CODEWITHSUNDEEP
androidlistfirebasefirebase-realtime-database
Elior B.Y.
Elior B.Y.

Reputation: 301

Android Firebase - Removing a single value from a list

I have this Firebase database (the list was created by the push() method):

enter image description here

I want to remove only "Place 1".

I tried the following code, but it deletes the entire "Restaurants" node...

databaseReference.child("Restaurants").orderByValue().equalTo("Place 1")
                    .addListenerForSingleValueEvent(new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot dataSnapshot) {
        dataSnapshot.getRef().setValue(null);
    }

    @Override
        public void onCancelled(DatabaseError databaseError) {
    }
});

How do I remove only "Place 1"?

Upvotes: 2

Views: 882

Answers (2)

Priya singh
Priya singh

Reputation: 87

TRY This:

    databaseReference.child("Restaurants").orderByValue().equalTo("Place 1")
                    .addListenerForSingleValueEvent(new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot dataSnapshot) {
         dataSnapshot.getRef().removeValue(); //used this
}

    @Override
        public void onCancelled(DatabaseError databaseError) {
    }
});

Upvotes: 0

koceeng
koceeng

Reputation: 2165

You are using databaseReference.child("Restaurant") as a base reference. Please note that orderByValue() does not change the reference, only filter the values. In short, dataSnapshot object produced by onDataChange() will always point to your reference.

You should do it like this:

... onDataChange(DataSnapshot dataSnapshot) {
    // dataSnapshot object will point at "Restaurant"
    for (DataSnapshot snapshot : dataSnapshot.getChildren()) {
        // snapshot object is every child of "Restaurant" that match the filter
        snapshot.getRef().setValue(null);
    }
}

Hope this helps.

Upvotes: 1

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