I wrote the following code.
const salons = query(collection(db, "user"), orderBy('', "desc"), limit(5))
const querySnapshot = await getDocs(salons)
await querySnapshot.forEach((doc) => {
console.log(new Date(doc._document.version.timestamp.seconds * 1000).toString())
//processing
})
I want to sort the orderby by "doc._document.version.timestamp.seconds" which is displayed in console.log.
But I don't know how to do it.
This date is automatically registered by firestore, so the position of the value is different.
Does anyone know how I can sort it?
Firestore can only order/filter data on values that it has indexes for. Indexes are only created for fields in your document (and the document ID), not for implicit metadata such as the timestamp it keeps internally.
There is no way to order Firestore results based on the internal timestamp. If you want to be able to order documents on a timestamp, you'll have to store that timestamp as a field in the document, and then pass that field name to orderBy.
Let's consider a social network to be built by NodeJS and MongoDB.
So, if a user creates a new post, it should be saved to his/her followers feed.
The straightforward implementation of this operation as follow (simplified):
var newPost="New Post";
//get list of followers of user 1
var listOfFollowers = followersCollection.find({u:"1"});
for(var i=0;i<listOfFollowers.length;i++){
var followerId = listOfFollowers[i]._id;
//insert new post of user 1 to every follower feed
feedsCollection.insertOne(
{ownerId:followerId,authorId:"1",content:newPost}
);
}
This, of course, has very bad performance in case of big numbers in followers count. How can do this with a single fast performing MongoDB query?
MongoDB provides bulk document insert functionality, check out this link - https://docs.mongodb.com/manual/reference/method/db.collection.initializeUnorderedBulkOp/
db.collection.initializeUnorderedBulkOp() It creates an unordered list of operations and mongodb executes this list in parallel, so it's fast and you don't have to take extra care of performance as mongo handles it.
For ordered insertions, you can use db.collection.initializeOrderedBulkOp().
e.g.
var newPost="New Post";
var bulk = db.followersCollection.initializeUnorderedBulkOp();
//get list of followers of user 1
var listOfFollowers = followersCollection.find({u:"1"});
for(var i=0;i<listOfFollowers.length;i++){
var followerId = listOfFollowers[i]._id;
//insert new post of user 1 to every follower feed
bulk.insert( {ownerId:followerId,authorId:"1",content:newPost});
}
bulk.execute();
If you are using Mongoose then checkout Mongoose docs for the same. In the above example, I have just trying to explain how you can do it using plain MongoDB.
Insert Many Read this document I think you will get the answer
Check this:
var newPost="New Post";
//Object Array
var collection = []
//get list of followers of user 1
var listOfFollowers = followersCollection.find({u:"1"});
for(var i=0;i<listOfFollowers.length;i++){
var followerId = listOfFollowers[i]._id;
collection.push({ownerId:followerId,authorId:"1",content:newPost})
}
feedsCollection.insert(collection); //Or use insertMany()
You can create your object array and insert it at once.
Check documentation :- https://docs.mongodb.com/manual/reference/method/db.collection.insert/#insert-multiple-documents
Even though this is a simple answer for your question, If the collection array has a large number of elements, there still might be performance issues. So the best way to handle this is using triggers. https://docs.mongodb.com/stitch/triggers/
At the moment, Firebase doesn't stores the data in the order it is entered. For ex - There could be 2 existing childs and a new child is inserted, that new child goes and sits in the middle of the already existing ones! I guess this works by alphabet order but I want to store data in the correct order only.
firebase.database().ref('someChildRef').set({
//Could insert anywhere
});
I think push could insert it in the correct order but push also generates a unique key which I don't want, I want to insert data only using set and in the correct order.
It is not possible to insert data into Firebase into a desired position. Firebase data is sorted alphabetically and all newly inserted data will be rearranged accordingly.
Use query rules to return the data in the order you want.
Such as:
orderByChild()
orderByKey()
orderByValue()
equalTo()
You are correct about push(), it will always create a unique key and add to the end, just as pushing to an array will add to the end.
It sounds like you're looking for Firebase's push() method, which generates a unique key based on the timestamp:
firebase.database().ref('someChildRef').set({
//Will result at "the end"
});
var messageListRef = new Firebase('https://samplechat.firebaseio-demo.com/message_list');
var newMessageRef = messageListRef.push();
newMessageRef.set({ 'user_id': 'fred', 'text': 'Yabba Dabba Doo!' });
// We've appended a new message to the message_list location.
var path = newMessageRef.toString();
Try using the push() method instead of set. According to their docs:
The unique name generated by push() is prefixed with a client-generated timestamp so that the resulting list will be chronologically-sorted.
Reference: https://www.firebase.com/docs/web/api/firebase/push.html
Actually i am using mongoDB, and i am able to update document with single ObjectID, so right now i want to perform update on multiple documents having different ObjectID, i did considerable search, came to know {multi:true} will help to update multiple documents.
Let's say i have
var uid = {'Userid': ObjectId(..)};
var id = {'Newid': Object(..)};
now i want to perform update on both of these different documents having different ObjectId' s total separatley, so far i tried this considering "xyz" and "abc" array's in Schema just for demo
var update_uid = {$push:{'xyz': some_id}};
var update_id = {$push:{'abc': some_other_id}};
Test.update(uid,id,update_uid,update_id,{multi:true},function(){..});
this is wrong i know, i mentioned it for the records just in case, Any help and suggestion would be much appreciated
If you have multiple ids that you want to match you can use the keywork $in which checks if there's any matches in the array. Now that you also have multiple properties that you want to check you can user the $or keyword, which will find a match if either of the objects in the array matches a document in the database.
var userIds = [{'Userid':ObjectId(..)}, {'Userid':ObjectId(..)}];
var newIds = [{'Newid': Object(..)}, {'Newid': Object(..)}];
var query = {
$or: [{$in: userIds}, {$in: newIds}]
}
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.