Delete Firestore documents based on where condtiion [duplicate] - firebase

Question

var jobskill_ref = db.collection('job_skills').where('job_id','==',post.job_id);
jobskill_ref.delete();
Error thrown
jobskill_ref.delete is not a function

Answer 1

You can only delete a document once you have a DocumentReference to it. To get that you must first execute the query, then loop over the QuerySnapshot and finally delete each DocumentSnapshot based on its ref.
var jobskill_query = db.collection('job_skills').where('job_id','==',post.job_id);
jobskill_query.get().then(function(querySnapshot) {
querySnapshot.forEach(function(doc) {
doc.ref.delete();
});
});

Answer 2

I use batched writes for this. For example:
var jobskill_ref = db.collection('job_skills').where('job_id','==',post.job_id);
let batch = firestore.batch();
jobskill_ref
.get()
.then(snapshot => {
snapshot.docs.forEach(doc => {
batch.delete(doc.ref);
});
return batch.commit();
})
ES6 async/await:
const jobskills = await store
.collection('job_skills')
.where('job_id', '==', post.job_id)
.get();
const batch = store.batch();
jobskills.forEach(doc => {
batch.delete(doc.ref);
});
await batch.commit();

Answer 3

//The following code will find and delete the document from firestore
const doc = await this.noteRef.where('userId', '==', userId).get();
doc.forEach(element => {
element.ref.delete();
console.log(`deleted: ${element.id}`);
});

Answer 4

the key part of Frank's answer that fixed my issues was the .ref in doc.ref.delete()
I originally only had doc.delete() which gave a "not a function" error. now my code looks like this and works perfectly:
let fs = firebase.firestore();
let collectionRef = fs.collection(<your collection here>);
collectionRef.where("name", "==", name)
.get()
.then(querySnapshot => {
querySnapshot.forEach((doc) => {
doc.ref.delete().then(() => {
console.log("Document successfully deleted!");
}).catch(function(error) {
console.error("Error removing document: ", error);
});
});
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});

Answer 5

or try this, but you must have the id beforehand
export const deleteDocument = (id) => {
return (dispatch) => {
firebase.firestore()
.collection("contracts")
.doc(id)
.delete()
}
}

Answer 6

You can now do this:
db.collection("cities").doc("DC").delete().then(function() {
console.log("Document successfully deleted!");
}).catch(function(error) {
console.error("Error removing document: ", error);
});

Answer 7

And of course, you can use await/async:
exports.delete = functions.https.onRequest(async (req, res) => {
try {
var jobskill_ref = db.collection('job_skills').where('job_id','==',post.job_id).get();
jobskill_ref.forEach((doc) => {
doc.ref.delete();
});
} catch (error) {
return res.json({
status: 'error', msg: 'Error while deleting', data: error,
});
}
});
I have no idea why you have to get() them and loop on them, then delete() them, while you can prepare one query with where to delete in one step like any SQL statement, but Google decided to do it like that. so, for now, this is the only option.

Answer 8

If you're using Cloud Firestore on the Client side, you can use a Unique key generator package/module like uuid to generate an ID. Then you set the ID of the document to the ID generated from uuid and store a reference to the ID on the object you're storing in Firestore.
For example:
If you wanted to save a person object to Firestore, first, you'll use uuid to generate an ID for the person, before saving like below.
const uuid = require('uuid')
const person = { name: "Adebola Adeniran", age: 19}
const id = uuid() //generates a unique random ID of type string
const personObjWithId = {person, id}
export const sendToFireStore = async (person) => {
await db.collection("people").doc(id).set(personObjWithId);
};
// To delete, get the ID you've stored with the object and call // the following firestore query
export const deleteFromFireStore = async (id) => {
await db.collection("people").doc(id).delete();
};
Hope this helps anyone using firestore on the Client side.

Answer 9

The way I resolved this is by giving each document a uniqueID, querying on that field, getting the documentID of the returned document, and using that in the delete. Like so:
(Swift)
func rejectFriendRequest(request: Request) {
DispatchQueue.global().async {
self.db.collection("requests")
.whereField("uniqueID", isEqualTo: request.uniqueID)
.getDocuments { querySnapshot, error in
if let e = error {
print("There was an error fetching that document: \(e)")
} else {
self.db.collection("requests")
.document(querySnapshot!.documents.first!.documentID)
.delete() { err in
if let e = err {
print("There was an error deleting that document: \(e)")
} else {
print("Document successfully deleted!")
}
}
}
}
}
}
The code could be cleaned up a bit, but this is the solution I came up with. Hope it can help someone in the future!

Answer 10

const firestoreCollection = db.collection('job_skills')
var docIds = (await firestoreCollection.where("folderId", "==", folderId).get()).docs.map((doc => doc.id))
// for single result
await firestoreCollection.doc(docIds[0]).delete()
// for multiple result
await Promise.all(
docIds.map(
async(docId) => await firestoreCollection.doc(docId).delete()
)
)

Answer 11

delete(seccion: string, subseccion: string)
{
const deletlist = this.db.collection('seccionesclass', ref => ref.where('seccion', '==', seccion).where('subseccion', '==' , subseccion))
deletlist.get().subscribe(delitems => delitems.forEach( doc=> doc.ref.delete()));
alert('record erased');
}

Answer 12

The code for Kotlin, including failure listeners (both for the query and for the delete of each document):
fun deleteJobs(jobId: String) {
db.collection("jobs").whereEqualTo("job_id", jobId).get()
.addOnSuccessListener { documentSnapshots ->
for (documentSnapshot in documentSnapshots)
documentSnapshot.reference.delete().addOnFailureListener { e ->
Log.e(TAG, "deleteJobs: failed to delete document ${documentSnapshot.reference.id}", e)
}
}.addOnFailureListener { e ->
Log.e(TAG, "deleteJobs: query failed", e)
}
}