Develop Reference

How to get multiple documents by using where_not_in operator in Firebase Firestore database?

How to get multiple documents from a Collection in firebase if we want to filter using not_in operator with a uniqueId list of documents.
I have a arrayList like this:
ArrayList<String> idList = new ArrayList();
idList.addAll(uniqueIdList);
// now idList have more than 500 uniqueId
Query query = db.collection("my_collection")
.whereEqualTo("status", "DONE")
.whereNotIn("uniqueId", idList)
.orderBy("uniqueId", Query.Direction.DESCENDING)
.orderBy("createdOn", Query.Direction.DESCENDING);
/* FIREBASE DOCUMENTATION SAYS
------------------------------
Use the in operator to combine up to 10 equality (==)
clauses on the same field with a logical OR */
If the idList object have more than 10 items. It crashes the android application due to FirestoreException.
So, should we not use where_not_in operator? But I have specific demand of this for the query.

So, if you want to query using where_not_in operator in Firebase then, you have to do some part from client side also. The query has a serious limitation. So here is a solution.
// Assume idList contains the uniqueId of documents that you don't want
// Assume status can be DONE or PENDING
// Assume list_objects is the ArrayList you have to pass to Recycler view or list view in your app
if (idList.size() > 0 && idList.size() <= 10) {
query = db.collection("my_collection")
.whereEqualTo("status", "DONE")
.whereNotIn("uniqueId", idList)
.orderBy("uniqueId", Query.Direction.DESCENDING)
.orderBy("createdOn", Query.Direction.DESCENDING);
// your on success code here
} else {
query = db.collection("my_collection")
.whereEqualTo("status", "DONE")
.orderBy("createdOn", Query.Direction.DESCENDING);
// here we are fetching all data where status is done
query.get().addOnCompleteListener(task -> {
if (task.isSuccessful()) {
List<Object> list_toRemove = new ArrayList<>();
list_objects = task.getResult().toObjects(ClassName.class);
for (int i = 0; i < list_objects.size(); i++) {
Object item = list_objects.get(i);
if (idList.contains(item.getUniqueId())) {
list_toRemove.add(list_objects.get(i));
}
}
list_objects.removeAll(list_toRemove);
}
}
// remove the data manually here and we are now good. There is no other way for now.
}
So this is a limitation in firebase, but if we look at the advantages of using firebase firestore DB then it's a trade off.
The rule is that if you cannot filter data using query then fetch with applying filters that are possible then again filter using the Collection Framework (if you are using Java). All things are possible in this DB.

how can i count the number of document from firestore and return it in a text [duplicate]

I want to calculate how many documents are in a collection, not the length of the document. I have tried it with some code but what appears is the length of the character from my document name.
this my code :
StreamSubscription<DocumentSnapshot> userpost;
final DocumentReference documentReference =
Firestore.instance.document("product/$documentPost");
userpost = documentReference.snapshots().listen((datasnapshot) {
if (datasnapshot.exists) {
for (int i = 0; i < datasnapshot.data.length; i++){
print(datasnapshot.data.length);
}

An Example Function to fetch Documents Count.
void countDocuments() async {
QuerySnapshot _myDoc = await Firestore.instance.collection('product').getDocuments();
List<DocumentSnapshot> _myDocCount = _myDoc.documents;
print(_myDocCount.length); // Count of Documents in Collection
}

You can use the count() function which was added in cloud_firestore version 4.0.0
Accepted answer might be a bad solution because you have to fetch all the documents just to count the number of documents. As per Firestore pricing, every document read is taken as 1 read count.
So a better solution is to use the count() function instead.
AggregateQuerySnapshot query = FirebaseFirestore.instance.collection('random_collection').count().get();
int numberOfDocuments = query.count;
count() is an Aggregation Query
PS: You might need to update your firebase plugins in pubspec.yaml.

With Cloud Firebase 2.0, there is a new way to count documents in a collection. According to reference notes, the count does not count as a read per document but a metaData request:
"[AggregateQuery] represents the data at a particular location for retrieving metadata without retrieving the actual documents."
Example:
final CollectionReference<Map<String, dynamic>> userList = FirebaseFirestore.instance.collection('users');
Future<int> countUsers() async {
AggregateQuerySnapshot query = await userList.count().get();
debugPrint('The number of users: ${query.count}');
return query.count;
}

How to count the number of documents firestore on flutter?

I want to calculate how many documents are in a collection, not the length of the document. I have tried it with some code but what appears is the length of the character from my document name.
this my code :
StreamSubscription<DocumentSnapshot> userpost;
final DocumentReference documentReference =
Firestore.instance.document("product/$documentPost");
userpost = documentReference.snapshots().listen((datasnapshot) {
if (datasnapshot.exists) {
for (int i = 0; i < datasnapshot.data.length; i++){
print(datasnapshot.data.length);
}

An Example Function to fetch Documents Count.
void countDocuments() async {
QuerySnapshot _myDoc = await Firestore.instance.collection('product').getDocuments();
List<DocumentSnapshot> _myDocCount = _myDoc.documents;
print(_myDocCount.length); // Count of Documents in Collection
}

You can use the count() function which was added in cloud_firestore version 4.0.0
Accepted answer might be a bad solution because you have to fetch all the documents just to count the number of documents. As per Firestore pricing, every document read is taken as 1 read count.
So a better solution is to use the count() function instead.
AggregateQuerySnapshot query = FirebaseFirestore.instance.collection('random_collection').count().get();
int numberOfDocuments = query.count;
count() is an Aggregation Query
PS: You might need to update your firebase plugins in pubspec.yaml.

With Cloud Firebase 2.0, there is a new way to count documents in a collection. According to reference notes, the count does not count as a read per document but a metaData request:
"[AggregateQuery] represents the data at a particular location for retrieving metadata without retrieving the actual documents."
Example:
final CollectionReference<Map<String, dynamic>> userList = FirebaseFirestore.instance.collection('users');
Future<int> countUsers() async {
AggregateQuerySnapshot query = await userList.count().get();
debugPrint('The number of users: ${query.count}');
return query.count;
}

Firebase query on the basis of attribute exist(key)

As I am trying to short my data. So I decided to save userLikeCollection as
questionID = true/false -> where true means like and false means dislike. And if this document not exist it means it is neither like nor disliked...
and there ID comprises of (userID + questionID)
Now I want to query whether this question is like by a user. Is there a way to achieve it.
db.collection("userLikeCollection").where(questionID ).exist()
or
read that document which has property string name questionID
db.collection("userLikeCollection").whereStringProperty(questionID).exist()
where userLikeDocument look like below
(userID + questionID) -> Document Unique ID
- questionID = true/false //where questionID = any unique id for each document.
Edit Question
UserLikeDocument - 1
blahblahQuestionIdOne = true;
UserLikeDocument - 2
blahblahQuestionIdTwo = true;
UserLikeDocument - 3
blahblahQuestionIdThree = true;
UserLikeDocument - 4
blahblahQuestionIdFour = true;

Now I want to query whether this question is like by a user. Is there a way to achieve it.
db.collection("userLikeCollection").where(questionID ).exist()
Yes there is. To solve this, you should use a query and a get() call. In code it looks like this:
Query query = db.collection("userLikeCollection").whereEqualTo("questionID", true);
query.get().addOnCompleteListener(new OnCompleteListener<QuerySnapshot>() {
#Override
public void onComplete(#NonNull Task<QuerySnapshot> task) {
if (task.isSuccessful()) {
for (QueryDocumentSnapshot document : task.getResult()) {
if (document.exists()) {
Boolean questionID = document.getBoolean("questionID");
}
}
}
}
});
In this code, the query will help you find all documents in which the questionID property will hold the value of true and the use of document.exists() will help find if that document actually exists.
or read that document which has property string name questionID
db.collection("userLikeCollection").whereStringProperty(questionID).exist()
In this case you should also use a query but instead of passing a boolean value to the whereEqualTo() method, you should pass a literal String:
Query query = db.collection("userLikeCollection").whereEqualTo("questionID", id);
In which the id holds a value of type String which is the actual id of the question that you are looking for.
(userID + questionID) -> Document Unique ID - questionID = true/false //where questionID = any unique id for each document.
This is actually possible in Firestore but only if the name of the properties are different, let's say questionID that holds a boolean value and and a id property that holds a String value. This can be done by chaining two whereEqualTo() call like this:
Query query = db.collection("userLikeCollection")
.whereEqualTo("questionID", true)
.whereEqualTo("id", id);
Edit:
According to your comment:
whereEqualTo("questionID", true) comes only if it is liked but when it is dislike .It will not retrieved
That's correct. To solve your problem, according to the official documentation regarding Query limitations:
Cloud Firestore does not support the following types of queries:
Queries with a != clause. In this case, you should split the query into a greater-than query and a less-than query. For example,
although the query clause where("age", "!=", "30") is not supported,
you can get the same result set by combining two queries, one with
the clause where("age", "<", "30") and one with the clause
where("age", ">", 30).
So with other words, there is no != (not equal to) operator in Firestore. As they say, the option that you have is to split your query into a greater-than and a less-than query and then it will work perfectly fine.
You can also create another query that looks like this:
Query query = db.collection("userLikeCollection").whereEqualTo("questionID", false);
In this case, you get the question that have the questionID property false.
Edit2:
According to your edited question, there is no way in Firestore to filter documents based on a dynamic id. There is also no way to use wildcars. The properties should have the same name.

Firebase Cloud Firestore Query whereEqualTo for reference

I am trying to create a query which only selects documents whose reference is equal to a given reference, using Java for Android development. A document which it would match contains the reference for the path "/users/someUser". I am creating the reference like so:
DocumentReference ref = mDatabase.document("users/someUser");
I have also tried:
DocumentReference ref = mDatabase.document("/users/someUser");
Then the query:
Query query = mDatabase.collection("myCollection").whereEqualTo("refField", ref).limit(10);
However, when I run the query and check the task.isSuccessful() in the onComplete method, it's not passing, i.e. it didn't work, whereas when I remove the .whereEqualTo(), it passes and the task's result isn't empty. How can I properly use .whereEqualTo() to check for all documents containing a specific reference?
An example of a document that should match my query would be:
/myCollection/GDpojS5koac2C7YlIqxS which contains the field:
refField: /users/someUser (value of type reference)
And an example of a document that should not match my query would be:
/myCollection/J5ZcVAMYU1nI5XZmh6Bv which contains the field:
refField: /users/wrongUser (value of type reference)

I think you need to add a get() method to run the query and add an onCompletionListener.
Something like this should work:
mDatabase.collection("myCollection")
.whereEqualTo("refField", ref)
.limit(10)
.get()
.addOnCompleteListener({task ->
if(task.isSuccessful){
val result = task.result
})
The above example is in kotlin, but i guess in java it is something similar

You need not to worry about the documents, if you create a query based on your fields then all the documents will be returned in the "QuerySnapshot" object,
for eg,
CollectionReference collectionReference = db.collection(FIRESTORE_USERS);
DocumentReference documentReference = collectionReference.document(userID);
CollectionReference notificationCollection = documentReference.collection(FIRESTORE_NOTIFICATIONS);
notificationCollection.whereEqualTo(USER_TYPE, userType)
.whereGreaterThanOrEqualTo(SEND_AT, calendar.getTime())
.get().addOnSuccessListener(new OnSuccessListener<QuerySnapshot>() {
#Override
public void onSuccess(QuerySnapshot documentSnapshots) {
List<DocumentSnapshot> snapshotsList = documentSnapshots.getDocuments();
ArrayList<NotificationCollections> notificationCollectionsArrayList = new ArrayList<>();
for (DocumentSnapshot snapshot : snapshotsList) {
// each document having that particular field based on query
}
}});
in the above example I am fetching all those documents which match a particular user id and also having time greater than or equal to supplied time (time will not be used in your case)
I hope this helps...
Happy coding :)