Checking if a particular value exists in the Firebase database

To check the existence of a user, please use the below code:

DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference();
DatabaseReference userNameRef = rootRef.child("Users").child("Nick123");
ValueEventListener eventListener = new ValueEventListener() {
    @Override
    public void onDataChange(DataSnapshot dataSnapshot) {
        if(!dataSnapshot.exists()) {
            //create new user
        }
    }

    @Override
    public void onCancelled(DatabaseError databaseError) {
        Log.d(TAG, databaseError.getMessage()); //Don't ignore errors!
    }
};
userNameRef.addListenerForSingleValueEvent(eventListener);

You can also use a Query to achieve the same thing like this:

DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference();
Query query = rootRef.child("Users").orderByChild("userName").equalTo("Nick123");
query.addValueEventListener(/* ... */);

This is another approach which is looping through the entire Users node but is not just using a direct reference to a single user. This option is more likely to be used when you are using as a unique identifier beteeen users the uid instead of the user name (as you do right now). So if your database structure might looks similar to this:

Firebase-root
   |
   --- Users
        |
        --- uid
             |
             --- userName: "Test User"
             |
             --- emailAddress: "[email protected]"

The second solution is the recommended one.

There is also another solution which involves you to create another node named userNames, in which you can hold only the unique user names. Please also find below the corresponding security rules:

"Users": {
  "$uid": {
    ".write": "auth !== null && auth.uid === $uid",
    ".read": "auth !== null && auth.provider === 'password'",
    "userName": {
      ".validate": "
        !root.child('userNames').child(newData.val()).exists() ||
        root.child('userNames').child(newData.val()).val() == $uid"
    }
  }
}

But since in this case, your user name is already the name of the node, I recommend you go ahead with the first one.

Try this:

DatabaseReference ref=FirebaseDatabase.getInstance().getReference().child("Users");
ref.orderByChild("username").equalTo(Nick123).addValueEventListener(new ValueEventListener(){
  @Override
  public void onDataChange(DataSnapshot dataSnapshot){
      if(dataSnapshot.exist() {
         //username exist
          }
        }

You have to use orderbychild and equalto to check the value if it is there. If you dont use orderbychild and equalto then it will just check if username child node is there and doesnt care about the value.

this orderByChild("username").equalTo(Nick123) is like saying:

WHERE username=Nick123

Works Perfectly for me

DatabaseReference reference = FirebaseDatabase.getInstance().getReference();
        Query query = reference
                .child(getString(R.string.dbname_users))
                .orderByChild("username")
                .equalTo(username);
        query.addListenerForSingleValueEvent(new ValueEventListener() {
            @Override
            public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
                if(dataSnapshot.getChildrenCount()>0) {
                    //username found

                }else{
                    // username not found
                }

            }
            @Override
            public void onCancelled(DatabaseError databaseError) {

            }
        });