Question

这允许我通过Gmail登录并显示其显示的图片，我的姓名等，但不能与其他随机人的ID一起使用。

我在developer.google.com上添加了更多所有者，并且允许他们，但我不允许。

public class MainActivity extends AppCompatActivity implements View.OnClickListener, GoogleApiClient.OnConnectionFailedListener{

    private LinearLayout Prof_Section;
    private Button SignOut;
    private SignInButton  SignIn;
    private TextView Name,Email;
    private ImageView Prof_Pic;
    private GoogleApiClient googleApiClient;
    private static final int REQ_CODE=9001;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        Prof_Section = (LinearLayout)findViewById(R.id.Prof_Section);
        SignOut = (Button)findViewById(R.id.bn_logout);
        SignIn = (SignInButton)findViewById(R.id.bn_login);
        Name=(TextView)findViewById(R.id.name);
        Email=(TextView)findViewById(R.id.email);
        Prof_Pic=(ImageView) findViewById(R.id.prof_pic);
        SignIn.setOnClickListener(this);
        SignOut.setOnClickListener(this);
        Prof_Section.setVisibility(View.GONE);
        GoogleSignInOptions signInOptions = new GoogleSignInOptions.Builder(GoogleSignInOptions.DEFAULT_SIGN_IN).requestEmail().build();
        googleApiClient = new GoogleApiClient.Builder(this).enableAutoManage(this,this).addApi(Auth.GOOGLE_SIGN_IN_API,signInOptions).build();
    }
    @Override
    public void onClick(View v) {
        switch(v.getId())
        {
            case R.id.bn_login:
                signIn();
                break;
            case R.id.bn_logout:
                signOut();
                break;
        }
    }

    @Override
    public void onConnectionFailed(@NonNull ConnectionResult connectionResult) {

    }
    private void signIn()
    {
        Intent intent = Auth.GoogleSignInApi.getSignInIntent(googleApiClient);
        startActivityForResult(intent,REQ_CODE);
    }
    private void signOut()
    {
        Auth.GoogleSignInApi.signOut(googleApiClient).setResultCallback(new ResultCallback<Status>() {
            @Override
            public void onResult(@NonNull Status status) {
                updateUI(false);
            }
        });

    }
    private void handleResult(GoogleSignInResult result)
    {if(result.isSuccess())
    {
        GoogleSignInAccount account = result.getSignInAccount();
        String name = account.getDisplayName();
        String email = account.getEmail();
        String img_url = account.getPhotoUrl().toString();
        Name.setText(name);
        Email.setText(email);
       // Glide.with(this).load(img_url).into(Prof_Pic);
      //  Glide.with(this)
             //   .load(img_url)
           // //    .override(300, 200)
             //   .into(Prof_Pic);
      //  Glide.with(this).load(img_url).into(Prof_pic);
        updateUI(true);

    }
    else
    {updateUI(false);}

    }
    private void updateUI(boolean isLogin){
        if(isLogin){
            Prof_Section.setVisibility(View.VISIBLE);
            SignIn.setVisibility((View.GONE));
        }
        else
        {
            Prof_Section.setVisibility(View.GONE);
            SignIn.setVisibility(View.VISIBLE);
        }
    }
    @Override
    protected void onActivityResult(int requestCode, int resultCode,Intent data)
    {
        super.onActivityResult(requestCode,resultCode,data);
        if(requestCode==REQ_CODE)
        {
            GoogleSignInResult result = Auth.GoogleSignInApi.getSignInResultFromIntent(data);
            handleResult(result);
        }

    }
}

它记录我并显示我的详细信息，但不显示其他人的详细信息。我希望他们尝试登录时也能提供详细信息

Answer 1

SignInButton登录= findViewById（R.id。）;

findViewById（R.id.bn_login）.setOnClickListener（this）;

findViewById（R.id.bn_login）.setVisibility（View.GONE）;

我的Android应用程序中的Gmail登录问题

1 个答案: