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'm using drupal(7.38), civicrm(4.6.5), ubercart(7.x-3.8) and ubercart_civicrm(7.x-4.x-dev). Whenever a user(registered/anonymous) is purchase a product from the site, a contribution will be added into the civcrm user. If the user is anonymous, system will create the user by using email both in drupal as well as civicrm also. I notice some of the products status is 'Abandoned' in ubercart, but under the corresponding user contribution tab, the product is listing as Completed. I gone throw the uc_civicrm module at last I found some piece of codes in uc_civicrm/uc_civicrm.module
function _uc_civicrm_map_contribution_status($order_status) {
// NOTE: There may be a more "correct" way to do this.
$map = array(
"completed" => 1,
"payment_received" => 1,
"pending" => 2,
"processing" => 5,
"canceled" => 3,
"in_checkout" => 5,
);
if (array_key_exists($order_status, $map)) {
$id = $map[$order_status];
}
else {
// Oh no.
$id = 1;
watchdog('uc_civicrm', 'There is no status that maps to %order_status, marking as "Complete"', array('%order_status' => $order_status), WATCHDOG_WARNING);
}
return $id;
}
I want to synchronize ubercart order status with civicrm contributions status. Please help me out this. Thanks in advance
If you want to map 'abandoned' to the 'canceled' status in CiviCRM (there is no "abandoned" status in CiviCRM) you could revise the above like so:
$map = array(
"completed" => 1,
"payment_received" => 1,
"pending" => 2,
"processing" => 5,
"canceled" => 3,
"abandoned" => 3,
"in_checkout" => 5,
);
Then I'd recommend making a bug report/patch on the issue you filed at the project page so it gets into the module core if the maintainers agree.
On the one site where we had CiviCRM and Ubercart well (it was Drupal 6) we ended up needing to use both
- UC CiviCRM,
- Ubercart CiviCRM Products
but I see the latter only was released for D6
More recently for Commerce integration we have switched from using the module to using CiviCRM Entity and Drupal Rules.
Not sure how much the above helps but thought I should mention it.
I would like to query some offers by skills from my database with laravel.
The relationship between an offer and its skills is a belongsToMany relationship.
How to get all offers that match an array of skills from the database ?
To be clearer here is some code I started to write :
$skills = Input::get('skills');
$offers = Offer::with(
array(
'skills' => function($query){
// here I'd like to select skills from array $skills
$query->where(?????);
}
)
)->get();
Thanks in advance !
Use whereIn with array
http://laravel.com/docs/queries#selects
$skills = Input::get('skills');
$offers = Offer::with(array('skills' => function($query) use ($skills){
$query->whereIn('skill', $skills);
}))->get();
I have read and worked with the other posts about this and it appears the version of Laravel 4 I just downloaded has more changes made to the way the JSON input is handled by a controller.
$input = Input::json()->all(); gives me errors as if I am referring to something that does not exist when I request some part of the payload after doing a PUT request. And without ->all(); I get a symfony error.
Does anyone know how to get good JSON from backbone in Laravel 4's latest version?
Currently, I am doing the long way around to get my data, ie:
$input_title = Input::get('title');
$input_completed = Input::get('completed');
$task = Task::find($id);
$task->title = $input_title;
$task->completed = $input_completed;
$task->save();
Yes, I am doing the tutorial on tutsplus to learn laravel/backbone, so a little noob patience is apreciated.
The error I get when using Input::get(); is:
{"error":{"type":"UnexpectedValueException","message":"The Response content must be a string or object implementing __toString(), \"array\" given.","file":"/Users/brentlawson23/Sites/laravel4App/bootstrap/compiled.php","line":16858}}
I really want to get the Laravel-specific answer instead of using straight php to stringify the payload.
I get same error using just Input::json();
For the current beta of Laravel 4, Input::json(); is not getting a stringified version of the request payload that can be used to create a new row in a table, nor does Input::json()->all(); (hoping to play nice with the ParameterBag from symfony). I have tried json_encode among other hacks and basically every step of the way in this tut, I hit some brick wall. Anyone have a suggestion based on what I have presented here?
Today I got this when simply trying to echo the result of $input = Input::json(); :
{"error":{"type":"ErrorException","message":"Catchable Fatal Error: Object of class Symfony\Component\HttpFoundation\ParameterBag could not be converted to string in /Users/brentlawson23/Sites/laravel4App/app/controllers/TasksController.php line 45","file":"/Users/brentlawson23/Sites/laravel4App/app/controllers/TasksController.php","line":45}}
Yes, I have studied the Symfony API.
I had a similar problem. Input from Backbone is converted to array in Laravel. On tutsplus, Jeffrey Way is using object. So I was trying to do this (like in tutorial):
return $input->title // using object,but got an error.
If I change that line to:
return $input["title"] // everything works fine with array.
I'm also working through the Backbone tutorial on tuts+. If I'm right in assuming are you stuck on the Creating New Contacts section? Below is how I got it to work for me, in ContactController.php:
public function store()
{
$input = Input::all();
Contact::create(array(
'first_name' => $input['first_name'],
'last_name' => $input['last_name'],
'email_address' => $input['email_address'],
'description' => $input['description']
));
}
And then also needed to update app/models/Contact.php with the below:
class Contact extends Eloquent {
protected $fillable = array('first_name', 'last_name', 'email_address', 'description');
}
That should get it working for you and insert the contact into the database. If I've misread let me know and I can have another look.
Cheers,
Sean
I've read that, to be able to rank search results you may query MySQL like this:
SELECT * ,
MATCH (title, body) AGAINST ('$search') AS rating
FROM posts
WHERE MATCH (title, body) AGAINST ('$search')
ORDER BY rating DESC
Is there a way to do this in CakePHP 2.X?
Also, I need to do this while paginating at the same time. So I think I would need to write condition for the paginator, not a direct 'query'.
Thanks for your help!
Use like this it will prevent mysql injection too
array("MATCH(User.current_position) AGAINST(? IN BOOLEAN MODE)" => $srch_arr['text'])
Ok, it took me some time... Since, the key issue was to get a rating on the resulting matches, the complicated part in this query was the specific field:
MATCH (title, body) AGAINST ('$search') AS rating
I figured that I should just write that field in the "field" option, in the pagination array.
The resulting code was the following:
$this->paginate = array(
'limit' => 15,
'fields' => array('*', "MATCH (data) AGAINST ('$q') AS rating"),
'conditions' => "MATCH(SearchIndex.data) AGAINST('$q' IN BOOLEAN MODE)",
'order' => array(
'rating' => 'desc',
),
);
$paginatedResults = $this->paginate('SearchIndex');
And that worked seamlessly!
I think this is the best way to achieve real search results using Cake. Unless someone has a better alternative :)
Searching phrases in between double quotes will give you the results you should expect!
I have used the above database call by Thomas (thank you) and it does work seamlessly.
However the code:
'conditions' => "MATCH(SearchIndex.data) AGAINST('$q' IN BOOLEAN MODE)",
removes the Data Abstraction Layer and opens up your site to SQL injection.
It's probably not quite as good (haven't fully tested it) but try:
'SearchIndex.data LIKE'=>'%'.$search.'%'
I hope this is helpful in someway.