test("Comparison") {
val list: List[String] = List("Thing", "Entity", "Variable")
val expected: List[String] = List("Thingg", "Entityy", "Variablee")
var expectedPosition = 0
for (item <- list) {
assert(list(expectedPosition) == expected(expectedPosition))
expectedPosition += 1
}
}
In Scala, in order to make my low-level code tests more readable, I thought it would be a good idea to just use one assertion and have it iterate through a loop and increase an accumulator at the end.
This is to test more than one input at a time when the multiple inputs would more or less have similar attributes. When the assertion fails, it comes back as "Thing[]" did not equal "Thing[g]". Instead of it reporting the item in the list that failed, is there a way to get it to directly state the list position without concatenating the list position before the assertion or using a conditional statement that returns the list position? Like I would rather keep it all contained within the assert() error report.
val expected: LazyList[String] = ...
expected.zip(list)
.zipWithIndex
.foreach{case ((exp,itm),idx) =>
assert(exp == itm, s"off at index $idx")
}
//java.lang.AssertionError: assertion failed: off at index 0
expected is lazy so that it is traversed only once even though it gets zipped twice.
I am making a search request on the List with the Provider pattern.
List<Device> _devices = [
Device(one: 'Apple', two: 'iphone'),
Device(one: 'Samsung', two: 'Galaxy')
];
And Query is like this
List<Device> queryQuery(String value) {
return _devices
.where((device) => device.one.toLowerCase().contains(value.toLowerCase()))
.toList();
the result I expect to get is iphone when I passed the value Apple.
But the result on the screen that I got is [instance of ‘Device’]
when I code like this
child: Text('${deviceData.getDevice('Apple')}'
I do know I should be using some kind of key using two... but I have no idea :-(
You serialized the wrong object.
What you did end-up being similar to:
Text(Device(one: 'Apple', two: 'iphone').toString());
But you don't want to do Device.toString(). What you want instead is to pass Device.two to your Text.
As such your end result is:
Text('${chordData.chordExpand('Apple').two}')
By the look of [Instance of 'Device'], it seems the function is returning a list so it is a good idea to check if the list is empty or not. if it is not empty, one of the elements is still needed to be selected. I guess it should be Text('${chordData.chordExpand('Apple')[0].two}') in case the list is not empty.
To summarize, use something like this to handle the case when list is empty
// Inside your build method before returning the widget
var l = chordData.chordExpand('Apple'); // Returns a list of devices
String textToWrite; // Here we will store the text that needs to be written
if(l.isEmpty) textToWrite = 'No results'; // If the filter resulted in an empty list
else textToWrite = l[0].two; // l[0] is an instance of a device which has a property called two. You can select any instance from the list provided it exists
return <Your Widget>(
.....
Text(textToWrite),
.....
);
Scala is pretty new for me and I have problems as soon as a leave the gatling dsl.
In my case I call an API (Mailhog) which responds with a lot of mails in json-format. I can’t grab all the values.
I need it with “jsonPath” and I need to “regex” as well.
That leads into a map and a list which I need to iterate through and save each value.
.check(jsonPath("$[*]").ofType[Map[String,Any]].findAll.saveAs("id_map"))
.check(regex("href=3D\\\\\"(.*?)\\\\\"").findAll.saveAs("url_list"))
At first I wanted to loop the “checks” but I did’nt find any to repeat them without repeating the “get”-request too. So it’s a map and a list.
1) I need every value of the map and was able to solve the problem with the following foreach loop.
.foreach("${id_map}", "idx") {
exec(session => {
val idMap = session("idx").as[Map[String,Any]]
val ID = idMap("ID")
session.set("ID", ID)
})
.exec(http("Test")
.get("/{ID}"))
})}
2) I need every 3rd value of the list and make a get-request on them. Before I can do this, I need to replace a part of the string. I tried to replace parts of the string while checking for them. But it won’t work with findAll.
.check(regex("href=3D\\\\\"(.*?)\\\\\"").findAll.transform(raw => raw.replace("""=\r\n""","")).saveAs("url"))
How can I replace a part of every string in my list?
also how can I make a get-request on every 3rd element in the list.
I can't get it to work with the same foreach structure above.
I was abole to solve the problem by myself. At first I made a little change to my check(regex ...) part.
.check(regex("href=3D\\\\\"(.*?)\\\\\"").findAll.transform(_.map(raw => raw.replace("""=\r\n""",""))).saveAs("url_list"))
Then I wanted to make a Get-Request only on every third element of my list (because the URLs I extracted appeared three times per Mail).
.exec(session => {
val url_list =
session("url_list").as[List[Any]].grouped(3).map(_.head).toList
session.set("url_list", url_list)
})
At the end I iterate through my final list with a foreach-loop.
foreach("${url_list}", "urls") {
exec(http("Activate User")
.get("${urls}")
)
}
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.
I have a payload which is a lists of type bytes:
var payload : list of byte;
payload= {1;2;3;4;5;6};
var item1 :list of byte;
item = {3;4;5};
var item2 :list of byte;
item = {1;4};
I would like to implement a code that checks if a list is a sub-list of another. Using "if ..in.." doesn't quite work as it does not take into account the order of the items or if they appear successively or not. I want something that does the following:
if (item1 in payload) ...... should return TRUE. Items exist in payload in the same order.
if (item2 in payload) ...... should return FALSE because although each element in the list exists in the payload, but the item2 elements do not appear successively in the payload list.
Is there an easy way to achieve this? There must be a build -in function in specman for this.
Thanks
The following code should work:
if (item.size()==0) {return TRUE};
for i from 0 to payload.size()-item.size() {
if (item == payload[i..i+item.size()-1]) {
return TRUE;
};
};
return FALSE;
Note that this code is quite expensive memory-wise (the list[a..b] syntax creates a new list every time) so if you have memory considerations it should be modified.