Reputation: 195
I'm trying to figure it out how to remove a sensitive field on a firestore document. For example, my collection is a group information. The group is protected with a pin code field. Any one wants to join the group has to know the pin code.
In the meantime, I want to let users query what group is available to join. For query part, I don't want return group information with pin code information. Do we have anyway to remove sensitive fields from a document for Firestore for reading event?
Cloud function only supports write event. 1 possible solution is use cloud function on write event, and put pin code in a separate document. Is there a better solution? THanks.
My group schema is:
group: {
name: string,
pinCode: string
}
Upvotes: 3
Views: 1347
Reputation: 1
You can create a Firebase Function that returns only the fields that you need (non sensitive), here an example:
exports.getTopUsers = functions.https.onCall(async (data) => {
const users = [];
return db.collection('users').orderBy('bids').limit(data.limit).get()
.then((querySnapshot) => {
querySnapshot.forEach((user) => {
users.push({
diplayName: user.get('displayName'),
});
});
return {
topUsers: users,
};
})
.catch((err) => {
console.error(err);
});
});
So, you need to create a separate array (that will be returned) and filling it with only the field that you want while iterating your Firestore collection.
Upvotes: 0
Reputation: 599601
A user can either access a document, or they can't. There is no property-level access control in Firestore.
So to accomplish what you want, you will need to store the public and private information in separate documents.
You could either create a second document with the private information in the same collection and then secure them using:
match /databases/{database}/documents {
match /groups/{group} {
allow read: if resource.data.visibility != "private"
}
}
Alternatively (and simpler to secure) you could create a separate collection for the private documents.
Upvotes: 3