How I can paginate DB::select(...) result and get links() method? - laravel

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.

Related

Laravel simplePaginate() for Grouped Data

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()

Laravel Eloquent limit and offset

This is mine
$art = Article::where('id',$article)->firstOrFail();
$products = $art->products;
I just wanna take a limit 'product'
This is wrong way
$products = $art->products->offset($offset*$limit)->take($limit)->get();
Please give me a hand!
Thanks!
skip = OFFSET
$products = $art->products->skip(0)->take(10)->get(); //get first 10 rows
$products = $art->products->skip(10)->take(10)->get(); //get next 10 rows
From laravel doc 5.2 https://laravel.com/docs/5.2/queries#ordering-grouping-limit-and-offset
skip / take
To limit the number of results returned from the query, or to skip a
given number of results in the query (OFFSET), you may use the skip
and take methods:
$users = DB::table('users')->skip(10)->take(5)->get();
In laravel 5.3 you can write (https://laravel.com/docs/5.3/queries#ordering-grouping-limit-and-offset)
$products = $art->products->offset(0)->limit(10)->get();
Quick:
Laravel has a fast pagination method, paginate, which only needs to pass in the number of data displayed per page.
//use the paginate
Book::orderBy('updated_at', 'desc')->paginate(8);
how to customize paging:
you can use these method: offset,limit ,skip,take
offset,limit : where does the offset setting start, limiting the amount of data to be queried
skip,take: skip skips a few pieces of data and takes a lot of data
for example:
Model::offset(0)->limit(10)->get();
Model::skip(3)->take(3)->get();
//i use it in my project, work fine ~
class BookController extends Controller
{
public function getList(Request $request) {
$page = $request->has('page') ? $request->get('page') : 1;
$limit = $request->has('limit') ? $request->get('limit') : 10;
$books = Book::where('status', 0)->limit($limit)->offset(($page - 1) * $limit)->get()->toArray();
return $this->getResponse($books, count($books));
}
}
laravel have own function skip for offset and take for limit. just like below example of laravel query :-
Article::where([['user_id','=',auth()->user()->id]])
->where([['title','LIKE',"%".$text_val."%"]])
->orderBy('id','DESC')
->skip(0)
->take(2)
->get();
You can use skip and take functions as below:
$products = $art->products->skip($offset*$limit)->take($limit)->get();
// skip should be passed param as integer value to skip the records and starting index
// take gets an integer value to get the no. of records after starting index defined by skip
EDIT
Sorry. I was misunderstood with your question. If you want something like pagination the forPage method will work for you. forPage method works for collections.
REf : https://laravel.com/docs/5.1/collections#method-forpage
e.g
$products = $art->products->forPage($page,$limit);
Maybe this
$products = $art->products->take($limit)->skip($offset)->get();
$products = Article::orderBy('id', 'asc')->limit($limit)->get();
Try this sample code:
$art = Article::where('id',$article)->firstOrFail();
$products = $art->products->take($limit);
$collection = collect([1, 2, 3, 4, 5, 6, 7, 8, 9]);
$chunk = $collection->forPage(2, 3);
$chunk->all();
Please try like this,
return Article::where('id',$article)->with(['products'=>function($products, $offset, $limit) {
return $products->offset($offset*$limit)->limit($limit);
}])
Laravel 5.1< (current v 9)
$art->products->skip($offset)->take($limit)->all();
https://laravel.com/docs/9.x/collections#method-take

filtering a paginated eloquent collection

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();

Returning array of objects in Laravel

I need to be able to change objects on server, save those, and return results back to frontend part of application.
So basically, I have some code that does manipulation of data, than Eloquent that does saving, and than I want to return Eloquent object back. Problem is that I have more than one object, that I'll manipulate, and right now, I'm putting all of them in array. When it comes back to my frontend all it has is this:
[{"incrementing":true,"timestamps":true,"exists":true}]
Here is the simplified code:
$results = array();
foreach ($tasks as $task){
//some manipulation
$result = Task::find($task['id']);
$result->order = $task['order'];
$result->save();
$results[] = $result;
}
return Response::json($results);
Ok, the solution was to call toArray() method before putting element to array.
$results[] = $result->toArray();

laravel 4 paginate collection

I cant create a proper pagination system using laravel 4. I have the following models and function that return collections:
Model Restaurant:
public function fooditem()
{
return $this->hasMany('Fooditem','rest_id');
}
public function get_rest_foods($id){
return Restaurant::find($id)->fooditem->toArray();
}
The second function returns all food items for a certain restaurant as an array. I also use this in an API call.
in my controller i have this:
$food = $food->get_rest_foods($id);
$paginator = Paginator::make($food, 10, 5);
I pass the paginator to the view and it shows the links ok but also shows all my item from the food array.
I tried using
public function get_rest_foods($id){
return Restaurant::find($id)->fooditem->paginate(5);
}
but i get an error:
FatalErrorException: Error: Call to undefined method Illuminate\Database\Eloquent\Collection::paginate()
I searched this and many other sites but cant understant how to paginate a collection.
Thanks for your help
The paginator must get the items that it would normally get from a database query with an offset/limit statement.
So when you have a collection with all items, you should do the offset/limit yourself.
$food = $food->get_rest_foods($id);
$page = 1;
if( !empty(Input::get['page']) ) {
$page = Input::get['page'];
}
$perPage = 15;
$offset = (($page - 1) * $perPage);
$food = Paginator::make($food->slice($offset,$perPage, true)->all(), $food->count(), $perPage);
I created a subclass of Collection and implemented my own paginate method
public function paginate($perPage) {
$pagination = App::make('paginator');
$count = $this->count();
$page = $pagination->getCurrentPage($count);
$items = $this->slice(($page - 1) * $perPage, $perPage)->all();
$pagination = $pagination->make($items, $count, $perPage);
return $pagination;
}
The Laravel paginator does not do any of the splicing for you. Typically the paginator should be used hand in hand with Eloquent/Fluent. In your second example you should be doing it like this.
return Restaurant::find($id)->fooditem()->paginate(5);
Without calling the actual method you'll just be getting a collection of results and not the query builder instance.
If you want to use the paginator manually you need to pass in the spliced array (the correct items for the current page). This usually involves determining the current page, the total results and total pages. That's why, where possibly, it's best to use it with Eloquent or Fluent.

Resources