How I can Delete Record from Firebase?-CodePudding

I want to delete this record ?

Because I don't know what exactly it is, I want to know what it is... I expect that it is a special key for every data I upload it creates it automatically How do I delete and what is this key?

 DatabaseReference reference = FirebaseDatabase.getInstance().getReference("poll_post").child(firebaseUser.getUid());
        HashMap<String, Object> hashMap = new HashMap<>();
        
        String postid = reference.push().getKey();

        hashMap.put("postid", postid);
        hashMap.put("time_post", ServerValue.TIMESTAMP);
        hashMap.put("stopcomment", checked);

        hashMap.put("tv_question", addcomment.getText().toString());
        hashMap.put("tvoption1", Answer1.getText().toString());
        hashMap.put("tvoption2", Answer2.getText().toString());
        hashMap.put("vote1","0");
        hashMap.put("vote2", "0");

        hashMap.put("publisher", FirebaseAuth.getInstance().getCurrentUser().getUid());

        reference.push().setValue(hashMap, new DatabaseReference.CompletionListener() {
            @Override
            public void onComplete(DatabaseError databaseError, DatabaseReference databaseReference) {
                //Problem with saving the data
                if (databaseError != null) {


                    Toast.makeText(Write_poll.this, "Error", Toast.LENGTH_SHORT).show();
                    myLoadingButton.showErrorButton();

                } else {


                    myLoadingButton.showDoneButton();
              finish();
                }

            }
        });

CodePudding user response：

refer to firebase documentation and read about the function called push , they clearly said that

push : Add to a list of data in the database. Every time you push a new node onto a list, your database generates a unique key, like messages/users//

so that's means that every time you are pushing to the database , a new record with a unique ID will be generated every you push and you will not be able to override a record using the function push as every time it's called , it will generate a unique ID that you don' want in your case.

to avoid that , instead of

reference.push().setValue(hashMap, new DatabaseReference.CompletionListener(){ . . . }

write :

reference.setValue(hashMap, new DatabaseReference.CompletionListener(){ . . . }

CodePudding user response：

If you want to delete a record from the Realtime Database, you have to create a reference that points to that node. In your particular case it would be:

String uid = FirebaseAuth.getInstance().getCurrentUser().getUid();
DatabaseReference db = FirebaseDatabase.getInstance().getReference();
db.child("poll_post").child("-NBcO...y8cX").removeValue();

If you don't know the key, then you have to create a query to find that post based on something that uniquely identifies it, for example, the time_post. So here is the code:

DatabaseReference db = FirebaseDatabase.getInstance().getReference();
Query queryByTime = db.child("poll_post").orderByChild("time_post").equalTo(1662830174410);
queryByTime.get().addOnCompleteListener(new OnCompleteListener<DataSnapshot>() {
    @Override
    public void onComplete(@NonNull Task<DataSnapshot> task) {
        if (task.isSuccessful()) {
            for (DataSnapshot ds : task.getResult().getChildren()) {
                ds.getRef().removeValue();
            }
        } else {
            Log.d("TAG", task.getException().getMessage()); //Never ignore potential errors!
        }
    }
});

You can also attach a listener to the removeValue() operation to see if something goes wrong.