I would like to store information such as rowid (key) and orderId (value) in MutableMap. Then iterate through the MutableMap and get the current item key and value and the next item key and value.
The reason for that is that while I am iterating through the Map I am changing the value of the first item using the second item value or I am changing the value of the second item value using the first item value.
Example:
val mutableMap: MutableMap<Int, Int> = mutableMapOf<Int, Int>()
mutableMap[350] = 0
mutableMap[351] = 1
mutableMap[352] = 2
mutableMap[353] = 3
mutableMap.put(354, 4)
mutableMap.put(355, 5)
mutableMap.put(356, 6)
for (key in mutableMap.keys) {
Log.d("print data", "print data, Key = ${key}, Value = ${mutableMap[key]}")
}
How to get this item: myLinkedHashMap[450] = 0
and this item myLinkedHashMap[453] = 3 information at the same time so I can swap the value between them ?
Although the Map and MutableMap interfaces do not claim to be ordered, the specific implementation used by mutableMapOf is ordered anyway. This is guaranteed because it is declared in the documentation of the function. So if you are working with a MutableMap that you declared locally with mutableMapOf, then you know it is ordered. If you are working with a MutableMap that was passed to you from some other class, then you cannot be sure it is ordered. But that should be a pretty rare occurrence since it is poor encapsulation to pass mutable collections between classes.
To iterate a map that is ordered in pairs, you can use zipWithNext on the keys. This call creates a copy of the list of keys, so you can safely mutate the map inside the loop. This is like a moving window of pairs--each element besides the first and last will appear in two different pairs, once as the second element and then once as the first element.
for ((keyA, keyB) in mutableMap.keys.zipWithNext()) {
}
If you want to iterate exclusive pairs, you can use chunked(2) instead of zipWithNext(), but you would need to be sure you only chunk an even number of elements:
for ((keyA, keyB) in mutableMap.keys.take(mutableMap.size - mutableMap.size % 2).chunked(2)) {
}
// or
for (keys in mutableMap.keys.chunked(2)) {
if (keys.size == 1) continue
val (keyA, keyB) = keys
}
I'm deserializing an entire document, to keep things in handy and prevent me to check for each value and construct the object I use
myList.add(documentSnapshot.toObject(House::class.java))
Now, lets say House is this
data class House(val name:String,val:address:String)
Now, if I want to also get the House document Id and put it inside my document I do this
data class House(val houseId:String,val name:String,val:address:String)
But after doing that , the first line of code transforms into this
val houseId = documentSnapshot.id
val houseName = docuementSnapshot.getString("name")
val houseAddress = documentSnapshot.getString("address")
myList.add(House(houseId,houseName,houseAddress))
What I want to do is use .toObject() to also map that extra field that is the document id inside of it because if the House object expands in size, I will need to hand write again each property, and now think that house has 100 properties and I just need the id of it inside the object. I will need to write off 99 get fields to just place the document Id inside that house object.
Is there a way to map that id to the object without doing the above and just placing .toObject ?
Thanks
You need just add annotation #DocumentId
data class House(#DocumentId val houseId:String,val name:String,val:address:String)
What I want to do is use .toObject() to also map that extra field that is the document id
This will be possible only if the document already contains the id property that holds the document id as a value. If you only have the name and the address, then you cannot map the id, because it doesn't exist in the document. To be able to map all those three properties you should update each and every document in that collection so it contains the document id. If you have lots of documents in the collection, I recommend you use batch writes.
Right after that, you'll be able to use:
val house = documentSnapshot.toObject(House::class.java)
And now this house object will contain the id too.
I am creating an app where users can upload photos and tag them, and then tags that they have uses are added to the list of that user's interests.
I am making a transition from Realtime database to cloud firestore and having a bit of an issue.
I want to store all these tags in an array inside a document for each user. Whenever a user uploads a photo, these tags should be added to this array.
I am using the update method as I don't want the list to be overwritten, but the data doesn't seem to be created at all. Could it be because for a new user the list doesn't actually exist?
I need the function to create the document and array if it is the first time the user adds any interests, or just update the list and add items if the array already exists.
Also, the tags are added to a complete list of all the tags across the app. For this one I've tried to use the set method but then the list just gets completely overwritten, even when I added SetOptions.merge()
val db = FirebaseFirestore.getInstance()
val userInterestsDoc = db.collection("interests").document(currentUser.uid)
val allTagsListDoc = db.collection("all_tags").document("all_tags")
for (tag in imageTagsList) {
userInterestsDoc.update("interests_list", FieldValue.arrayUnion(tag))
.addOnSuccessListener {
allTagsListDoc.set(
mapOf(
"all_tags_list" to arrayListOf(tag)
), SetOptions.merge()
)
}
Is it possible to get the values in a map associated to a key, without using the key?
I have this code.
val map = prices.associateBy({it.productName},{it.productPrice})
which gives me this
{Coffee=3.0, Gum=0.5, Beer=18.0}
I want to be able to just grab all the prices (3.0, 0.5, etc.) and save them to a list.
Any help is appreciated!
So to get all the values from a map, you can just use the built-in values property on the map like this.
val map = prices.associateBy({it.productName},{it.productPrice})
val values = map.values
It returns a read-only Collection of all values in this map. Note that this collection may contain duplicate values.
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.