Firebase return output from function

Question

I am new in firebase and I'm trying to pass a $variable in a function to check if the $variable is exists.

function ifExistWaybillNo(waybill_no)
{
  var databaseRef = firebase.database().ref('masterlist');
  databaseRef.orderByChild("waybill_no").equalTo(waybill_no).on('value', function(snapshot){
    alert(snapshot.exists()); //Alert true or false
  });
}

The above function work's fine but when I changed alert(snapshot.exists()); to return snapshot.exists(); it doesn't working. It just return undefined, which should return true or false.

How can I do this? thanks in advance

Community · Accepted Answer · 2017-05-23 12:34:30Z

2

Almost everything Firebase does is asynchronous. When you call the function ifExistWaybillNo it expects an immediate return, not to wait. So before your databaseRef.orderByChild("waybill_no") is finished the statement that called the function has already decided the return is undefined.

The way to fix this is by passing a callback function and using the return there. An exact explanation of this is done very well here: return async call.

You just need to rename some of the functions and follow syntax used there.

To start:

function(waybill_no, callback) { 
    databaseRef.orderByChild("waybill_no").equalTo(waybill_no).on('value', function(snapshot) {
    var truth = snapshot.exists();
    callback(truth); // this will "return" your value to the original caller
  });
}

Remember, almost everything Firebase is asynchronous.

edited May 23, 2017 at 12:34

CommunityBot

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answered May 31, 2016 at 13:54

theblindprophet

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1 Comment

theblindprophet Over a year ago

No problem, happy coding.

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