How to delete document from firestore using where clause

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

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 2

or try this, but you must have the id beforehand

export const deleteDocument = (id) => {
return (dispatch) => {
    firebase.firestore()
    .collection("contracts")
    .doc(id)
    .delete()
}

}

Answer 3

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 4

//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 5

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 6

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 7

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 8

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 9

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 10

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 11

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);
});