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 a somewhat complex Mongoose query I want to make and while I can find the different parts around the internet (including here) I can't seem to find exactly what I need and piecing the existing info together has not worked so I am looking for specific how to do this.
I want to findOneAndUpdate to add data to blockSection array in my document.
Block.findOneAndUpdate({
section: {$elemMatch: {section_code: sectionID}}
},
{"$push": {blockSection.section: blocSecData}}, // blocSecData is defined elsewhere and is an object
(err, result) => {
if(err) {
return res.json({success: false, err})
}
... // more code here to processstuff
})
Rough example of a sample Block with only one item per array
{
"section_code":"<abc>",
"blockSection": [
{
section: [{"text":"hello world"}], // add the data here
"_id":"<foobar>",
"blockSecID":"<a uuid>"
}
]
}
I am attempting to add a help request system which allows the requestor to make only one request for help on each topic from an expert. If the expert lists multiple topics which they can help, I want to limit each requestor to one help request per topic per expert.
I am using node.js and mongoose.js with a self-hosted mongodb instance
I have tried using the $and operator to find the ._id of the expert as long as they don't already have an existing request from the same requestor on the same topic. It works for one update but after the experts document has a subdocument inserted with either the topic_id or the requestor_id the filter is applied and no expert is returned.
// Schema
ExpertSchema = new mongoose.Schema({
expert_id: String,
helpRequests: [
requestor_id: String,
topic_id: String
]
});
//query
const query = {
$and:[
{expert_id: req.body.expert_id},
{'helpRequests.requestor_id': {$ne: req.body.requestor_id}},
{'helpRequests.topic_id': {$ne: req.body.topic_id}}
]
};
// desired update
const update = {
$push: {
helpRequests: {
requestor_id: req.body.requestor_id,
topic_id: req.body.topic_id
}
}
Expert.findOneAndUpdate(query, update, {new: true}, (err, expert) =>{
// handle return or error...
});
The reason you are not getting any expert is condition inside your query.
Results always returned based on the condition of your query if your condition inside query get satisfied you will get your result as simple as that.
Your query
{'helpRequests.requestor_id': {$ne: req.body.requestor_id}},
{'helpRequests.topic_id': {$ne: req.body.topic_id}}
you will get your expert only if requestor_id and topic_id is not exists inside helpRequests array. thats you are querying for.
Solution
As per you schema if helpRequests contains only requestor_id and topic_id then you can achieve what you desire by below query.
Expert.findOneAndUpdate(
{
expert_id: req.body.expert_id,
}, {
$addToSet: {
helpRequests: {
requestor_id: req.body.requestor_id,
topic_id: req.body.topic_id
}
}
}, { returnNewDocument: true });
I am new to NodeJs and I am trying to create a web application using express framework and MySql. I get that in MVC architecture the views are for example the *.ejs files. The controllers are supposed to have the logic and the models should focus on the database.
But still I am not quite sure what is supposed to be inside the model. I have the following code in my controller (probably wrong, not following mvc design):
const mysql = require('mysql');
const db = mysql.createConnection(config);
db.query(query, (err, result) => {
if (err) {
return res.redirect('/');
}
res.render('index.ejs', {
users: result
});
});
Now from what I've read the controller should ask the model to execute the query to the database, get the results and render the view (index.ejs').
My question is this: What should be inside the model.js file? Can I make something like this?
controller.js
const db = require('./models/model.js');
db.connect();
const results = db.query(query);
if(results != null) {
res.render('index.ejs'){
users: result
});
}
model.js will make a query to mysql handle any errors and return the result.
From what I've read I have two options. Option1: pass callback function to model and let the model render the view (I think that's wrong, model should not communicate with view, or not?) Option2: possible use of async/await and wait for model to return the results but I am not sure if this is possible or how.
The model is the programmatic representation of the data stored in the database. Say I have an employees table with the following schema:
name: string
age: number
company: company_foreign_key
And another table called companies
name: string
address: string
I therefore have two models: Company and Employee.
The model's purpose is to load database data and provide a convenient programmatic interface to access and act upon this data
So, my controller might look like this:
var db = require('mongo');
var employeeName = "bob";
db.connect(function(err, connection){
const Employee = require('./models/Employee.js'); // get model class
let employeeModel = new Employee(connection); // instantiate object of model class
employee.getByName(employeeName, function(err, result){ // convenience method getByName
employee.getEmployeeCompany(result, function(err, companyResult){ // convenience method getEmployeeCompany
if(companyResultl) { // Controller now uses the results from model and passes those results to a view
res.render('index.ejs')
company: companyResult
});
})
})
}
})
Basically, the model provides a convenient interface to the raw data in the database. The model executes the queries underneath, and provides convenient methods as a public interface for the controller to access. E.g., the employee model, given an employee object, can find the employee's company by executing that query.
The above is just an example and, given more thought, a better interface could be thought up. In fact, Mongoose provides a great example of how to set up model interfaces. If you look at how Mongoose works, you can use the same principles and apply them to your custom model implementation.
I am new to couchdb,I have a design with two key. I am using node to view the list. Is there anyway I can pass only one key from node and if it matches any of the two key from couchdb and fetch me result.
My design in couchdb:
function(doc) {
if(doc.doc_type==="messages")
emit([doc.from, doc.to], doc);
}
Passing key using node
db.view('message','fetch_msg',{key:"user1"}, function(err, body) {
if(err)console.log(JSON.stringify(err))
console.log(body.rows.length)
console.log(JSON.stringify(body));
})
Like 'user1' is one of the key.
message/fetch_msg is my design.
Any Help would be appreciated.
You can emit multiple keys per doc - every key will return the same doc.
function (doc) {
if (doc.doc_type !== 'messages')
return
emit(doc.from, null)
emit(doc.to, null)
}
If you want to fetch the whole doc of a row it's recommended for performance reasons to not emit it as value of the row. Use ?include_docs=true instead when you request the view.
I think you want something like this.
You can specify the other parameter as {} in the key. So that the key becomes ["user1", {}] or [{}, "user1"]. You can use the start key, endkey concept of couchdb for your case.