删除对Firebase数据中child_added的引用

Question

I'm writing a Firebase web app (Javascript). If I write a reference to child_added like this:

dataRef.child('bids').child(auction).child(lotno).on('child_added', function(data){...}

"auction" and "lotno" are variables and will change as the app runs. I am writing the script to close each old reference (using OFF) whenever this happens, and then I open a new reference. My OFF looks like this:

dataRef.child('bids').child(auction).child(lotno).off('child_added');

First of all, am I doing this correctly? And second, is there a more universal way I can close child_added references, to avoid the possibility of one being left open accidentally?

Meaning, can I simply write Javascript that will close all child_added within 'bids'? Wondering if something like this is valid:

dataRef.child('bids').off('child_added');

Is there a wildcase that would cover all children? Would that close everything, or must each specific path through child elements be written out?

Answer 1

Since the path is built with two variables, and only one listener is open at any given time, I decided to solve this problem by simply storing the two values separately, in their own variables, such as:

    var storedPath1;
    var storedPath2;

    dataRef.child('bids').child(auction).child(lotno).on('child_added', function(data){
        if (storedPath1!=undefined && storedPath2!=undefined{
            dataRef.child('bids').child(storedPath1).child(storedPath2).off('child_added');
        }
        storedPath1=auction;
        storedPath2=lotno;
    }

If the listener was previously established, this will close it when opening the new one.