android - 如何从 Firestore 任务中获取响应代码

Question

在 com.google.android.gms.tasks 中，我想知道有哪些类型的异常以及如何在错误响应时过滤它们？

这展示了我希望看到的内容:

   db.collection("users")
            .get()
            .addOnCompleteListener(new OnCompleteListener<QuerySnapshot>() {
                @Override
                public void onComplete(@NonNull Task<QuerySnapshot> task) {
                     if (task.isSuccessful()) {
                         for (DocumentSnapshot document : task.getResult()) {
                             s(document.getId() + " => " + document.getData());
                         }
                     } else {
// Demonstrating here what I like to see using the okhttp respons code as en example.. 
                         switch (task.getException().getCause()) {// obviously not working only for demonstration.. 
                            case "OK":
                               return Status.OK;
                            case "ZERO_RESULTS":
                               return Status.ZERO_RESULTS;
                            case "OVER_QUERY_LIMIT":
                               return Status.OVER_QUERY_LIMIT;
                            case "REQUEST_DENIED":
                               return Status.REQUEST_DENIED;
                            case "INVALID_REQUEST":
                               return Status.INVALID_REQUEST;
                            default:
                               return null;
                         }
                     }
                 }
             });

如果可以的话就太好了。

Answer 1

我自己还没有尝试过，但是通过阅读引用文档，您应该可以调用 FirebaseFirestoreException.getCode() :

((FirebaseFirestoreException)task.getException()).getCode()

这为您提供了此处记录的代码之一:https://firebase.google.com/docs/reference/android/com/google/firebase/firestore/FirebaseFirestoreException.Code