In my chat app, I have private chat between the two users. I intend to set the chat document's id using these two user's docId/uid in such a way that it doesn't depend on the order they're combined and I can determine the chat document's docId using the uid of users irrespective of the order of uid.
I know, I can use where clauses to get the chat doc as well. Is there any major flaw with my approach of generating the chat document's docId? Should I let it be generated automatically and use normal where clauses supported by firestore and limit(1) to get the chat?
basically, it seems I'm looking for is to encrypt uid1 in such a way that it returns a number only and then same with uid2 and then add them together to create the ChatId. This way it'll not depend on the order I use to add them and I can get the chatId and maybe convert that number back to a string using Base64 encode. This way, if I know the users participating in the chat, I can generate the same ChatId. Will that work or is there any flaw to it?
Converting each user ID to a number and then adding them together will likely lead to collisions. As a simple example, think of the many ways you can add up to the number 5: 0+5, 1+4, 2+3.
This answer builds upon #NimnaPerera's answer.
Method 1: <uid>_<uid>
If your app doesn't plan on using large groups, you can make use of the <uid>_<uid> format. To make sure the two user IDs are ordered in the same way, you can sort them first and then combine them together using some delimiter.
A short way to achieve this is to use:
const docId = [uid1, uid2].sort().join("_");
If you wanted to have a three-way group chat, you'd just add the new userID in the array:
const docId = [uid1, uid2, uid3].sort().join("_");
You could also turn this into a method for readability:
function getChatIdForMembers(userIds) {
return userIds.sort().join("_");
}
Here's an example of it in action:
const uid1 = "apple";
const uid2 = "banana";
const uid3 = "carrot";
[uid1, uid2].sort().join("_"); // returns "apple_banana"
[uid1, uid3].sort().join("_"); // returns "apple_carrot"
[uid2, uid1].sort().join("_"); // returns "apple_banana"
[uid2, uid3].sort().join("_"); // returns "banana_carrot"
[uid3, uid1].sort().join("_"); // returns "apple_carrot"
[uid3, uid2].sort().join("_"); // returns "banana_carrot"
// chats to yourself are permitted
[uid1, uid1].sort().join("_"); // returns "apple_apple"
[uid2, uid2].sort().join("_"); // returns "banana_banana"
[uid3, uid3].sort().join("_"); // returns "carrot_carrot"
// three way chat
[uid1, uid2, uid3].sort().join("_"); // returns "apple_banana_carrot"
[uid1, uid3, uid2].sort().join("_"); // returns "apple_banana_carrot"
[uid2, uid1, uid3].sort().join("_"); // returns "apple_banana_carrot"
[uid2, uid3, uid1].sort().join("_"); // returns "apple_banana_carrot"
[uid3, uid1, uid2].sort().join("_"); // returns "apple_banana_carrot"
[uid3, uid2, uid1].sort().join("_"); // returns "apple_banana_carrot"
Method 2: Member list properties
If you intend on supporting group chats, you should use automatic document IDs (see CollectionReference#add()) and store a list of chat members as one of it's fields as introduced in #NimnaPerera's answer for better use of queries.
I recommend two fields:
"members" - an array containing each chat member's ID. This allows you to query the /chats collection for chats that contain the given user.
"membersAsString" - a string, built from sorting "members" and joining them using "_". This allows you to query the /chats collection for chats that contain the exact list of members.
"chats/{chatId}": {
"members": string[], // list of users in this chat
"membersAsString": string, // sorted list of users in this chat, delimited using "_"
/* ... */
}
To find all chats that I am a part of:
const myUserId = firebase.auth().currentUser.uid;
const myChatsQuery = firebase.firestore()
.collection("chats")
.where("members", "array-contains", myUserId);
myChatsQuery.onSnapshot(querySnapshot => {
// do something with list of chat documents
});
To find all three-way chats between Apple, Banana and I:
const myUserId = firebase.auth().currentUser.uid;
const members = [myUserId, "banana", "apple"];
const membersAsString = members.sort().join("_");
const groupChatsQuery = firebase.firestore()
.collection("chats")
.where("membersAsString", "==", membersAsString);
groupChatsQuery.onSnapshot(querySnapshot => {
// do something with list of chat documents
// normally this would return 1 result, but you may get
// more than one result if a user gets added/removed a chat
});
A normal flow, would be to:
Get a list of the relevant chats
For each chat, get the most recent message
Based on the most recent message, sort the chats in your UI
You can very well use a combination of two users uids to define your Firestore document IDs, as soon as you respect the following constraints:
Must be valid UTF-8 characters
Must be no longer than 1,500 bytes
Cannot contain a forward slash (/)
Cannot solely consist of a single period (.) or double periods (..)
Cannot match the regular expression __.*__
What I'm not sure to understand in your question is "in such a way that it doesn't depend on the order they're combined". If you combine the uids of two users you need to combine them in a certain order. For example, uid1_uid2 is not equal to ui2_uid1.
As you are asking #lightsaber you can follow following methods to achieve your objective. But my personal preference is using an where clause, because firestore is supporting that compound queries which cannot be done in real time database.
Method 1
Create a support function to generate a chatId and check whether document is exist from that id. Then you can create chat document or retrieve the document using that id.
const getChatId = (currentUserId: string, guestUserId: string) => {
/* In this function whether you changed the order of the values when passing as parameters
it will always return only one id using localeCompare */
const comp = currentUserId.localeCompare(guestUserId);
if (comp === 0) {
return null;
}
if (comp === -1) {
return currentUserId + '_' + guestUserId;
} else {
return guestUserId + '_' + currentUserId;
}
}
Method 2
Use where clause with array-contains query for retrieving the chat document. And when creating add two user Ids to array and set the array with a relevant field name.
Firestore docs for querying arrays
I need to compare some values in my Firebase Database using JavaScript, but I'm having a hard time doing that. See image above.
The way that is suppose to work is I need to check if the user likes the same id as the other user. So for example:
User XbsX0IskrHVcaEmEBgyeok9isiM2 liked 4 items with unique ID's. Now I need to check if user jBc2Ls32DgMUSgzKUkVSw38UjQD2 liked the same thing to see if it's a match.
I have this code:
var check = ref.child('likes').child(uid2).child(uid2);
but it's not working.
You could iterate over the likes of the first user and make sure that everyone exists on the list of likes of the second user:
// Gets the list of likes from both users
var likes_user_1 = ref.child('likes').child(uid1);
var likes_user_2 = ref.child('likes').child(uid2);
// Iterates over the list of likes of the first user
var isAMatch = Object.keys(likes_user_1).every(function(like) {
// Returns true if user_2 has this like
return likes_user_2[like];
});
CONTEXT
I am trying to create a search functionality allowing users to fill in multiple fields, submit, and see a list of matching items from one collection. I do this using a form on the front end, which updates session variables on back-end, which are then passed as query to a mongodb collection.
HOW IT SHOULD WORK
If a user submits a venue size, then venues of that size are shown. If only a location is typed in, then venues within that location are shown. If both a size and a location are submitted, then venues that match both criteria are shown.
HOW IT ACTUALLY WORKS
If nothing is filled in, pressing search yields all items in the collection. Submitting both location and size yields venues that match both criteria. However, filling in only one field and leaving the other empty yields nothing in results. I'm wondering why this might be - it's almost as if the query is searching for a field that literally contains ''... but then why don't I see this behavior when leaving both fields empty? Help much appreciated!
CODE SNIPPET
//Search Form Helper
Template.managevenues.helpers({
venue: function () {
var venueNameVar = Session.get('venueNameVar');
var venueLocationVar = Session.get('venueLocationVar');
if(venueNameVar || venueLocationVar){
console.log(venueNameVar);
console.log(venueLocationVar);
return Venues.find({
venueName: venueNameVar,
'venueAddress.neighbourhood': venueLocationVar
});
} else {
return Venues.find({});
}
});
The answer lies in your query
Venues.find({
venueName: venueNameVar,
'venueAddress.neighbourhood': venueLocationVar
});
If you don't have one of your vars set it will look like this...
{
venueName: undefined,
'venueAddress.neighbourhood':'someVal'
}
So it would match any venue that doesn't have a name and is in some neighborhood.
A better approach would be to only set query criteria if there's a value to search...
var query = {};
if(Session.get('venueNameVar')) {
query.venueName = Session.get('venueNameVar');
}
if(Session.get('venueLocationVar') {
query.venueAddress = {
neighbourhood : Session.get('venueLocationVar');
}
}
return Venues.find(query);
I think this will work a bit better for you!
I've run into a bit of an issue with some data that I'm storing in my MongoDB (Note: I'm using mongoose as an ODM). I have two schemas:
mongoose.model('Buyer',{
credit: Number,
})
and
mongoose.model('Item',{
bid: Number,
location: { type: [Number], index: '2d' }
})
Buyer/Item will have a parent/child association, with a one-to-many relationship. I know that I can set up Items to be embedded subdocs to the Buyer document or I can create two separate documents with object id references to each other.
The problem I am facing is that I need to query Items where it's bid is lower than Buyer's credit but also where location is near a certain geo coordinate.
To satisfy the first criteria, it seems I should embed Items as a subdoc so that I can compare the two numbers. But, in order to compare locations with a geoNear query, it seems it would be better to separate the documents, otherwise, I can't perform geoNear on each subdocument.
Is there any way that I can perform both tasks on this data? If so, how should I structure my data? If not, is there a way that I can perform one query and then a second query on the result from the first query?
Thanks for your help!
There is another option (besides embedding and normalizing) for storing hierarchies in mongodb, that is storing them as tree structures. In this case you would store Buyers and Items in separate documents but in the same collection. Each Item document would need a field pointing to its Buyer (parent) document, and each Buyer document's parent field would be set to null. The docs I linked to explain several implementations you could choose from.
If your items are stored in two separate collections than the best option will be write your own function and call it using mongoose.connection.db.eval('some code...');. In such case you can execute your advanced logic on the server side.
You can write something like this:
var allNearItems = db.Items.find(
{ location: {
$near: {
$geometry: {
type: "Point" ,
coordinates: [ <longitude> , <latitude> ]
},
$maxDistance: 100
}
}
});
var res = [];
allNearItems.forEach(function(item){
var buyer = db.Buyers.find({ id: item.buyerId })[0];
if (!buyer) continue;
if (item.bid < buyer.credit) {
res.push(item.id);
}
});
return res;
After evaluation (place it in mongoose.connection.db.eval("...") call) you will get the array of item id`s.
Use it with cautions. If your allNearItems array will be too large or you will query it very often you can face the performance problems. MongoDB team actually has deprecated direct js code execution but it is still available on current stable release.
I'm trying to test out Firebase to allow users to post comments using push. I want to display the data I retrieve with the following;
fbl.child('sell').limit(20).on("value", function(fbdata) {
// handle data display here
}
The problem is the data is returned in order of oldest to newest - I want it in reversed order. Can Firebase do this?
Since this answer was written, Firebase has added a feature that allows ordering by any child or by value. So there are now four ways to order data: by key, by value, by priority, or by the value of any named child. See this blog post that introduces the new ordering capabilities.
The basic approaches remain the same though:
1. Add a child property with the inverted timestamp and then order on that.
2. Read the children in ascending order and then invert them on the client.
Firebase supports retrieving child nodes of a collection in two ways:
by name
by priority
What you're getting now is by name, which happens to be chronological. That's no coincidence btw: when you push an item into a collection, the name is generated to ensure the children are ordered in this way. To quote the Firebase documentation for push:
The unique name generated by push() is prefixed with a client-generated timestamp so that the resulting list will be chronologically-sorted.
The Firebase guide on ordered data has this to say on the topic:
How Data is Ordered
By default, children at a Firebase node are sorted lexicographically by name. Using push() can generate child names that naturally sort chronologically, but many applications require their data to be sorted in other ways. Firebase lets developers specify the ordering of items in a list by specifying a custom priority for each item.
The simplest way to get the behavior you want is to also specify an always-decreasing priority when you add the item:
var ref = new Firebase('https://your.firebaseio.com/sell');
var item = ref.push();
item.setWithPriority(yourObject, 0 - Date.now());
Update
You'll also have to retrieve the children differently:
fbl.child('sell').startAt().limitToLast(20).on('child_added', function(fbdata) {
console.log(fbdata.exportVal());
})
In my test using on('child_added' ensures that the last few children added are returned in reverse chronological order. Using on('value' on the other hand, returns them in the order of their name.
Be sure to read the section "Reading ordered data", which explains the usage of the child_* events to retrieve (ordered) children.
A bin to demonstrate this: http://jsbin.com/nonawe/3/watch?js,console
Since firebase 2.0.x you can use limitLast() to achieve that:
fbl.child('sell').orderByValue().limitLast(20).on("value", function(fbdataSnapshot) {
// fbdataSnapshot is returned in the ascending order
// you will still need to order these 20 items in
// in a descending order
}
Here's a link to the announcement: More querying capabilities in Firebase
To augment Frank's answer, it's also possible to grab the most recent records--even if you haven't bothered to order them using priorities--by simply using endAt().limit(x) like this demo:
var fb = new Firebase(URL);
// listen for all changes and update
fb.endAt().limit(100).on('value', update);
// print the output of our array
function update(snap) {
var list = [];
snap.forEach(function(ss) {
var data = ss.val();
data['.priority'] = ss.getPriority();
data['.name'] = ss.name();
list.unshift(data);
});
// print/process the results...
}
Note that this is quite performant even up to perhaps a thousand records (assuming the payloads are small). For more robust usages, Frank's answer is authoritative and much more scalable.
This brute force can also be optimized to work with bigger data or more records by doing things like monitoring child_added/child_removed/child_moved events in lieu of value, and using a debounce to apply DOM updates in bulk instead of individually.
DOM updates, naturally, are a stinker regardless of the approach, once you get into the hundreds of elements, so the debounce approach (or a React.js solution, which is essentially an uber debounce) is a great tool to have.
There is really no way but seems we have the recyclerview we can have this
query=mCommentsReference.orderByChild("date_added");
query.keepSynced(true);
// Initialize Views
mRecyclerView = (RecyclerView) view.findViewById(R.id.recyclerView);
mManager = new LinearLayoutManager(getContext());
// mManager.setReverseLayout(false);
mManager.setReverseLayout(true);
mManager.setStackFromEnd(true);
mRecyclerView.setHasFixedSize(true);
mRecyclerView.setLayoutManager(mManager);
I have a date variable (long) and wanted to keep the newest items on top of the list. So what I did was:
Add a new long field 'dateInverse'
Add a new method called 'getDateInverse', which just returns: Long.MAX_VALUE - date;
Create my query with: .orderByChild("dateInverse")
Presto! :p
You are searching limitTolast(Int x) .This will give you the last "x" higher elements of your database (they are in ascending order) but they are the "x" higher elements
if you got in your database {10,300,150,240,2,24,220}
this method:
myFirebaseRef.orderByChild("highScore").limitToLast(4)
will retrive you : {150,220,240,300}
In Android there is a way to actually reverse the data in an Arraylist of objects through the Adapter. In my case I could not use the LayoutManager to reverse the results in descending order since I was using a horizontal Recyclerview to display the data. Setting the following parameters to the recyclerview messed up my UI experience:
llManager.setReverseLayout(true);
llManager.setStackFromEnd(true);
The only working way I found around this was through the BindViewHolder method of the RecyclerView adapter:
#Override
public void onBindViewHolder(final RecyclerView.ViewHolder holder, int position) {
final SuperPost superPost = superList.get(getItemCount() - position - 1);
}
Hope this answer will help all the devs out there who are struggling with this issue in Firebase.
Firebase: How to display a thread of items in reverse order with a limit for each request and an indicator for a "load more" button.
This will get the last 10 items of the list
FBRef.child("childName")
.limitToLast(loadMoreLimit) // loadMoreLimit = 10 for example
This will get the last 10 items. Grab the id of the last record in the list and save for the load more functionality. Next, convert the collection of objects into and an array and do a list.reverse().
LOAD MORE Functionality: The next call will do two things, it will get the next sequence of list items based on the reference id from the first request and give you an indicator if you need to display the "load more" button.
this.FBRef
.child("childName")
.endAt(null, lastThreadId) // Get this from the previous step
.limitToLast(loadMoreLimit+2)
You will need to strip the first and last item of this object collection. The first item is the reference to get this list. The last item is an indicator for the show more button.
I have a bunch of other logic that will keep everything clean. You will need to add this code only for the load more functionality.
list = snapObjectAsArray; // The list is an array from snapObject
lastItemId = key; // get the first key of the list
if (list.length < loadMoreLimit+1) {
lastItemId = false;
}
if (list.length > loadMoreLimit+1) {
list.pop();
}
if (list.length > loadMoreLimit) {
list.shift();
}
// Return the list.reverse() and lastItemId
// If lastItemId is an ID, it will be used for the next reference and a flag to show the "load more" button.
}
I'm using ReactFire for easy Firebase integration.
Basically, it helps me storing the datas into the component state, as an array. Then, all I have to use is the reverse() function (read more)
Here is how I achieve this :
import React, { Component, PropTypes } from 'react';
import ReactMixin from 'react-mixin';
import ReactFireMixin from 'reactfire';
import Firebase from '../../../utils/firebaseUtils'; // Firebase.initializeApp(config);
#ReactMixin.decorate(ReactFireMixin)
export default class Add extends Component {
constructor(args) {
super(args);
this.state = {
articles: []
};
}
componentWillMount() {
let ref = Firebase.database().ref('articles').orderByChild('insertDate').limitToLast(10);
this.bindAsArray(ref, 'articles'); // bind retrieved data to this.state.articles
}
render() {
return (
<div>
{
this.state.articles.reverse().map(function(article) {
return <div>{article.title}</div>
})
}
</div>
);
}
}
There is a better way. You should order by negative server timestamp. How to get negative server timestamp even offline? There is an hidden field which helps. Related snippet from documentation:
var offsetRef = new Firebase("https://<YOUR-FIREBASE-APP>.firebaseio.com/.info/serverTimeOffset");
offsetRef.on("value", function(snap) {
var offset = snap.val();
var estimatedServerTimeMs = new Date().getTime() + offset;
});
To add to Dave Vávra's answer, I use a negative timestamp as my sort_key like so
Setting
const timestamp = new Date().getTime();
const data = {
name: 'John Doe',
city: 'New York',
sort_key: timestamp * -1 // Gets the negative value of the timestamp
}
Getting
const ref = firebase.database().ref('business-images').child(id);
const query = ref.orderByChild('sort_key');
return $firebaseArray(query); // AngularFire function
This fetches all objects from newest to oldest. You can also $indexOn the sortKey to make it run even faster
I had this problem too, I found a very simple solution to this that doesn't involved manipulating the data in anyway. If you are rending the result to the DOM, in a list of some sort. You can use flexbox and setup a class to reverse the elements in their container.
.reverse {
display: flex;
flex-direction: column-reverse;
}
myarray.reverse(); or this.myitems = items.map(item => item).reverse();
I did this by prepend.
query.orderByChild('sell').limitToLast(4).on("value", function(snapshot){
snapshot.forEach(function (childSnapshot) {
// PREPEND
});
});
Someone has pointed out that there are 2 ways to do this:
Manipulate the data client-side
Make a query that will order the data
The easiest way that I have found to do this is to use option 1, but through a LinkedList. I just append each of the objects to the front of the stack. It is flexible enough to still allow the list to be used in a ListView or RecyclerView. This way even though they come in order oldest to newest, you can still view, or retrieve, newest to oldest.
You can add a column named orderColumn where you save time as
Long refrenceTime = "large future time";
Long currentTime = "currentTime";
Long order = refrenceTime - currentTime;
now save Long order in column named orderColumn and when you retrieve data
as orderBy(orderColumn) you will get what you need.
just use reverse() on the array , suppose if you are storing the values to an array items[] then do a this.items.reverse()
ref.subscribe(snapshots => {
this.loading.dismiss();
this.items = [];
snapshots.forEach(snapshot => {
this.items.push(snapshot);
});
**this.items.reverse();**
},
For me it was limitToLast that worked. I also found out that limitLast is NOT a function:)
const query = messagesRef.orderBy('createdAt', 'asc').limitToLast(25);
The above is what worked for me.
PRINT in reverse order
Let's think outside the box... If your information will be printed directly into user's screen (without any content that needs to be modified in a consecutive order, like a sum or something), simply print from bottom to top.
So, instead of inserting each new block of content to the end of the print space (A += B), add that block to the beginning (A = B+A).
If you'll include the elements as a consecutive ordered list, the DOM can put the numbers for you if you insert each element as a List Item (<li>) inside an Ordered Lists (<ol>).
This way you save space from your database, avoiding unnecesary reversed data.