How to get an array from Firestore?
Solution 1
When you call DocumentSnapshot.getData(), it returns a Map. You're just calling toString() on that map, which is going to give you a dump of all the data in the document, and that's not particularly helpful. You need to access the dungeon_group
field by name:
DocumentSnapshot document = task.getResult();
List<String> group = (List<String>) document.get("dungeon_group");
- edit: syntax error in typecasting
Solution 2
There are two solutions for your problem, one, you can cast the value from your document in the next way:
DocumentSnapshot document = task.getResult();
List<String> dungeonGroup = (List<String>) document.get("dungeon_group");
Or, and I would recommend you this solution because there is always a possibility that your model will change when you are developing your app . This solution is just model everything in Firebase POJO's even if they have just one parameter:
public class Dungeon {
@PropertyName("dungeon_group")
private List<String> dungeonGroup;
public Dungeon() {
// Must have a public no-argument constructor
}
// Initialize all fields of a dungeon
public Dungeon(List<String> dungeonGroup) {
this.dungeonGroup = dungeonGroup;
}
@PropertyName("dungeon_group")
public List<String> getDungeonGroup() {
return dungeonGroup;
}
@PropertyName("dungeon_group")
public void setDungeonGroup(List<String> dungeonGroup) {
this.dungeonGroup = dungeonGroup;
}
}
Remember that you can use the Annotation @PropertyName to avoid call your variables in the same way that your value in the database. Doing it in this way finally you can just do:
DocumentSnapshot document = task.getResult();
Dungeon dungeon= toObject(Dungeon.class);
Hope that it will help you! Happy coding!
Solution 3
If you want to get the entire dungeon_group
array you need to iterate over a Map
like this:
Map<String, Object> map = documentSnapshot.getData();
for (Map.Entry<String, Object> entry : map.entrySet()) {
if (entry.getKey().equals("dungeon_group")) {
Log.d("TAG", entry.getValue().toString());
}
}
But note, even if the dungeon_group
object is stored in the database as an array, entry.getValue()
returns an ArrayList
and not an array.
A better approach for you would be if you consider this alternative database structure, in which each group
is the key in a Map
and all values are set to the boolean true
:
dungeon_group: {
3P: true,
Urgent: true,
Mission Chalange: true
//and so on
}
Using this structure you'll also be able to query it based on the property that exists within the dungeon_group
map, otherwise as in the official documentation:
Although Cloud Firestore can store arrays,
it does not support
querying array members or updating single array elements.
Edit 13 Jan 2021:
If instead of an array of String values you would have had an array of objects, then you can map that array of objects to a List of custom objects, as explained in the following article:
Edit 13 Aug 2018:
According to the updated documentation regarding array membership, now it is possible to filter data based on array values using whereArrayContains()
method. A simple example would be:
CollectionReference citiesRef = db.collection("cities");
citiesRef.whereArrayContains("regions", "west_coast");
This query returns every city document where the regions field is an array that contains west_coast. If the array has multiple instances of the value you query on, the document is included in the results only once.
Solution 4
Since Your document looks like this "dongeon_group=[SP, urgent, missinon challenge,...]
when you convert it to string, say via String.valueOf(document.getData())
I think another way to simply achieve this is by unpacking the document right away into a string of array as follows:
String[] unpackedDoc = document.getData().entrySet().toArray()[0].toString().split("=")[1].split(",");
Explanation
The document
is gotten from documentSnapshot.getDocuments()
which return a list containing you documents.
Since your documents, from the firestore, seem to be nested, calling the document.getData()
will return the list, a list with single element of course, of your document.
document.getData().entrySet()
will make the document ready to be converted to array (you can't do that with getData() alone) also containing single element.
Accessing the single element document.getData().entrySet().toArray()[0]
and then converting it to string, document.getData().entrySet().toArray()[0].toString()
will leave with a string that you can then split (using the =
found in the string) and then take the second part.
The second part can also be split into an array containing the values of your document.
Since this solution convert one element at a time, you can wrap it within a loop so you can convert all the documents available.
For example:
for(DocumentSnapshot document :documentSnapshot.getDocuments()){
String[] unpackedDoc = document.getData().entrySet().toArray()[0].toString().split("=")[1].split(",");
//do something with the unpacked doc
}
Simon Ho
Updated on August 25, 2021Comments
-
Simon Ho over 2 years
I have the data structure illustrated below stored in Cloud Firestore. I want to save the
dungeon_group
which is an array of strings stored in Firestore.I have difficulty in getting the data and stored as an array. I am just able to get a weird string but any method to store as a string array? Below is the code I used.
I am able to achieve this in Swift as follow, but not sure how to do the same in Android.
Swift:
Firestore.firestore().collection("dungeon").document("room_en").getDocument { (document, error) in if let document = document { let group_array = document["dungeon_group"] as? Array ?? [""] print(group_array) } }
Java Android:
FirebaseFirestore.getInstance().collection("dungeon") .document("room_en").get() .addOnCompleteListener(new OnCompleteListener<DocumentSnapshot>() { @Override public void onComplete(@NonNull Task<DocumentSnapshot> task) { DocumentSnapshot document = task.getResult(); String group_string= document.getData().toString(); String[] group_array = ???? Log.d("myTag", group_string); } });
Console output as the follow:
{dungeon_group=[3P, Urgent, Mission Challenge, Descended, Collaboration, Daily, Technical, Normal]}