I thought I read that you can query subcollections with the new Firebase Firestore, but I don't see any examples. For example I have my Firestore setup in the following way:
Dances [collection]
danceName
Songs [collection]
songName
How would I be able to query "Find all dances where songName == 'X'"
Update 2019-05-07
Today we released collection group queries, and these allow you to query across subcollections.
So, for example in the web SDK:
db.collectionGroup('Songs')
.where('songName', '==', 'X')
.get()
This would match documents in any collection where the last part of the collection path is 'Songs'.
Your original question was about finding dances where songName == 'X', and this still isn't possible directly, however, for each Song that matched you can load its parent.
Original answer
This is a feature which does not yet exist. It's called a "collection group query" and would allow you query all songs regardless of which dance contained them. This is something we intend to support but don't have a concrete timeline on when it's coming.
The alternative structure at this point is to make songs a top-level collection and make which dance the song is a part of a property of the song.
UPDATE
Now Firestore supports array-contains
Having these documents
{danceName: 'Danca name 1', songName: ['Title1','Title2']}
{danceName: 'Danca name 2', songName: ['Title3']}
do it this way
collection("Dances")
.where("songName", "array-contains", "Title1")
.get()...
#Nelson.b.austin Since firestore does not have that yet, I suggest you to have a flat structure, meaning:
Dances = {
danceName: 'Dance name 1',
songName_Title1: true,
songName_Title2: true,
songName_Title3: false
}
Having it in that way, you can get it done:
var songTitle = 'Title1';
var dances = db.collection("Dances");
var query = dances.where("songName_"+songTitle, "==", true);
I hope this helps.
UPDATE 2019
Firestore have released Collection Group Queries. See Gil's answer above or the official Collection Group Query Documentation
Previous Answer
As stated by Gil Gilbert, it seems as if collection group queries is currently in the works. In the mean time it is probably better to use root level collections and just link between these collection using the document UID's.
For those who don't already know, Jeff Delaney has some incredible guides and resources for anyone working with Firebase (and Angular) on AngularFirebase.
Firestore NoSQL Relational Data Modeling - Here he breaks down the basics of NoSQL and Firestore DB structuring
Advanced Data Modeling With Firestore by Example - These are more advanced techniques to keep in the back of your mind. A great read for those wanting to take their Firestore skills to the next level
What if you store songs as an object instead of as a collection? Each dance as, with songs as a field: type Object (not a collection)
{
danceName: "My Dance",
songs: {
"aNameOfASong": true,
"aNameOfAnotherSong": true,
}
}
then you could query for all dances with aNameOfASong:
db.collection('Dances')
.where('songs.aNameOfASong', '==', true)
.get()
.then(function(querySnapshot) {
querySnapshot.forEach(function(doc) {
console.log(doc.id, " => ", doc.data());
});
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});
NEW UPDATE July 8, 2019:
db.collectionGroup('Songs')
.where('songName', isEqualTo:'X')
.get()
I have found a solution.
Please check this.
var museums = Firestore.instance.collectionGroup('Songs').where('songName', isEqualTo: "X");
museums.getDocuments().then((querySnapshot) {
setState(() {
songCounts= querySnapshot.documents.length.toString();
});
});
And then you can see Data, Rules, Indexes, Usage tabs in your cloud firestore from console.firebase.google.com.
Finally, you should set indexes in the indexes tab.
Fill in collection ID and some field value here.
Then Select the collection group option.
Enjoy it. Thanks
You can always search like this:-
this.key$ = new BehaviorSubject(null);
return this.key$.switchMap(key =>
this.angFirestore
.collection("dances").doc("danceName").collections("songs", ref =>
ref
.where("songName", "==", X)
)
.snapshotChanges()
.map(actions => {
if (actions.toString()) {
return actions.map(a => {
const data = a.payload.doc.data() as Dance;
const id = a.payload.doc.id;
return { id, ...data };
});
} else {
return false;
}
})
);
Query limitations
Cloud Firestore does not support the following types of queries:
Queries with range filters on different fields.
Single queries across multiple collections or subcollections. Each query runs against a single collection of documents. For more
information about how your data structure affects your queries, see
Choose a Data Structure.
Logical OR queries. In this case, you should create a separate query for each OR condition and merge the query results in your app.
Queries with a != clause. In this case, you should split the query into a greater-than query and a less-than query. For example, although
the query clause where("age", "!=", "30") is not supported, you can
get the same result set by combining two queries, one with the clause
where("age", "<", "30") and one with the clause where("age", ">", 30).
I'm working with Observables here and the AngularFire wrapper but here's how I managed to do that.
It's kind of crazy, I'm still learning about observables and I possibly overdid it. But it was a nice exercise.
Some explanation (not an RxJS expert):
songId$ is an observable that will emit ids
dance$ is an observable that reads that id and then gets only the first value.
it then queries the collectionGroup of all songs to find all instances of it.
Based on the instances it traverses to the parent Dances and get their ids.
Now that we have all the Dance ids we need to query them to get their data. But I wanted it to perform well so instead of querying one by one I batch them in buckets of 10 (the maximum angular will take for an in query.
We end up with N buckets and need to do N queries on firestore to get their values.
once we do the queries on firestore we still need to actually parse the data from that.
and finally we can merge all the query results to get a single array with all the Dances in it.
type Song = {id: string, name: string};
type Dance = {id: string, name: string, songs: Song[]};
const songId$: Observable<Song> = new Observable();
const dance$ = songId$.pipe(
take(1), // Only take 1 song name
switchMap( v =>
// Query across collectionGroup to get all instances.
this.db.collectionGroup('songs', ref =>
ref.where('id', '==', v.id)).get()
),
switchMap( v => {
// map the Song to the parent Dance, return the Dance ids
const obs: string[] = [];
v.docs.forEach(docRef => {
// We invoke parent twice to go from doc->collection->doc
obs.push(docRef.ref.parent.parent.id);
});
// Because we return an array here this one emit becomes N
return obs;
}),
// Firebase IN support up to 10 values so we partition the data to query the Dances
bufferCount(10),
mergeMap( v => { // query every partition in parallel
return this.db.collection('dances', ref => {
return ref.where( firebase.firestore.FieldPath.documentId(), 'in', v);
}).get();
}),
switchMap( v => {
// Almost there now just need to extract the data from the QuerySnapshots
const obs: Dance[] = [];
v.docs.forEach(docRef => {
obs.push({
...docRef.data(),
id: docRef.id
} as Dance);
});
return of(obs);
}),
// And finally we reduce the docs fetched into a single array.
reduce((acc, value) => acc.concat(value), []),
);
const parentDances = await dance$.toPromise();
I copy pasted my code and changed the variable names to yours, not sure if there are any errors, but it worked fine for me. Let me know if you find any errors or can suggest a better way to test it with maybe some mock firestore.
var songs = []
db.collection('Dances')
.where('songs.aNameOfASong', '==', true)
.get()
.then(function(querySnapshot) {
var songLength = querySnapshot.size
var i=0;
querySnapshot.forEach(function(doc) {
songs.push(doc.data())
i ++;
if(songLength===i){
console.log(songs
}
console.log(doc.id, " => ", doc.data());
});
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});
It could be better to use a flat data structure.
The docs specify the pros and cons of different data structures on this page.
Specifically about the limitations of structures with sub-collections:
You can't easily delete subcollections, or perform compound queries across subcollections.
Contrasted with the purported advantages of a flat data structure:
Root-level collections offer the most flexibility and scalability, along with powerful querying within each collection.
I have one counter left to create. And for this one I don't really know which direction to take.
I believe counting the number of members with the boost role will be wrong as members can have boosted the server twice.
module.exports = async(client) => {
const guild = client.guilds.cache.get('912706237806829598');
setInterval(async () => {
const boostCount = (await guild.members.fetch()).filter(????????).size;
const channel = guild.channels.cache.get('960832075349499925');
channel.setName(`╭🌕・Boosts: ${boostCount.toLocaleString()}`);
console.log('Updating Boost Count');
console.log(boostCount);
}, 600000);
}
I don't know if I get achieve this by putting something inside filter. Looking at the members is probably not correct as well.
Really need some assistance for this one.
Making use of guild.premiumSubscriptionCount would help you out. It will count all the boosts in total including double boosts.
I want to create an array of strings by getting the names of the channels in a guild, but the methods I've tried haven't worked. This is my code, but the console.log doesn't send the array.
if (message.content.startsWith(`${prefix}command`)){
var channel_list = Array.from(message.guild.channels.name);
console.log(channel_list);
}
You're going to want to use Collection.prototype.map()
Discord.js v11.x
console.log(message.guild.channels.map((c) => c.name))
Discord.js v12.x
console.log(message.guild.channels.cache.map((c) => c.name))
I want to create a bot that can tell me the list of the first 300 members of my server. Is this possible with Discord.js? Any help would be appreciated. I just don't know where to start
Assuming you want to get the first 300 members that joined the guild, you can use the GuildMember.joinedAt property. You would go through these steps:
Fetch all the members in the guild: you can use Guild.members.fetch() for that
Sort them by the date you get from the joinedAt property
Get the first 300
Here's how I would do it:
guild.members.fetch() // Fectch all the members in the guild
.then(members => {
let first300 = members
.sort((a, b) => a.joinedAt - b.joinedAt) // Order them by they date the joined the guild
.first(300) // Take the first 300
})
Like syntle asked it depends which kinda but, here's the first 300 to fetch:
Might not work since there might be an already set limit of members you can fetch
const members = await <Guild>.members.fetch({ limit: 300 });
//or if you are fine with cached
const members = <Guild>.members.cache.array();
members.length = 300;
So I've been trying to create a ?members command which lists all the users with a role.
So far I've got this:
if (message.content.startsWith("?members")) {
let roleName = message.content.split(" ").slice(1).join(" ");
let membersWithRole = message.guild.members.filter(member => {
return member.roles.find("name", roleName);
}).map(member => {
return member.user.username;
})
const embed = new Discord.RichEmbed({
"title": `Members in ${roleName}`,
"description": membersWithRole.join("\n"),
"color": 0xFFFF
});
return message.channel.send(embed);
}
So, it works if you type the exact name of the role, but not when you ping it or type the first word. I've been trying for hours to figure out how to do it, and I figured I should ask for help.
Thanks in advance!
Pings get translated into a code as they come through, there is a lot of information on how to parse them in the official guide After it's parsed into a role id you can just use members.roles.get() because that is what they are indexed by.
As for finding a partial name, for that you are going to have to run a function on your find and use String.includes.
return member.roles.find(role => role.name.includes(roleName));
This will also work for find the whole name of course, so it can replace your existing line.
However, this may result in more than one role. This is also true for searching by the entire role name, however, as there are no restrictions on duplicate named roles. For this you may want to invert you design and search through message.guild.roles first for any matching roles, then search for members with the roles found.