I want to sort a collection by name column. Using sortBy() method is not working.
$collection = \App\Http\Resources\MyResource::collection($test);
return $collection->sortBy('name');
For me only worked example exactly as in the docs
$statisticCollection = collect($statistics);
$sorted = $statisticCollection->sortByDesc('date');
return $sorted->values()->all();
So the problem was that I did not save the sorted result to a variable before returning it.
I found that I had to use values() after sorting, as suggested in the documentation.
The thing you want to do is:
$collection = \App\Http\Resources\MyResource::collection($test);
return $collection->sortBy('name')->all();
Check how it is done in the docs.
The default sortBy for Laravel is defined as ascending:
public function sortBy($callback, $options = SORT_REGULAR, $descending = false)
If you were looking to sort descending use sortByDesc:
public function sortByDesc($callback, $options = SORT_REGULAR)
{
return $this->sortBy($callback, $options, true);
}
Where the first parameter $callback can be a string or a callable, so your code could look something like this with items sorted descending:
$collection = \App\Http\Resources\MyResource::collection($test);
return $collection->sortByDesc('name')->all();
If the Laravel default sorting doesn't work for your purpose, try to do this:
$collection = \App\Http\Resources\MyResource::collection->sortBy(function ($element, $key) {
return yourFunctionForSorting($element['name']);
});
Replace obviously "yourFunctionForSorting", with your sorting criteria.
Related
I have a big and difficult SQL query. It works fine for me. And I have the following code in the controller:
public function index(OpenRegDepDataReportInterface $openRegDepDataReport, Request $request): Renderable
{
$applications = $openRegDepDataReport->getReport($advertisers, $category);
return view('applications.index', compact('applications'));
}
So, the method getReport gives me a result of DB::select('<here is my big diffecult SQL>'), and, as well known it's an array.
But as you can see I'm trying to pass the result on a view. And, of course, I would like to call $applications->links() in the view, like for eloquent collection. Which is proper and faster way to do that?
doc
To display pagination at the table, you must call the select and then the pagination method.
in Controller:
$test = DB::table('users')->select('id')->paginate(10);
in View:
$test->links();
So based on the documentation if your $applications returns a Query Builder result, then just append ->paginate(10); to it.
https://laravel.com/docs/master/pagination#paginating-query-builder-results
$applications = $openRegDepDataReport->getReport($advertisers, $category)->paginate(10);
Simple answer, use paginate() method:
$basicQuery = DB::select(DB::raw("<here is the big diffcult SQL query>"));
However, paginate() works only on collections, and since you have an array of objects, you need to turn it into a collection first, using the forPage() method:
The forPage method returns a new collection containing the items that would be present on a given page number. The method accepts the page number as its first argument and the number of items to show per page as its second argument:
$collection = collect($basicQuery);
$chunk = $collection->forPage(2, 3);
$chunk->all();
Complicated answer: build a paginator instance yourself:
$perPage = 10;
$page = $request->input("page", 1);
$skip = $page * $perPage;
if($take < 1) { $take = 1; }
if($skip < 0) { $skip = 0; }
$basicQuery = DB::select(DB::raw("<here is the big diffcult SQL query>"));
$totalCount = $basicQuery->count();
$results = $basicQuery
->take($perPage)
->skip($skip)
->get();
$paginator = new \Illuminate\Pagination\LengthAwarePaginator($results, $totalCount, $take, $page);
return $paginator;
I would recommend using the paginate() method.
You can read more about Laravel's pagination.
I have the following query.
$projects = Project::orderBy('created_at', 'desc');
$data['sorted'] = $projects->groupBy(function ($project) {
return Carbon::parse($project->created_at)->format('Y-m-d');
})->simplePaginate(5);
When I try to paginate with the simplePaginate() method I get this error.
stripos() expects parameter 1 to be string, object given
How can I paginate grouped data in this case?
The created_at attribute is already casted as a Carbon Object (by default in laravel models). that's why you are getting that error. Try this:
$projects = Project::orderBy('created_at', 'desc')->get();
$data['sorted'] = $projects->groupBy(function ($project) {
return $project->created_at->format('Y-m-d');
})->simplePaginate(5);
this answer is just for the error you're getting. now if you want help with the QueryBuilder, can you provide an example of the results you're expecting to have and an example of the database structure ?
The pagination methods should be called on queries instead of collection.
You could try:
$projects = Project::orderBy('created_at', 'desc');
$data['sorted'] = $projects->groupBy('created_at');
The problem was solved. I was create custom paginator via this example:
https://stackoverflow.com/a/30014621/6405083
$page = $request->has('page') ? $request->input('page') : 1; // Use ?page=x if given, otherwise start at 1
$numPerPage = 15; // Number of results per page
$count = Project::count(); // Get the total number of entries you'll be paging through
// Get the actual items
$projects = Project::orderBy('created_at', 'desc')
->take($numPerPage)->offset(($page-1)*$numPerPage)->get()->groupBy(function($project) {
return $project->created_at->format('Y-m-d');
});
$data['sorted'] = new Paginator($projects, $count, $numPerPage, $page, ['path' => $request->url(), 'query' => $request->query()]);
simplePaginate Method is exist in the path below:
Illuminate\Database\Eloquent\Builder.php::simplePaginate()
I need to paginate the results and sort them using sortBy(), without losing the pagination links. I also need to use a resource to return the results.
$sorted = Model::paginate(10)->sortBy('name');
$results = \App\Http\Resources\MyResource::collection($sorted);
Doing this breaks the pagination links (I get only the data part).
$paginated = Model::paginate(10);
$results = \App\Http\Resources\MyResource::collection($paginated);
return $results->sortBy('name');
This also doesn't work.
Any ideas?
Using the getCollection() and setCollection() method in the paginator class, You can update the pagination result without losing the meta data.
$result = Post::orderBy('another_key')->paginate();
$sortedResult = $result->getCollection()->sortBy('key_name')->values();
$result->setCollection($sortedResult);
return $result;
I think you can sort the results first, and then paginate
$sorted = Model::orderBy('name')->paginate(10);
I've solved the problem with this solution:
$sorted = Model::get()
->sortBy('example_function') //appended attribute
->pluck('id')
->toArray();
$orderedIds = implode(',', $sorted);
$result = DB::table('model')
->orderByRaw(\DB::raw("FIELD(id, ".$orderedIds." )"))
->paginate(10);
I've appended example_function attribute to the model, to be used by sortBy. With this solution I was able to use orderBy to order by an appended attribute of the model. And also I was able to use pagination.
I need to get only the roomnumber arrays returned from the following query:
$roomnumbers = Room::with(['floorroomcount' => function($query){
$query->with('roomnumber')->get();
}])->where('roomtype_id', $roomtype_id)->get();
Tried:
The follow pluck is returning floorroomcount
$roomnumbers->pluck('floorroomcount');
but i need roomnumber array, how can i get?
This gives you all roomnumber results in one collection:
$roomnumbers->pluck('floorroomcount')->collapse()->pluck('roomnumber')->collapse();
You may shorten #Jonas Staudenmeir's answer like so:
$roomnumbers->pluck('floorroomcount.*.roomnumber.*')->collapse();
pluck('*') is essentially the same as collapse() in this particular context.
This is working, but with many loop and echoing directly, if anything can be simplified please let me know :
$roomnumbers = Room::with(['floorroomcount.roomnumber'])->where('roomtype_id', $roomtype_id)->get();
$floorroomcounts = $roomnumbers->pluck('floorroomcount');
$records = $floorroomcounts->map(function($floorroomcount, $value){
return $floorroomcount->pluck('roomnumber')->flatten();
})->values()->all();
foreach($records as $record){
foreach($record as $row){
echo '<option value='.$row->id.'>'.$row->roomnumber.'</option>';
}
}
//return response()->json($roomnumbers);
Try,
$roomnumbers = Room::with(['floorroomcount' => function($query){
$query->with('roomnumber')->get();
}])
->where('roomtype_id', $roomtype_id)
->get();
$records = $roomnumbers->map(function($element, $value){
return $element->map(function($e, $v){
return $e->roomnumber;
});
})->values()->all();
map() is a Laravel collection method so you need to import the collection facade on the top of the controller like: use Illuminate\Support\Collection;
In Laravel 5.1 and + you can use flatten() on collection.
method flattens a multi-dimensional collection into a single dimension:
$roomnumbers->flatten()->pluck('floorroomcount');
I am trying to filter a paginated eloquent collection, but whenever I use any of the collection methods, I lose the pagination.
$models = User::orderBy('first_name','asc')->paginate(20);
$models = $models->each(function($model) use ($filters) {
if(!is_null($filters['type'])) {
if($model->type == $filters['type'])
return $model;
}
if(!is_null($filters['state_id'])) {
if($model->profile->state_id == $filters['state_id'])
return $model;
}
if(!is_null($filters['city_id'])) {
if($model->profile->city_id == $filters['city_id'])
return $model;
}
});
return $models;
I am working with Laravel 4.2, is there any way to persist the pagination?
An answer to the titular question, which is possible in Laravel 5.2+:
How to filter the underlying collection of a Paginator without losing the Paginator object?
You can eject, modify and inject the collection as follows:
$someFilter = 5;
$collection = $paginator->getCollection();
$filteredCollection = $collection->filter(function($model) use ($someFilter) {
return $model->id == $someFilter;
});
$paginator->setCollection($filteredCollection);
The Paginator is built on an underlying collection, but indeed when you use any of the inherited Collection methods they return the underlying collection and not the full Paginator object: collection methods return collections for chaining together collection calls.
Expanding on mininoz's answer with your specific case:
//Start with creating your object, which will be used to query the database
$queryUser = User::query();
//Add sorting
$queryUser->orderBy('first_name','asc');
//Add Conditions
if(!is_null($filters['type'])) {
$queryUser->where('type','=',$filters['type']);
}
if(!is_null($filters['state_id'])) {
$queryUser->whereHas('profile',function($q) use ($filters){
return $q->where('state_id','=',$filters['state_id']);
});
}
if(!is_null($filters['city_id'])) {
$queryUser->whereHas('profile',function($q) use ($filters){
return $q->where('city_id','=',$filters['city_id']);
});
}
//Fetch list of results
$result = $queryUser->paginate(20);
By applying the proper conditions to your SQL query, you are limiting the amount of information that comes back to your PHP script, and hence speeding up the process.
Source: http://laravel.com/docs/4.2/eloquent#querying-relations
I have a scenario that requires that I filter on the collection, I cannot rely on the Query Builder.
My solution was to instantiate my own Paginator instance:
$records = Model::where(...)->get()->filter(...);
$page = Paginator::resolveCurrentPage() ?: 1;
$perPage = 30;
$records = new LengthAwarePaginator(
$records->forPage($page, $perPage), $records->count(), $perPage, $page, ['path' => Paginator::resolveCurrentPath()]
);
return view('index', compact('records'));
Then in my blade template:
{{ $records->links() }}
paginate() is function of Builder. If you already have Collection object then it does not have the paginate() function thus you cannot have it back easily.
One way to resolve is to have different builder where you build query so you do not need to filter it later. Eloquent query builder is quite powerful, maybe you can pull it off.
Other option is to build your own custom paginator yourself.
You can do some query on your model before do paginate.
I would like to give you some idea. I will get all users by type, sort them and do paginate at the end. The code will look like this.
$users = User::where('type', $filters['type'])->orderBy('first_name','asc')->paginate(20);
source: http://laravel.com/docs/4.2/pagination#usage
This was suitable for me;
$users = User::where('type', $filters['type'])->orderBy('first_name','asc')->paginate(20);
if($users->count() < 1){
return redirec($users->url(1));
}
A workaround when mixing search parameters and pagination, since default pagination won't keep the search parameters, using GET:
$urlSinPaginado = url()->full();
$pos=strrpos(url()->full(), 'page=');
if ($pos) {
$urlSinPaginado = substr(url()->full(), 0, $pos-1);
}
[...]
->paginate(5)
->withPath($urlSinPaginado);
sample generated pagination link: http://myhost/context/list?filtro_1=5&filtro_2=&filtro_3=&filtro_4=&filtro_5=&filtro_6&page=8
You can simply use this format in views
{!! str_replace('/?', '?', $data->appends(Input::except('page'))->render()) !!}
For newer versions of Laravel, you can use this:
$data->paginate(15)->withQueryString();