If I create a document reference, and fetch it with getDoc, I get the document back just fine, but if I do a query for id == 'adsadasfsge' or id in ['adsadasfsge'] on the same database, I get nothing back. Here's the code I'm trying:
// Directly fetching one doc by id, using a docRef
const dr = doc(firestore, 'TestPrograms', id);
getDoc(dr).then((doc) => {
console.log(doc.exists());
});
// Fetching with an unambiguous == query
const q = query(collection(firestore, 'TestPrograms'), where('id', '==', id));
getDocs(q).then((docs) => {
console.log(docs.size);
});
// Fetching with an 'in' query
const q2 = query(collection(firestore, 'TestPrograms'), where('id', 'in', [id]));
getDocs(q2).then((docs) => {
console.log(docs.size);
});
Running this logs:
true
0
0
I'm baffled. What am I doing wrong here? Thanks!
Your query checks for a field named id inside the document with a specific value. If you want to check for documents whose document ID has a specific value, you need to specify the special documentId() marker as the field for the query.
I have user and group collections. Under the user collection, each document id is a user UID and each user document has an array field "userGroups" which contains the groups that user belongs to and those groups are the group's document ID under group collection.
I have been able to retrieve the userGroups array for the current user which i stored in groupRef (see code below). What I'm trying to do now is to map those array values into groups collection and retrieve only those documents that are in the groupRef. (Basically the goal is to just show in the UI the groups that the current user is a member of)
user collection
group collection
const [groupsList, setGroupList] = useState([]);
const [groupRef, setGroupRef] = useState([]);
const [track, setTrack] = useState('')
const handleSubmit = () => {
setTrack('start')
fire.firestore().collection('users').doc(fire.auth().currentUser.uid).get().then((value) => {
console.log("userGroups " + value.data().userGroups) // this returns an object
setGroupRef([value.data().userGroups])
})
}
useEffect(() => {
handleSubmit()
}, [track])
console.log("sample list " + groupRef)
fire.firestore().collection('groups').onSnapshot(snapshot => (
setGroupList(snapshot.docs.map(doc => doc.data()))
))
^ this returns all the documents under groups collection. any ideas how i can retrieve only those that the current user is a member of? any help would be much appreciated. (ive been stuck on this for a long time. im also new to firebase.)
#DougStevenson directed me to the right path of using array-contains which is a helper function on querying/getting data. the code below is the answer to my problem. this way is shorter and more efficient than the work i came up with.
fire.firestore().collection('groups').where("groupMembers", "array-contains", fire.auth().currentUser.uid).onSnapshot(snapshot => ( setGroupList(snapshot.docs.map(doc => doc.data()))
))
From the docs:
You can also chain multiple where() methods to create more specific queries (logical AND).
How can I perform an OR query?
Example:
Give me all documents where the field status is open OR upcoming
Give me all documents where the field status == open OR createdAt <= <somedatetime>
OR isn't supported as it's hard for the server to scale it (requires keeping state to dedup). The work around is to issue 2 queries, one for each condition, and dedup on the client.
Edit (Nov 2019):
Cloud Firestore now supports IN queries which are a limited type of OR query.
For the example above you could do:
// Get all documents in 'foo' where status is open or upcmoming
db.collection('foo').where('status','in',['open','upcoming']).get()
However it's still not possible to do a general OR condition involving multiple fields.
With the recent addition of IN queries, Firestore supports "up to 10 equality clauses on the same field with a logical OR"
A possible solution to (1) would be:
documents.where('status', 'in', ['open', 'upcoming']);
See Firebase Guides: Query Operators | in and array-contains-any
suggest to give value for status as well.
ex.
{ name: "a", statusValue = 10, status = 'open' }
{ name: "b", statusValue = 20, status = 'upcoming'}
{ name: "c", statusValue = 30, status = 'close'}
you can query by ref.where('statusValue', '<=', 20) then both 'a' and 'b' will found.
this can save your query cost and performance.
btw, it is not fix all case.
I would have no "status" field, but status related fields, updating them to true or false based on request, like
{ name: "a", status_open: true, status_upcoming: false, status_closed: false}
However, check Firebase Cloud Functions. You could have a function listening status changes, updating status related properties like
{ name: "a", status: "open", status_open: true, status_upcoming: false, status_closed: false}
one or the other, your query could be just
...where('status_open','==',true)...
Hope it helps.
This doesn't solve all cases, but for "enum" fields, you can emulate an "OR" query by making a separate boolean field for each enum-value, then adding a where("enum_<value>", "==", false) for every value that isn't part of the "OR" clause you want.
For example, consider your first desired query:
Give me all documents where the field status is open OR upcoming
You can accomplish this by splitting the status: string field into multiple boolean fields, one for each enum-value:
status_open: bool
status_upcoming: bool
status_suspended: bool
status_closed: bool
To perform your "where status is open or upcoming" query, you then do this:
where("status_suspended", "==", false).where("status_closed", "==", false)
How does this work? Well, because it's an enum, you know one of the values must have true assigned. So if you can determine that all of the other values don't match for a given entry, then by deduction it must match one of the values you originally were looking for.
See also
in/not-in/array-contains-in: https://firebase.google.com/docs/firestore/query-data/queries#in_and_array-contains-any
!=: https://firebase.googleblog.com/2020/09/cloud-firestore-not-equal-queries.html
I don't like everyone saying it's not possible.
it is if you create another "hacky" field in the model to build a composite...
for instance, create an array for each document that has all logical or elements
then query for .where("field", arrayContains: [...]
you can bind two Observables using the rxjs merge operator.
Here you have an example.
import { Observable } from 'rxjs/Observable';
import 'rxjs/add/observable/merge';
...
getCombinatedStatus(): Observable<any> {
return Observable.merge(this.db.collection('foo', ref => ref.where('status','==','open')).valueChanges(),
this.db.collection('foo', ref => ref.where('status','==','upcoming')).valueChanges());
}
Then you can subscribe to the new Observable updates using the above method:
getCombinatedStatus.subscribe(results => console.log(results);
I hope this can help you, greetings from Chile!!
We have the same problem just now, luckily the only possible values for ours are A,B,C,D (4) so we have to query for things like A||B, A||C, A||B||C, D, etc
As of like a few months ago firebase supports a new query array-contains so what we do is make an array and we pre-process the OR values to the array
if (a) {
array addObject:#"a"
}
if (b) {
array addObject:#"b"
}
if (a||b) {
array addObject:#"a||b"
}
etc
And we do this for all 4! values or however many combos there are.
THEN we can simply check the query [document arrayContains:#"a||c"] or whatever type of condition we need.
So if something only qualified for conditional A of our 4 conditionals (A,B,C,D) then its array would contain the following literal strings: #["A", "A||B", "A||C", "A||D", "A||B||C", "A||B||D", "A||C||D", "A||B||C||D"]
Then for any of those OR combinations we can just search array-contains on whatever we may want (e.g. "A||C")
Note: This is only a reasonable approach if you have a few number of possible values to compare OR with.
More info on Array-contains here, since it's newish to firebase docs
If you have a limited number of fields, definitely create new fields with true and false like in the example above. However, if you don't know what the fields are until runtime, you have to just combine queries.
Here is a tags OR example...
// the ids of students in class
const students = [studentID1, studentID2,...];
// get all docs where student.studentID1 = true
const results = this.afs.collection('classes',
ref => ref.where(`students.${students[0]}`, '==', true)
).valueChanges({ idField: 'id' }).pipe(
switchMap((r: any) => {
// get all docs where student.studentID2...studentIDX = true
const docs = students.slice(1).map(
(student: any) => this.afs.collection('classes',
ref => ref.where(`students.${student}`, '==', true)
).valueChanges({ idField: 'id' })
);
return combineLatest(docs).pipe(
// combine results by reducing array
map((a: any[]) => {
const g: [] = a.reduce(
(acc: any[], cur: any) => acc.concat(cur)
).concat(r);
// filter out duplicates by 'id' field
return g.filter(
(b: any, n: number, a: any[]) => a.findIndex(
(v: any) => v.id === b.id) === n
);
}),
);
})
);
Unfortunately there is no other way to combine more than 10 items (use array-contains-any if < 10 items).
There is also no other way to avoid duplicate reads, as you don't know the ID fields that will be matched by the search. Luckily, Firebase has good caching.
For those of you that like promises...
const p = await results.pipe(take(1)).toPromise();
For more info on this, see this article I wrote.
J
OR isn't supported
But if you need that you can do It in your code
Ex : if i want query products where (Size Equal Xl OR XXL : AND Gender is Male)
productsCollectionRef
//1* first get query where can firestore handle it
.whereEqualTo("gender", "Male")
.addSnapshotListener((queryDocumentSnapshots, e) -> {
if (queryDocumentSnapshots == null)
return;
List<Product> productList = new ArrayList<>();
for (DocumentSnapshot snapshot : queryDocumentSnapshots.getDocuments()) {
Product product = snapshot.toObject(Product.class);
//2* then check your query OR Condition because firestore just support AND Condition
if (product.getSize().equals("XL") || product.getSize().equals("XXL"))
productList.add(product);
}
liveData.setValue(productList);
});
For Flutter dart language use this:
db.collection("projects").where("status", whereIn: ["public", "unlisted", "secret"]);
actually I found #Dan McGrath answer working here is a rewriting of his answer:
private void query() {
FirebaseFirestore db = FirebaseFirestore.getInstance();
db.collection("STATUS")
.whereIn("status", Arrays.asList("open", "upcoming")) // you can add up to 10 different values like : Arrays.asList("open", "upcoming", "Pending", "In Progress", ...)
.addSnapshotListener(new EventListener<QuerySnapshot>() {
#Override
public void onEvent(#Nullable QuerySnapshot queryDocumentSnapshots, #Nullable FirebaseFirestoreException e) {
for (DocumentSnapshot documentSnapshot : queryDocumentSnapshots) {
// I assume you have a model class called MyStatus
MyStatus status= documentSnapshot.toObject(MyStatus.class);
if (status!= null) {
//do somthing...!
}
}
}
});
}
I developing a simple chat applicaiton for my website using firebase firestore. where every chat session has an id
provided i have an array of ids
chat_sessions = ["q93XhadA9QQLu6X8yfZB", "YXEYISEI8b2iixCcP3VO", "GXrSZbtrmN5NcrtvjFYp"]
I want to get all document whose id is equal to any of the id's in the chat_sessions object using the code below.
return this.afs
.collection('chats', ref => ref.where('uid','in_array',chat_sessions)).snapshotChanges...
but I am not getting any results.
I come from a PHP/MYSQL background
the PHP equivalent of what i am trying to achieve will be sth like this in PHP
if(in_array(uid,chat_sessions)){
get(doc/uid)
}
can anyone help with the right query where we check for document id against a list of ids in an array? Thank You!
Thank you #frank van Puffelen. You were almost right. I should have used in instead of in_array
ref.where(firebase.firestore.FieldPath.documentId(),'in_array',chat_sessions)
did not work. Instead I replaced in_array with in :
ref.where(firebase.firestore.FieldPath.documentId(),'in',chat_sessions)
This worked! Thank you
Your query is:
ref.where('uid','in_array',chat_sessions)
This checks a field called uid in each document against the values of the chat_sessions.
It seems that instead you want to the check the ID of each document against the array, which you can do with:
ref.where(firebase.firestore.FieldPath.documentId(),'in_array',chat_sessions)
I found something else on firestore i.e "array-contains-any" for this case.
Maybe it's updated now.
UPDATE
Hi, firebase did some update recently, so for do it I found out this method
`
const [products, setProduct] = useState([]);
const ids = ['H11LlJsh3sObwORZhA0b','om9m0lU9HYWyOJZKvEdi','1AoHyHuSFcF01zoyXyTD','6xoBlxsRXUoyzBUcWl0F',
'GJqthlmBGZaFAJqtC2jK','QNT3PxMfhNGg1RZnuqcq','RZgGoFZHyDAYaVZJWxGk','g4UO5P0EgtEqJnawwhXX','gyrZm8p0cEgJdDvTuB1g','mrscldfeYlkaSF151MpI',]
useEffect(() => {
const saveFirebaseTodos = [];
ids.forEach((element) => {
fetchMyAPI()
async function fetchMyAPI() {
const q = query(collection(db, "a"), where('__name__', '==', element));
const querySnapshot = await getDocs(q);
querySnapshot.forEach((doc) => {
saveFirebaseTodos.push(({id: doc.id, ...doc.data()}));
/*console.log(doc.id, " => ", doc.data());*/
if (ids.length == saveFirebaseTodos.length) {
setProduct(saveFirebaseTodos)
}
});
}})
}, [])`
In this way, you can check how many element in arrays you want (bypassing 10 elements limit of firebase).
Hope can help anyone :D
I am building a web application using JavaScript and firebase.
everytime I click on the "remove" paragraph tag, I can only remove the most recent added item, but cannot remove the rest.
Example: if I have 9 items, and I just added a 10th item, I can only remove the 10th item, but not the other 9 items. I basically can only remove items in a backwards order, as opposed to being able to remove items in any order of my choice.
Here is my code:
function displayFav(){
const dbRef = firebase.database().ref();
dbRef.on("value", (firebaseData) => {
// empty out element before adding new updated firebase data so there are no repeated data
document.getElementById("displayUsername").innerHTML = "";
let accounts = [];
const accountData = firebaseData.val();
for (let itemKey in accountData) {
accountData[itemKey].key = itemKey;
accounts.push(accountData[itemKey])
const key = accountData[itemKey]["key"];
const password = accountData[itemKey]["password"];
let user = accountData[itemKey]["username"];
// code where I try to render results from page
document.getElementById('displayUsername').innerHTML += `
<li> Username: ${user} Password: ${password}</li>
<p id=${key}>Remove</p>
`;
// code where I try to remove the item
document.getElementById(key).addEventListener("click", function(){
removeItem(key)
})
}
});
}
This is my function to remove the firebase data:
function removeItem(itemToRemove){
const dbRef = firebase.database().ref(`${itemToRemove}`);
dbRef.remove();
};
What can I change or add to my code that will allow me to remove items in any order I want, instead of letting me delete only the most recent items?
Not trying to be rude, this is just a tip: Please indent/format your code well so that people can understand it without having to format it themselves. Usually when people see large code that's not well formatted, it throws them off (sometimes including me), and therefore they wouldn't want to continue to read or help you with the question.
Suppose you have a table called posts, and the structure looks like this:
that is:
{
<postId>: {
date: 1518925839059,
message: 'some message'
},
<postId>: {
date: 151892583967,
message: 'some message'
},
...
...
}
Suppose you want to delete the extra property of the first post, as circled in the picture, you must know the full path of the data:
firebase.database().ref('posts/-L5b1EZ1L_FycluzTIn8/extra').remove();
If you want to delete everything in post 1:
firebase.database().ref('posts/-L5b1EZ1L_FycluzTIn8').remove();
Or, if you want to delete every single post:
firebase.database().ref('posts').remove();