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.
Related
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 doc in couchDB:
{
"id":"avc",
"type":"Property",
"username":"user1",
"password":"password1",
"server":"localhost"
}
I want to write a view that returns a map of all these fields.
The map should look like this: [{"username","user1"},{"password","password1"},{"server","localhost"}]
Here's pseudocode of what I want -
HashMap<String,String> getProperties()
{
HashMap<String,String> propMap;
if (doc.type == 'Property')
{
//read all fields in doc one by one
//get value and add field/value to the map
}
return propMap;
}
I am not sure how to do the portion that I have commented above. Please help.
Note: right now, I want to add username, password and server fields and their values in the map. However, I might keep adding more later on. I want to make sure what I do is extensible.
I considered writing a separate view function for each field. Ex: emit("username",doc.username).
But this may not be the best way to do this. Also needs updates every time I add a new field.
First of all, you have to know:
In CouchDB, you'll index documents inside a view with a key-value pair. So if you index the property username and server, you'll have the following view:
[
{"key": "user1", "value": null},
{"key": "localhost", "value": null}
]
Whenever you edit a view, it invalidates the index so Couch has to rebuild the index. If you were to add new fields to that view, that's something you have to take into account.
If you want to query multiple fields in the same query, all those fields must be in the same view. If it's not a requirement, then you could easily build an index for every field you want.
If you want to index multiple fields in the same view, you could do something like this:
// We define a map function as a function which take a single parameter: The document to index.
(doc) => {
// We iterate over a list of fields to index
["username", "password", "server"].forEach((key, value) => {
// If the document has the field to index, we index it.
if (doc.hasOwnProperty(key)) {
// map(key,value) is the function you call to index your document.
// You don't need to pass a value as you'll be able to get the macthing document by using include_docs=true
map(doc[key], null);
}
});
};
Also, note that Apache Lucene allows to make full-text search and might fit better your needs.
I want to obtain all the fields of a schema in mongoose. Now I am using the following code:
let Client = LisaClient.model('Client', ClientSchema)
let query = Client.findOne({ 'userclient': userclient })
query.select('clientname clientdocument client_id password userclient')
let result = yield query.exec()
But I want all the fields no matter if they are empty. As always, in advance thank you
I'm not sure if you want all fields in a SQL-like way, or if you want them all in a proper MongoDB way.
If you want them in the proper MongoDB way, then just remove the query.select line. That line is saying to only return the fields listed in it.
If you meant in a SQL-like way, MongoDB doesn't work like that. Each document only has the fields you put in when it was inserted. If when you inserted the document, you only gave it certain fields, that document will only have those fields, even if other documents in other collections have different fields.
To determine all available fields in the collection, you'd have to find all the documents, loop through them all and build an object with all the different keys you find.
If you need each document returned to always have the fields that you specify in your select, you'll just have to transform your object once it's returned.
const fields = ['clientname', 'clientdocument', 'client_id', 'password', 'userclient'];
let Client = LisaClient.model('Client', ClientSchema)
let query = Client.findOne({ 'userclient': userclient })
query.select(fields.join(' '))
let result = yield query.exec()
fields.forEach(field => result[field] = result[field]);
That forEach loop will set all the fields you want to either the value in the result (if it was there) or to undefined if it wasn't.
MongoDB is schemaless and does not have tables, each collection can have different types of items.Usually the objects are somehow related or have a common base type.
Retrive invidual records using
db.collectionName.findOne() or db.collectionName.find().pretty()
To get all key names you need to MapReduce
mapReduceKeys = db.runCommand({
"mapreduce": "collection_name",
"map": function() {
for (var key in this) {
emit(key, null);
}
},
"reduce": function(key, stuff) {
return null;
},
"out": "collection_name" + "_keys"
})
Then run distinct on the resulting collection so as to find all the keys
db[mapReduceKeys.result].distinct("_id") //["foo", "bar", "baz", "_id", ...]
I have a work-around, however I'm wondering why this is failing - is it a JavaScript error on my part?
My app uses two models. I'm using Forerunner to persist a JSON model that is previously created in my MVC.
I want to store the Forerunner primary key in an object, to make it clearer (in my first model) that the key is used only in the Forerunner model.
The key will need to be saved in both models, so that when the first model changes, I can immediately update the Forerunner model - it is the link between two models.
Here is a code sample to illustrate:
var games = db.collection("games", {
primaryKey: "gamedb._id"
});
games.insert({
gamedb : {
_id: 1,
},
userDesc : "original 11x11",
Size : [11,11],
Plays : []
.
.
}, function (result) {
console.log(result);
});
I did try it out, and got confusing results, so for now I'm just leaving the key as property of the collection.
Here is the console.log result of my attempt using the above code sample. The forerunner insert() did not fail.
I'm confused about the property 'gamedb._id : "225b...".
Thanks for any suggestions or insight.
gamedb : Object
_id: 1
gamedb._id: "225b3c25aeb38a0",
}
.
.
Primary keys in ForerunnerDB must currently be in the root of the object. When you tell it to make the primary key "gamedb._id" it is not parsing that string as a path, it is assuming you are passing that as a root key e.g. {"gamedb._id": myId}
There are plans to update ForerunnerDB to allow nested primary keys, but this is not currently possible.
Source: I wrote ForerunnerDB
In Mongoose, I can use a query populate to populate additional fields after a query. I can also populate multiple paths, such as
Person.find({})
.populate('books movie', 'title pages director')
.exec()
However, this would generate a lookup on book gathering the fields for title, pages and director - and also a lookup on movie gathering the fields for title, pages and director as well. What I want is to get title and pages from books only, and director from movie. I could do something like this:
Person.find({})
.populate('books', 'title pages')
.populate('movie', 'director')
.exec()
which gives me the expected result and queries.
But is there any way to have the behavior of the second snippet using a similar "single line" syntax like the first snippet? The reason for that, is that I want to programmatically determine the arguments for the populate function and feed it in. I cannot do that for multiple populate calls.
After looking into the sourcecode of mongoose, I solved this with:
var populateQuery = [{path:'books', select:'title pages'}, {path:'movie', select:'director'}];
Person.find({})
.populate(populateQuery)
.execPopulate()
you can also do something like below:
{path:'user',select:['key1','key2']}
You achieve that by simply passing object or array of objects to populate() method.
const query = [
{
path:'books',
select:'title pages'
},
{
path:'movie',
select:'director'
}
];
const result = await Person.find().populate(query).lean();
Consider that lean() method is optional, it just returns raw json rather than mongoose object and makes code execution a little bit faster! Don't forget to make your function (callback) async!
This is how it's done based on the Mongoose JS documentation http://mongoosejs.com/docs/populate.html
Let's say you have a BookCollection schema which contains users and books
In order to perform a query and get all the BookCollections with its related users and books you would do this
models.BookCollection
.find({})
.populate('user')
.populate('books')
.lean()
.exec(function (err, bookcollection) {
if (err) return console.error(err);
try {
mongoose.connection.close();
res.render('viewbookcollection', { content: bookcollection});
} catch (e) {
console.log("errror getting bookcollection"+e);
}
//Your Schema must include path
let createdData =Person.create(dataYouWant)
await createdData.populate([{path:'books', select:'title pages'},{path:'movie', select:'director'}])