i have my typeorm column like this, what i want is an array of JSON object which i manage to get.
#Column({
type: 'jsonb',
array: false,
default: () => "'[]'",
nullable: false,
})
tokens!: Array<{ token: string }>;
this is how the field looks, and am fine with it, what i want is to find a document with a particular token, so i came up with the below code, but it returns an empty array.
const user = await User.find({ where: { _id: decoded._id, tokens: { token: token } } });
normally when am using mongooe i can get this working using
const user = await User.findOneBy({_id: decoded._id, "tokens.token": token,}); and this returns a particular user, with the id and token string passed.
i want help on how to find a user using the id and the token string inside in array of object, thanks.
TypeORM does not natively support queries on PostreSQL jsonb columns. Performing a query on the data in a jsonb column would require you to either issue a raw query or write your own WHERE clause in.where or .addWhere of a query builder (doc).
For reference, the jsonb query syntax documentation can be found here.
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.
While using sails as ORM (version 1.0), I notice that there is a function called Model.avg (as well as sum). - However there is not a maximum or minimum function to get the maximum or minimum from a column in a model; so it seems this is not necessary because it is covered by other functions already?
Now in my database I need to get the "maximum id" in a list; and I have it working for postgresql by using a native query:
const maxnum = await Order.getDatastore().sendNativeQuery('SELECT MAX(\"orderNr\") FROM \"order\"')
While this isn't the most difficult thing, it is not what I truly want: it is limited to only sql-based datastores (so we wouldn't be able to move easily to mongodb); and the syntax might actually be even different for another sql database type.
So I wonder - can this be transformed in such a way it doesn't rely on sendNativeQuery?
You can try .query() to execute a raw SQL query using the specified model's datastore and if u want u can try pg , an NPM package used for communicating with PostgreSQL databases:
Pet.query('SELECT pet.name FROM pet WHERE pet.name = $1', [ 'dog' ]
,function(err, rawResult) {
if (err) { return res.serverError(err); }
sails.log(rawResult);
// (result format depends on the SQL query that was passed in, and
the adapter you're using)
// Then parse the raw result and do whatever you like with it.
return res.ok();
});
You can use the limit and order options waterline provides to get a single Model with a maximal value (then just extract that value).
const orderModel = await Order.find({
where: {},
select: ['orderNr'],
limit: 1,
sort: 'orderNr DESC'
});
console.log(orderModel.orderNr);
Like most things in Waterline, it's probably not as efficient as an SQL SELECT MAX query (or some equivalent in mongo, etc), but it should allow swapping out the database with no maintenance. Last note, don't forget to handle the case of no models found.
How can I execute a bulk insert and continue in case of duplicate key error?
I have a collection with an unique index on the id field (not _id) and some data in it. Then I get more data and I want to add only the non-present documents to the collection.
I have the following code:
let opts = {
continueOnError: true, // Neither
ContinueOnError: true, // of
keepGoing: true, // this
KeepGoing: true, // works
};
let bulk = collection.initializeUnorderedBulkOp( opts );
bulk.insert( d1 );
bulk.insert( d2 );
bulk.insert( d3 );
...
bulk.insert( dN );
let result = yield bulk.execute( opts ); // this keep throwing duplicate key error
And I just want to ignore the errors and let the bulk finish with all the queued operations.
I searched in npm module api and in the MongoDB api for Bulk, initializeUnorderedBulkOp and the docs for Bulk write with no luck.
Also in the docs for Unordered Operations they say:
Error Handling
If an error occurs during the processing of one of the write operations, MongoDB will continue to process remaining write operations in the list.
Which is not true (at least in my case)
You can use db.collection.insertMany(), (new in version 3.2.) with:
ordered:false
With ordered to false, and in case of duplicate key error, the insert operation would continue with any remaining documents.
Here is link to documentation:
https://docs.mongodb.com/v3.2/reference/method/db.collection.insertMany/
Ordered Insert in MongoDB
db.hobbies.insertMany([{_id: "yoga", name: "Yoga"}, {_id: "cooking", name: "Cooking"}, {_id: "hiking", name: "Hiking"}], {ordered: true})
{ordered: true} is the default behaviour of insert statements
Unordered Insert in MongoDB
If you want mongodb to continue trying to insert other documents even after one or more failing due to any reason, you must set ordered to false. See example below:
db.hobbies.insertMany([{_id: "yoga", name: "Yoga"}, {_id: "cooking", name: "Cooking"}, {_id: "hiking", name: "Hiking"}], {ordered: false})
Not sure when this issue cropped up but I am not able to fetch items from mongo consistently. I have 4000+ items in the db. Here's the schema.
var Order = new Schema({
code: {
type: String,
unique: true
},
...
});
Now run some queries:
Order.find().exec(function(err, orders) {
console.log(orders.length); // always 101
})
Order.find().limit(100000).exec(function(err, orders) {
console.log(orders.length); // varies, sometimes 1150, 1790, 2046 - never more
})
Now if I remove the 'unique: true' from schema it will always return the total amount:
Order.find().exec(function(err, orders) {
console.log(orders.length); // always 4213 (correct total)
})
Any idea as to why this behavior occurs? afaik the codes are all unique (orders from a merchant). This is tested on 3.8.6, 3.8.8
Ok issue was indeed unique index being not there/corrupted. I a guilty of adding the unique index later on in the game and probably had some dups already which prevented Mongo from creating indexes.
I removed the duplicates and then in the mongo shell did this:
db.orders({name: 1}, {unique: true, dropDubs: true});
I would think the above would remove dups but it would just die because of the dups. I am sure there is a shell way to do this but I just did it with some js code then ran the above to recreate the index which can be verified with:
db.orders.getIndexes()