[英]How to connect 2 onActivityResult?
我在添加 Facebook 身份验证时遇到问题,因为我也在使用 Google 身份验证,这就是我出错的原因
public class SignUpActivity extends AppCompatActivity {
callbackManager.onActivityResult(requestCode, resultCode, data);
super.onActivityResult(requestCode, resultCode, data);
//Google
@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
// Result returned from launching the Intent from GoogleSignInApi.getSignInIntent(...);
if (requestCode == RC_SIGN_IN) {
Task<GoogleSignInAccount> task = GoogleSignIn.getSignedInAccountFromIntent(data);
try {
// Google Sign In was successful, authenticate with Firebase
GoogleSignInAccount account = task.getResult(ApiException.class);
firebaseAuthWithGoogle(account);
} catch (ApiException e) {
// Google Sign In failed, update UI appropriately
Log.w(TAG, "Google sign in failed", e);
// ...
}
}
}
//and Facebook
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
callbackManager.onActivityResult(requestCode, resultCode, data);
super.onActivityResult(requestCode, resultCode, data);
}
你怎么看我有两个 onActivityResult 方法。 有什么方法可以连接它们并消除错误吗? 这就是我的错误
方法 onActivityResult(int,int,Intent) 已在类 SignUpActivity 中定义
它只是现有两种相同方法的交流。 谢谢。
onActivityResult
是一个 Android 方法,它接收你从startActivityForResult
开始的活动的结果,并返回你提供的 request_code int。
解决方案是在startActivityForResult
使用不同的 REQUEST_CODES ,因此您可以在onActivityResult
进行比较
喜欢:
private static final int FACEBOOK_REQUEST_CODE = 1;
private static final int GOOGLE_REQUEST_CODE = 0;
startActivityForResult(googleLoginIntent, GOOGLE_REQUEST_CODE)
startActivityForResult(facebookLoginIntent, FACEBOOK_REQUEST_CODE)
public void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (requestCode == GOOGLE_REQUEST_CODE) {
//do the code for google result
} else if (requestCode == FACEBOOK_REQUEST_CODE) {
// do the code for facebook result
}
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.