignedIn()){
        mGoogleApiClient.connect();
    }
    mGoogleApiClient.connect();
    if(isSignedIn())
    startActivityForResult(Games.Leaderboards.getLeaderboardIntent(mGoogleApiClient,
            LEADERBOARD_TOP_SCORE_ID), 1);
    else
        Toast.makeText(this, getResources().getString(R.string.leaderboard_not_available), Toast.LENGTH_LONG).show();

}

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    // Check which request we're responding to
    if (requestCode == 1) {

        Log.i("GoogleSignInApi","Result: " + resultCode + " - Data: " + data.getData());
        if (resultCode == RESULT_OK) {

        }
    }
}



@Override
public void onConnected(Bundle bundle) {

}

@Override
public void onConnectionSuspended(int i) {
    mGoogleApiClient.connect();
}

@Override
public void onConnectionFailed(@NonNull ConnectionResult connectionResult) {
    Log.i("GoogleSignInApi","Problem: " + connectionResult.getErrorMessage() + " - Test: " + connectionResult.toString());
}

 }

Answer 1

AFAIK，您应该有SCOPE_PLUS_PROFILE - OAuth2.0范围来访问用户的个人资料数据。您需要启用Google+ API并使用SHA1和您的包创建凭据。 Oauth Consent屏幕显示您的电子邮件和产品名称。

确保您拥有所需的权限和所有包，例如：

applicationId in tag <application> 
package name in AndroidManifest.xml 
package name in Credentials in Google Developer Console

<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<uses-permission android:name="android.permission.GET_ACCOUNTS" />

此外，您需要致电onActivityResult。当客户端无法连接到Google Play服务时，需要此功能来满足所有可能的错误代码。

Answer 2

就我而言，即使用户以前从未使用过Google来登录我的应用程序，我仍试图从Google注销

const signInGoogle = async () => {
    try {
      await GoogleSignin.hasPlayServices();
      const userInfo = await GoogleSignin.signIn();
      console.log(userInfo);
      const payload = {
        token: userInfo.idToken,
        first_name: userInfo.user.givenName,
        last_name: userInfo.user.familyName,
        email: userInfo.user.email,
      };
      props.actions.socialLogin(payload);
      console.warn('google', payload);
    } catch (error) {
      if (error.code === statusCodes.SIGN_IN_CANCELLED) {
        // user cancelled the login flow
        alert('Cancel');
      } else if (error.code === statusCodes.IN_PROGRESS) {
        alert('Signin in progress');
        // operation (f.e. sign in) is in progress already
      } else if (error.code === statusCodes.PLAY_SERVICES_NOT_AVAILABLE) {
        alert('PLAY_SERVICES_NOT_AVAILABLE');
        // play services not available or outdated
      } else {
        console.log(error);
      }
    }
  }
  const handleGoogle = async () => {

    try{
      await GoogleSignin.revokeAccess();
      await GoogleSignin.signOut();
      signInGoogle()
    }catch(e){
      //HERE IT Throws the SIGN_IN_REQUIRED ERROR DUE TO signOut function
      signInGoogle()
    }
  };

SIGN_IN_REQUIRED - Google游戏

2 个答案: