How do I get documents where a specific field exists/does not exists in Firebase Cloud Firestore?

Question

In Firebase Cloud Firestore, I have "user_goals" in collections and goals may be a predefined goal (master_id: "XXXX") or a custom goal (no "master_id" key)

In JavaScript, I need to write two functions, one to get all predefined goals and other to get all custom goals.

I have got some workaround to get custom goals by setting "master_id" as "" empty string and able to get as below:

db.collection('user_goals')     .where('challenge_id', '==', '')  // workaround works     .get() 

Still this is not the correct way, I continued to use this for predefined goals where it has a "master_id" as below

db.collection('user_goals')     .where('challenge_id', '<', '')  // this workaround     .where('challenge_id', '>', '')  // is not working     .get() 

Since Firestore has no "!=" operator, I need to use "<" and ">" operator but still no success.

Question: Ignoring these workarounds, what is the preferred way to get docs by checking whether a specific field exists or does not exists?

Answer 1

As @Emile Moureau solution. I prefer

.orderBy(`field`) 

To query documents with the field exists. Since it will work with any type of data with any value even for null.

But as @Doug Stevenson said:

You can't query for something that doesn't exist in Firestore. A field needs to exist in order for a Firestore index to be aware of it.

You can't query for documents without the field. At least for now.

Answer 2

The preferred way to get docs where a specified field exists is to use the:

.orderBy(fieldPath) 

As specified in the Firebase documentation:

Thus the answer provided by @hisoft is valid. I just decided to provide the official source, as the question was for the preferred way.