I am working through a simple logical problem, but I cannot seem to have things work smoothly. Let me share my most convincing code experiment and then I'll share some thoughts.
useEffect(() => {
firebase
.firestore()
.collection("someCollection")
.orderBy("date", "desc")
.onSnapshot(docs => {
let documents = []
if (canGetUpdatesFromFirestore.current) {
docs.forEach((doc) => {
documents.push(doc.data())
})
if(documents.length > 3) {
documents.splice(4, 0, {questionPostId: 0})
documents.splice(5, 0, {questionPostId: 1})
}
setAllQuestions(documents)
setUsers(documents)
}
})
if (searchValue.length > 2) {
canGetUpdatesFromFirestore.current = false;
functions.searchForSearchVal(searchValue, "Sexuality")
.then((result) => {
setAllQuestions(result);
})
} else {
canGetUpdatesFromFirestore.current = true
}
}, [searchValue])
function setUsers(docs){
let arrFinal = []
let copyOfAllQuestions = ""
for(let i = 0; i< docs.length; i++) {
console.log("HERE")
if (docs[i].postedBy) {
docs[i].ref.get().then(userFire => {
copyOfAllQuestions = {
...allQuestions,
...{hasPremium: userFire.data().hasPremium}
}
})
arrFinal.push(copyOfAllQuestions)
}
}
setAllQuestions(arrFinal)
}
Let me share some of my current state and what I am trying to accomplish.
I have a that display allQuestions. Each question data has a ref to its user document in firestore. For each question I need to check if that user hasPremium. How should I go about doing that the correct way?
The problem currently is that I can get the data from my Users collection through the ref, but I have to refresh my state in order for it all to show.
Could someone help me get on the right path / think correctly on this one please?
One approach that I put forward is to embrace data denormalization. That is, rather than putting references to other documents (Users) inside of the Questions document, put all the relevant user information directly into the Questions document.
This is antithetical to SQL database approaches, but that's okay because Firestore is "NoSQL". Embrace the anti-SQL-idity!!
Essentially, in your Question document you want to copy in whatever information is required in your app when working with a Question, and avoid doing "joins" by fetching other documents. You don't need to copy in all of the User document into a Question document - just the elements needed when your app is working with a Question.
For example, maybe in the question all you need is:
question: {
name: ...,
type: ...,
lastUpdated: ...,
postedBy: {
email: ...,
displayName: ...,
avatarUrl: ...,
hasPremium: true,
}
}
With data duplicated, you often need a mechanism to keep duplicate data up-to-date from its "source". So you might consider a Cloud Function trigger for onUpdate() of User documents, and when a relevant value is modified (email, displayName, avatarUrl, and/or hasPremium) then you would loop through all questions that are postedBy that user and update accordingly.
The rules-of-thumb here are:
all data needed for one screen/function in your app goes into a SINGLE document
NoSQL document stores are used where reads are frequent and writes are infrequent
NoSQL data stores (typically) do not have "joins" - so don't design your app to require them (which is what your code above is doing: joining Question and Users)
often you don't care about updating ALL instances of duplicated data (e.g. if a user updates their displayName today, should you update a Question they posted 3 years ago? -- different apps/business needs will give different answers)
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 am currently working on a blogging app, in which users can create their own blogs and each blog has blogposts within that. I'm ideating about architecting a database that is scalable when each blog has a lot of blogposts.
So is it better to structure my database as this:
blog1 : {
blogname : 'blog1',
blogposts: [array of blogposts]
},
blog2 : {
blogname : 'blog2',
blogposts: [array of blogposts]
}
Or should I create a separate collection with all the blogposts, something like this:
blogpost1: {
id: 'blogpost1',
content: {blogpost content in json format}
},
blogpost2: {
id: 'blogpost2',
content: {blogpost content in json format}
}
and reference them in the blog collection.
I want to know which choice would be superior when there are a lot of blogposts. Because I remember reading somewhere in MongoDB docs that it's not recommended to have arrays within document that can grow beyond bounds, so approach #1 is not ideal, right?
When creating databases, I find it useful to think about the requests I would be making.
A blogging app user would want to search all blogs or find a blogger by some criteria.
In this case separate collections for bloggers and blogs would work best. Then structure your documents so that the bloggers link to their blogs and vice versa.
This can be done with Mongoose Schemas (https://mongoosejs.com/docs/index.html).
// models/blogger.js
const mongoose = require('mongoose')
const bloggerSchema = mongoose.Schema({
blogs: [
{
type: mongoose.Schema.Types.ObjectId,
ref: 'Blog'
}
],
name: String
})
bloggerSchema.set('toJSON', {
transform: (document, returnedObject) => {
const blogger = returnedObject
blogger.id = blogger._id.toString()
delete blogger._id
delete blogger.__v
}
})
module.exports = mongoose.model('Blogger', bloggerSchema)
Then use populate with your request:
// controllers/bloggers.js
const bloggersRouter = require('express').Router()
const Blogger = require('../models/blogger')
bloggersRouter.get('/', async (request, response) => {
const bloggers = await Blogger.find({}).populate(
'blogs', {
title: 1
}
)
response.json(bloggers.map(blogger => blogger.toJSON()))
})
module.exports = bloggersRouter
This way you don't have to add the blogs in their entirety to the blogger document, you can just include the title or anything else that you need on the bloggers initial view.
You could also think about limiting the length of a blog, so you can have more control over the data and then think about the options Joe suggested.
Why does it have to be one or the other?
Storing the blog posts in the same document as the blog is great as long as the individual posts are not very large, and there aren't very many of them.
Storing the posts in a separate collection is good for bigger posts and busy blogs but adds an additional query or lookup to retrieve.
I would think it is expected that your users' output will run the gamut from sparse to prolific, and individual posts will range from a few dozen bytes to many megabytes.
For small posts on not very active blogs, store the posts in the blog document for efficient retrieval.
For busy blogs, store them in an archive collection. Perhaps store the most recent couple of posts, or the most popular posts, in the blog document so you don't have to refer to the other collection every time.
You will also need to figure out how to split a post between documents. MongoDB has a 16MB limit on a single document, so if any of your users make huge posts, you'll need to be able to store them somewhere.
Your question as written seems to be asking whether it is better to follow a relation model or a strict document model. I think in reality neither is a perfect fit for this and a hybridized and flexible approach would work out better.
Closed. This question is opinion-based. It is not currently accepting answers.
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Closed 10 months ago.
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Im going to create a achievement system in Mongodb. But im not sure how i would format/store it in the database.
As of the users should have a progress (on each achievement they would have some progress value stored), im really confused what would be the best way to perform this, and without having an performence issue.
what should i do?, cause i dont know, what i had in mind, was maybe something like:
Should i store each achievement in an unique row in a Achievement collection, and an user array within that row, containing object with userid and achievement progress?
Would i then get an performance issue when its 1000+ achievements, that is beeing checked fairy often?
or should i do something else?
example schema for the option above:
{
name:{
type:String,
default:'Achievement name'
},
users:[
{
userid:{
type:String,
default:' users id here'
},
progress:{
type:Number,
default:0
}
}
]
}
Even though the question is specifically about the database design, I will give a solution for the tracking/awarding logic as well to establish more accurate context for the db design.
I would store the achievements progress separately from the already awarded achievements for cleaner tracking and discovery.
The whole logic is event based and has multiple layers of event handling. This gives you TONS of flexibility on how you track your data and also gives you a pretty good mechanism to track history. Basically, you can look at it as a form of logging.
Of course, your system design and contracts are highly dependent on the information you're gonna be tracking and its complexity. A simple progress field may not suffice for each case(you might want to track something more complex, not a simple number between X and Y). There is also the case of tracking data which updates quite frequently(as distance travelled in games, for example). You didn't give any context on the topic of your achievement system so we're gonna stick with a generic solution. It's just a couple of things that you should take a note about as it will affect the design.
Okay, so, let's start from the top and track the entire flow for a tracked piece of data and its eventual achievement progress. Let's say we're tracking consecutive days of user login and we're gonna award him with an achievement when he reaches [10].
Note that everything below is just a pseudo-code.
So, let's say today is [8th of July, 2017]. For now, our User entity looks like this:
User: {
id: 7;
trackingData: {
lastLogin: 7 of July, 2017 (should be full DateTime object, but using this for brevity),
consecutiveDays: 9
},
achievementProgress: [
{
achievementID: 10,
progress: 9
}
],
achievements: []
}
And our achievements collection contains the following entity:
Achievement: {
id: 10,
name: '10 Consecutive Days',
rewardValue: 10
}
The user tries to login(or visit the site). The application handler takes note of that and after handling the login logic fires an event of type ACTION:
ACTION_EVENT = {
type: ACTION,
name: USER_LOGIN,
payload: {
userID: 7,
date: 8 of July, 2017 (should be full DateTime object, but using this for brevity)
}
}
We have an ActionHandler which listens for events of type ACTION:
ActionHandler.handleEvent(actionEvent) {
subscribersMap = Map<eventName, handlers>;
subscribersMap[actionEvent.name].forEach(subscriber => subscriber.execute(actionEvent.payload));
}
subscribersMap gives us a collection of handlers that should respond to each specific action(this should resolve to USER_LOGIN for us). In our case we can have 1 or 2 that concern themselves with updating the user tracking information of lastLogin and consecutiveDays tracking properties in the user entity. The handlers in our case will update the tracking information and fire new events further down the line.
Once again, for brevity, we're gonna incorporate both into one:
updateLoginHandler: function(payload) {
user = db.getUser(payload.userID);
let eventType;
let eventValue;
if (date - user.trackingData.lastLogin > 1 day) {
user.trackingData = 1;
eventType = 'PROGRESS_RESET';
eventValue = 1;
}
else {
const newValue = user.trackingData.consecutiveDays + 1;
user.trackingData.consecutiveDays = newValue;
eventType = 'PROGRESS_INCREASE';
eventValue = newValue;
}
user.trackingData.lastLogin = payload.date;
/* DISPATCH NEW EVENT OF TYPE ACHIEVEMENT_PROGRESS */
AchievementProgressHandler.dispatch({
type: ACHIEVEMENT_PROGRESS
name: eventType,
payload: {
userID: payload.userID,
achievmentID: 10,
value: eventValue
}
});
}
Here, PROGRESS_RESET have the same contract as the PROGRESS_INCREASE but have a different semantic meaning and I would keep them separate for history/tracking purposes. If you wish, you can combine them into a single PROGRESS_UPDATE event.
Basically, we update the tracked fields that are dependent on the lastLogin date and fire a new ACHIEVEMENT_PROGRESS event which should be handled by a separate handler with the same pattern(AchievementProgressHandler). In our case:
ACHIEVEMENT_PROGRESS_EVENT = {
type: ACHIEVEMENT_PROGRESS,
name: PROGRESS_INCREASE
payload: {
userID: 7,
achievementID: 10,
value: 10
}
}
Then, in AchievementProgressHandler we follow the same pattern:
AchievementProgressHandler: function(event) {
achievementCheckers = Map<achievementID, achievementChecker>;
/* update user.achievementProgress code */
switch(event.name): {
case 'PROGRESS_INCREASE':
achievementCheckers[event.payload.achievementID].execute(event.payload);
break;
case 'PROGRESS_RESET':
...
}
}
achievementCheckers contains a checker function for each specific achievement that decides if the achievement has reached its desired value(a progress of 100%) and should be awarded. This enables us to handle all kinds of complex cases. If you only track a single X out of Y scenario, you can share the function between all achievements.
The handler basically does this:
achievementChecker: function(payload) {
achievementAwardHandler;
achievement = db.getAchievement(payload.achievementID);
if (payload.value >= achievement.rewardValue) {
achievementAwardHandler.dispatch({
type: ACHIEVEMENT_AWARD,
name: ACHIEVEMENT_AWARD,
payload: {
userID: payload.userID,
achievementID: achievementID,
awardedAt: [current date]
}
});
/* Here you can clear the entry from user.achievementProgress as you no longer need it. You can also move this inside the achievementAwardHandler. */
}
}
We once again dispatch an event and use an event handler - achievementAwardHandler. You can skip the event creation step and award the achievement to the user directly but we keep it consistent with the whole history logging flow.
An added benefit here is that you can use the handler to defer the achievement awarding to a specific later time thus effectively batching awarding for multiple users, which serve a couple of purposes including performance enhancement.
Basically, this pseudo code handles the flow from [a user action] to [achievement rewarding] with all intermediate steps included. It's not set in stone, you can modify it as you like but all in all, it gives you a clean separation of concerns, cleaner entities, it's performant, let's you add complex checks and handlers which are easy to reason about while in the same time provide a great history log of the user overall progress.
Regarding the DB schema entities, I would suggest the following:
User: {
id: any;
trackingData: {},
achievementProgress: {} || [],
achievements: []
}
Where:
trackingData is an object that contains everything you're willing
to track about the user. The beauty is that properties here are
independent from achievement data. You can track whatever and eventually use it for achievement purposes.
achievementProgress: a map of <key: achievementID, value: data> or
an array containing the current progress for each achievement.
achievements: an array of awarded achievements.
and Achievement:
Achievement: {
id: any,
name: any,
rewardValue: any (or any other field/fields. You have complete freedom to introduce any kind of tracking with the approach above),
users?: [
{
userID: any,
awardedAt: date
}
]
}
users is a collection of users who have been rewarded the given achievement. This is optional and is here only if you have the use for it and query for this data frequently.
What you might be looking for is a Badge style implementation. Just like Stack Overflow rewards it's users with badges for specific achievements.
Method 1: You can have flags in the user profile for each badge. Since you're doing it in NoSQL database, you just have to set a flag for each badge.
const badgeSchema = new mongoose.Schema({
badgeName: {
type: String,
required: true,
},
badgeDescription: {
type: String,
required: true,
}
});
const userSchema = new mongoose.Schema({
userName: {
type: String,
required: true,
},
badges: {
type: [Object],
required: true,
}
});
If your application architecture is event based, you can trigger awarding badges to users. And that operation is just inserting Badge object with progress in User badges array.
{
badgeId: ObjectId("602797c8242d59d42715ba2c"),
progress: 10
}
Update operation will be to find and update the badges array with progress percentage number
And while displaying user achievements on user interface, you can just loop over badges array to show the badges this user has achieved and their progress with it.
Method 2: Have a separate mongo collection for Badge and User Mapping. Whenever a user achieves a badge you insert a record in that collection. It will be one to one mapping of user _id and badge _id and progress value. But as the table will grow huge you will need to do indexing to efficiently query user and badge mapping.
You will have to do analysis on best approach according to your specific use case.
MongoDB is flexible enough to allow teams develop applications quickly, and involve their model with litter friction as the application needs it. In cases where you need a robust model from day one, theirs is a flexible methodology that can guide you through the process of modeling your data.
The methodology is composed of:
Workload: This stage is about gathering as much information as possible to understand your data. This will allow you formulate assumptions about, you data size the operations that will be performance against it (reads and writes), quantify operations and qualify operations.
You can get this by:
Scenarios
Prototype
Production Logs & Stats (if you are migrating).
Relationships: Identify the relationship between the different entities in your data, quantify those relationships and apply embedding or linking. In general you should prefer embedding by default, but remember that arrays should not grow without bound (6 Rules of Thumb for MongoDB Schema Design: Part 3).
Patterns: Apply schema design patterns. Take a look at Building with Patterns: A Summary, it presents a matrix that highlights the pattern that could be useful for a given use case.
Finally, the goal of this methodology is help you create a model, that can scale and perform well under stress.
If you design the achievement schema like this:
{
name: {
type: String,
default: "Achievement name",
},
userid: {
type: String,
default: " users id here",
},
progress: {
type: Number,
default: 0,
},
}
}
When an achievement is gained you just add another entry
for getting achievements Map-Reduce is a good candidate for running map reduce on the database. you can run them on a less regular basis, using them for offline computation of the data that you want.
based on documentation you can do like the following photo
React + Redux recommend saving data normalized and using selectors to get derived data. So in our store we save Users and Tags which have a many to many relationship.
type Store = {
entities: {
users: User[];
tags: Tag[];
userTagMapping: { userId: string, tagId: string }[]
}
}
And in our view we want to show some derived data for this many to many relation-ship. For example we want to calculate the total users with some tag or the online-users with some tag. Right now we solved this using rselect. The only problem is that calculating these things becomes quite tedious. For example we have a selector which returns a map from a tag-id to a list of users to show which users belong to this tag (and vice versa a selector from user-id to tag list).
const selectUsersPerTag = createSelector(
selectUsers, selectTags, selectUserTagMapping,
(users, tags, mapping) => {
let result = {};
for (const tag on tags) {
const isUserMappedToTag = user => ({userId, tagId}) => userId == user.id && tagId === tag.id
result[tag.id] = users.filter(user => mapping.some(isUserMappedToTag(user)))
}
return result
}
)
and in my opinion this looks quite ugly and is a bit too hard to read for my taste.
My questions are:
Are we understanding the recommendations correctly (to use normalization and selectors)?
Is using a map the correct way to process our data and show it in the view or is there a better way? I am asking because this basically copies our data (slightly modified) many times into the props of our React components
Is there a nicer way to do this mapping (which is basically a SQL like join)? Because I really don't like this imperative approach and would find a functional one much nicer.
I've got an app using mobx-state-tree that currently has a few simple stores:
Article represents an article, either sourced through a 3rd party API or written in-house
ArticleStore holds references to articles: { articles: {}, isLoading: bool }
Simple scenario
This setup works well for simple use-cases, such as fetching articles based on ID. E.g.
User navigates to /article/{articleUri}
articleStoreInstance.fetch([articleUri]) returns the article in question
The ID is picked up in render function, and is rendered using articleStoreInstance.articles.get(articleUri)
Complex scenario
For a more complex scenario, if I wanted to fetch a set of articles based on a complex query, e.g. { offset: 100, limit: 100, freeTextQuery: 'Trump' }, should I then:
Have a global SearchResult store that simply links to the articles that the user has searched for
Instantiate a one-time SearchResult store that I pass around for as long as I need it?
Keep queries and general UI state out of stores altogether?
I should add that I'd like to keep articles in the stores between page-loads to avoid re-fetching the same content over and over.
Is there a somewhat standardized way of addressing this problem? Any examples to look at?
What you need might be a Search store which keeps track of following information:
Query params (offset, limit, etc.)
Query results (results of the last search)
(Optional) Query state (isLoading)
Then to avoid storing articles in 2 places, the query results should not use Article model, but reference to Article model. Anytime you query, the actual result will be saved in existing store ArticleStore, and Search only holds references:
import { types, getParent, flow } from 'mobx-state-tree'
const Search = types.model({
params: // your own params info
results: types.array(types.reference(Article))
}).views(self => ({
get parent() {
return getParent(self) // get root node to visit ArticleStore
}
})).actions(self => ({
search: flow(function*(params) {
this.params = params // save query params
const result = yield searchByQuery(query) // your query here
this.parent.articleStore.saveArticles(result) // save result to ArticleStore
this.results = getArticleIds(result) // extract ids here for references
})
}))
Hope it's what you are looking for.