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.
Related
I'm trying to get the right data from the database, I'm retriving a model with a media relation via eloquent, but I want to return a photo that contains the 'main' tag stored in JSON, if this tag is missing, then I would like to return the first photo assigned to this model.
how i assign tags to media
I had 3 ideas:
Use orWhere() method, but i want more likely 'xor' than 'or'
$models = Model::with(['media' => function ($query) {
$query->whereJsonContains('custom_properties->tags', 'main')->orWhere();
}]);
return $models->paginate(self::PER_PAGE);
Raw SQL, but i don't really know how to do this i tried something with JSON_EXTRACT and IF/ELSE statement, but it was to hard for me and it was a disaster
Last idea was to make 2 queries and just add media from second query if there is no tag 'main'
$models = Model::with(['media' => function ($query) {
$query->whereJsonContains('custom_properties->tags', 'main');
}]);
$models_all_media = Model:: with(['media']);
return $models->paginate(self::PER_PAGE);
but i tried something like
for($i=0; $i<count($models); $i++) {
$models->media = $models_all_media
}
but i can't do this without get() method, beacuse i don't know how to change this to LengthAwarePaginator class after using get()
try using whereHas https://laravel.com/docs/9.x/eloquent-relationships
Model::with('media')
->whereHas('media',fn($media)=>$media->whereJsonContains('custom_properties->tags', 'main'))
->paginate(self::PER_PAGE);
as per your comment you can use
$models = Model::with(['media' => function ($query) {
$query->whereJsonContains('custom_properties->tags', 'main');
}])
->leftJoin('media', function ($join) {
$join->on('models.id', '=', 'media.model_id')
->whereNull('media.custom_properties->tags->main');
})
->groupBy('models.id')
->paginate(self::PER_PAGE);
return $models;
I want to remove duplicated value between 2 arrays. How can I do?
1, 3, 4, 6 are duplicated in a both arrays, I want to unique values.
I using map() to show arrays of item_id contains of quantity but it's duplicate value, I don't want it.
$deliveries = $pickupsGroupByDepartment->first()->map(function ($q) {
return $q->deliveries->groupBy('delivery_date')->map(function($r) {
return $r->mapToGroups(function ($item) {
return [$item['item_id'] => $item['quantity']];
});
});
});
You can use array_unique function then combine it using array_merge function.
$array = array_unique (array_merge ($array1, $array2));
if your data is came from object in Laravel. you can use code below.
$result = $object1->merge($object2)->unique();
If the situation is dynamic data you can do this..
$results = [];
foreach($dynamicArray as $key => $array){
$results = array_unique (array_merge ($results, $array));
}
return $results;
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 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 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();