I have a multiple query, but for some reason the pagination is not working, i dont no if the ussue is the way i construct it.
My code:
$query = \DB::table('products');
if ($request->has('type') && $type) {
$query->where('product_type_id', $type);
}
if ($request->has('material') && $material) {
$query->join("product_attribute_materials","products.id","=","product_attribute_materials.product_id")
->whereIn("product_attribute_materials.product_material_id", $material);
}
if($request->has('businessarea') && $area){
$query->join("product_attribute_businessareas","products.id","=","product_attribute_businessareas.product_id")
->whereIn("product_attribute_businessareas.product_businessarea_id", $area);
}
$products = $query->paginate(15);
You need to append the query string to the pagination using append() on $products
Passing array with the keys that you want to pass the query string, will return its value on the next request.
To confirm this, chick the href value from your browser before clicking on page number 2.
$products = $query->paginate(15)->appends(['type', 'material', 'businessarea']);
Related
I am trying to get all rows and distinct column from single query. but paginate method is only giving result but not pagination option like total prev next etc..
$offers = Offer::whereHas('users', function ($q) use ($authUser) {
$q->where('user_id', $authUser->parent_id);
$q->where('publisher_id', '=', $authUser->id);
});
and distinct column
$websites = $offers->distinct()->get(['website']);
with pivot columns (just wanted to show my full query)
$offers->orderBy($sortBy, $orderBy)->paginate($perPage)->map(function ($offer) {
if (!empty($offer->users)) {
$manager = $publisher = '';
foreach ($offer->users as $user) {
$manager = $user->pivot->user_id;
$publisher = $user->pivot->publisher_id;
}
$offer->manager = $manager;
$offer->publisher = $publisher;
}
return $offer;
});
Return
return response()->json([
'offers' => $offers,
'websites' => $websites
], 200);
hope my question will make sense.
Thanks.
You should run getCollection() before mapping to get the paginator's underlying collection.
(https://laravel.com/api/7.x/Illuminate/Pagination/LengthAwarePaginator.html#method_getCollection)
$offers->orderBy($sortBy, $orderBy)->paginate($perPage)
->getCollection()
->map(function ($offer) {
// ...
return $offer;
});
I'm answering based on it being $offers:
Your usage of map() is copying the modified results of your paginate() call to a new collection and that collection does not include the pagination information. That's why you no longer have pagination information.
Since there result of paginate() is already a usable collection, you can use each() instead of map() which will alter the objects in-place.
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 use laravel eloquent to get files of post that give me a collection
but I want an item from that collection by condition.
For example from that collection I want an item that type = 'product'.
I am using foreach and check every item that have my condition and return it, but isn't any better way?
I tested collection method like contain but it return null.
Files item have type filed that value is 'product' or 'blog'.
My code:
$post= Post::where('slug' , $slug)->first();
$cover = $post->files->contains(['type' , '=' , 'product']);
Use the collection filter method.
$cover = $post->files->filter(function($file) {
// show only the items that match this condition
return collect(['product', 'blog'])->contains($file->type);
});
The filter method filters the collection using the given callback, keeping only those items that pass a given truth test:
$filtered = $post->files->filter(function ($value, $key) {
return $value->type == 'product';
});
$filtered->all();
Collections - filter()
In Laravel project I am making a database query
$addresses = DB::table('residences')->where('user_id', '=', Auth::user()->id)->get();
I want to get one address from that query result, if it exists, where 'id' field equals to 10 for example. But I want to remove that address from $addresses query result simultaneously. Is there a simple way to do that?
You can use the reject() method on the $addresses collection, i.e:
$my_result = false;
$addresses = $addresses->reject(function ($value, $key) use (&$my_result) {
if($value->id == 10){
$my_result = $value;
return true
}
return false;
});
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();