So I am trying to get a list of all active clients that have active jobs but I'm at a loss of how to accomplish this. So here is what I have...
$query = Client::select( 'clients.*' )->where( 'is_enabled', 1 )->activeJobs();
Which throws an error
Call to undefined method Illuminate\Database\Eloquent\Builder::activeJobs()
in my client model i have the activeJobs() function as follows:
public function activeJobs()
{
return $this->hasMany( Job::class )->where( 'is_active', 1 );
}
Just to explain what I am after with my query in words, I'm trying to get a collection of all active client items (determined by is_enabled = 1) which have 1 or more active jobs (determined by is_active = 1)
thanks
What you are looking for is has/whereHas.
So, what you should do is
$query = Client::select('clients.*')
->whereHas('jobs', fn($query) => $query->where('is_active', 1))
->where('is_enabled', 1)
->get();
This will return you a collection of Clients whose Job's is_active column is 1
Try this
$query = Client::select('clients.*')
->where('is_enabled', 1)
->whereHas('activeJobs', function ($query) {
$query->where('is_active', 1);
})
->get();
Try this
$query = Client::has('activeJobs', '>', 1)->get();
Try This:
$query = Client::with('activeJobs')->select( 'clients.*' )->where( 'is_enabled', 1 );
This will return the clients with their relationship(Active Jobs), what you are doing in the code is trying to grab relationship data from collection..
You can either grab the clients, then foreach $client->activeJobs;
or use with which wil return it in query level.
Get active jobs and then collect related users:
$cliensWithActiveJobs = Job::with('client')
->where('is_active', 1)
->get(['id', 'client_id'])
->map(fn ($job) => $job->client)
->unique('id');
Related
I am using laravel eager loading to load data on the jquery datatables. My code looks like:
$columns = array(
0 => 'company_name',
1 => 'property_name',
2 => 'amenity_review',
3 => 'pricing_review',
4 => 'sqft_offset_review',
5 => 'created_at',
6 => 'last_uploaded_at'
);
$totalData = Property::count();
$limit = $request->input('length');
$start = $request->input('start');
$order = $columns[$request->input('order.0.column')];
$dir = $request->input('order.0.dir');
$query = Property::with(['company','notices']);
$company_search = $request->columns[0]['search']['value'];
if(!empty($company_search)){
$query->whereHas('company', function ($query) use($company_search) {
$query->where('name','like',$company_search.'%');
});
}
$property_search = $request->columns[1]['search']['value'];
if(!empty($property_search)){
$query->where('properties.property_name','like',$property_search.'%');
}
if(!Auth::user()->hasRole('superAdmin')) {
$query->where('company_id',Auth::user()->company_id);
}
$query->orderBy($order,$dir);
if($limit != '-1'){
$records = $query->offset($start)->limit($limit);
}
$records = $query->get();
With this method I received error: Column not found: 1054 Unknown column 'company_name' in 'order clause' .
Next, I tried with following order condition:
if($order == 'company_name'){
$query->orderBy('company.name',$dir);
}else{
$query->orderBy($order,$dir);
}
However, it also returns similar error: Column not found: 1054 Unknown column 'company.name' in 'order clause'
Next, I tried with whereHas condition:
if($order == 'company_name'){
$order = 'name';
$query->whereHas('company', function ($query) use($order,$dir) {
$query->orderBy($order,$dir);
});
}else{
$query->orderBy($order,$dir);
}
But, in this case also, same issue.
For other table, I have handled this type of situation using DB query, however, in this particular case I need the notices as the nested results because I have looped it on the frontend. So, I need to go through eloquent.
Also, I have seen other's answer where people have suggested to order directly in model like:
public function company()
{
return $this->belongsTo('App\Models\Company')->orderBy('name');
}
But, I don't want to order direclty on model because I don't want it to be ordered by name everytime. I want to leave it to default.
Also, on some other scenario, I saw people using join combining with, but I am not really impressed with using both join and with to load the same model.
What is the best way to solve my problem?
I have table like: companies: id, name, properties: id, property_name, company_id, notices: title, slug, body, property_id
The issue here is that the Property::with(['company','notices']); will not join the companies or notices tables, but only fetch the data and attach it to the resulting Collection. Therefore, neither of the tables are part of the SQL query issued and so you cannot order it by any field in those tables.
What Property::with(['company', 'notices'])->get() does is basically issue three queries (depending on your relation setup and scopes, it might be different queries):
SELECT * FROM properties ...
SELECT * FROM companies WHERE properties.id in (...)
SELECT * FROM notices WHERE properties.id in (...)
What you tried in the sample code above is to add an ORDER BY company_name or later an ORDER BY companies.name to the first query. The query scope knows no company_name column within the properties table of course and no companies table to look for the name column. company.name will not work either because there is no company table, and even if there was one, it would not have been joined in the first query either.
The best solution for you from my point of view would be to sort the result Collection instead of ordering via SQL by replacing $records = $query->get(); with $records = $query->get()->sortBy($order, $dir);, which is the most flexible way for your task.
For that to work, you would have to replace 'company_name' with 'company.name' in your $columns array.
The only other option I see is to ->join('companies', 'companies.id', 'properties.company_id'), which will join the companies table to the first query.
Putting it all together
So, given that the rest of your code works as it should, this should do it:
$columns = [
'company.name',
'property_name',
'amenity_review',
'pricing_review',
'sqft_offset_review',
'created_at',
'last_uploaded_at',
];
$totalData = Property::count();
$limit = $request->input('length');
$start = $request->input('start');
$order = $columns[$request->input('order.0.column')];
$dir = $request->input('order.0.dir');
$query = Property::with(['company', 'notices']);
$company_search = $request->columns[0]['search']['value'];
$property_search = $request->columns[1]['search']['value'];
if (!empty($company_search)) {
$query->whereHas(
'company', function ($query) use ($company_search) {
$query->where('name', 'like', $company_search . '%');
});
}
if (!empty($property_search)) {
$query->where('properties.property_name', 'like', $property_search . '%');
}
if (!Auth::user()->hasRole('superAdmin')) {
$query->where('company_id', Auth::user()->company_id);
}
if ($limit != '-1') {
$records = $query->offset($start)->limit($limit);
}
$records = $query->get()->sortBy($order, $dir);
I would like to order result of eloquent by field on the other related table.
I have users table. Every user has one profile. Profile has sponsored (which is boolean) field. So when I would like to get all users, I want to display first sponsored users, then non sponsored.
public function profile(){
return $this->hasOne('App\Doctor');
}
There are two ways:
1)You have to join tables,
User::join('profiles','users.id','=','profile.user_id')->orderBy('sponsored','DESC')->get()
2)Order by eager loading
User::with(array('profile' => function($query) {
$query->orderBy('sponsored', 'DESC');
}))
->get();
Try this one
User::leftJoin('profile', 'user.id', '=', 'profile.user_id')
->orderBy('profile.sponsored', 'ASC')
->get();
I highly recommend not using table joins as it would fail you on the scale.
A better solution is to get users, get their profiles and then sort them using laravel collection methods.
You can use this sample to achieve this solution.
//get all users
$users = User::all();
//extract your users Ids
$userIds = $users->pluck('id')->toArray();
//get all profiles of your user Ids
$profiles = Profile::whereIn('user_id', $userIds)->get()->keyBy('user_id');
//now sort users based on being sponsored or not
$users = $users->sort(function($item1, $item2) use ($profiles) {
if($profiles[$item1->id]->sponsored == 1 && $profiles[$item2->id]->sponsored == 1){
return 0;
}
if($profiles[$item1->id]->sponsored == 1) return 1;
return -1;
});
You can check this link which explains on laravel collection sorts.
$order = 'desc';
$users = User::join('profile', 'users.id', '=', 'profile.id')
->orderBy('profile.id', $order)->select('users.*')->get();
I have 2 models in my app:
1. Customer.php
2. Car.php
Now I would like to run a query that returns all customers that have less than 2 cars. Where 2 is a number that can be changed by the user.
I have tried this but it didn't work, it just returns all customer records:
$customers = Customer::whereHas("cars", function($query) {
$query->selectRaw("count(*) < ?", [2]);
})
->get();
Edit:
The two models are linked in a pivot table, meaning A customer can have more than 1 car and a Car can belong to more than 1 customer.
Use this:
$customers = Customer::withCount('cars')
->having('cars_count', '<', 2)
->get();
So , here is the result.
Relation in model Customer.php
public function cars()
{
return $this->belongsToMany('App\Car','car_customer','car_id','customer_id');
}
Query to get all customers with N cars:
$userInput = 2;
$data = Customer::with('cars')
->withCount('cars')
->has('cars', '<', $userInput)
->orderBy('cars_count', 'desc')
->get();
Where the $userInput is your 'N'.
This is the best way:
$customers = Customer::has('cars','<', 2)->get();
Have you tried this approach?
$input = 2;
$customers = Customer::whereHas("cars", function($query) use ($input) {
$query->where(DB::raw("count(cars.id)"), "<", DB::raw($input))
})->get();
I have the following relationship functions in my Job model:
public function resourceTypes(){
return $this->belongsToMany('ResourceType', 'job_requests');
}
public function resources(){
return $this->belongsToMany('Resource', 'jobs_resources')->withPivot('flow_type', 'resource_type_id');
}
I am able to get an object with data from both of the above relationships using:
$job = Job::findorfail($projectId);
$result = $job->with('resources.resourceTypes')->get();
I would like to put a where clause on the jobs_resources pivot table - specifically on the column flow_type.
How would I do this?
Try something like this:
$job = Job::with('resources' => function($q) {
$q->with('resourceTypes')->where('flow_type',2);
})->findorfail($projectId);
In above you will get only those resources with flow_type = 2
I ended up using the following statement:
Job::with(['resources' => function ($query){
$query->wherePivot('flow_type', '=', '1' );
}, 'resources.resourceTypes'])->where('id', $projectId)->firstOrFail();
$result = DB::table('job')
->join('job_resources', 'job.id', '=', 'job_resources.job_id')
->join('job_requests', 'job_resources.request_id', '=', 'job_requests.id')
->where('job_resources.flow_type', '=', CONDITION)
->get();
Your table data is not clear from your input, but this method (query builder) should work
I'd like to know the position of a user based on its creation date. How do I do that using Eloquent?
I'd like to be able to do something like this:
User::getRowNumber($user_obj);
I suppose you want MySQL solution, so you can do this:
DB::statement(DB::raw('set #row:=0'));
User::selectRaw('*, #row:=#row+1 as row')->get();
// returns all users with ordinal 'row'
So you could implement something like this:
public function scopeWithRowNumber($query, $column = 'created_at', $order = 'asc')
{
DB::statement(DB::raw('set #row=0'));
$sub = static::selectRaw('*, #row:=#row+1 as row')
->orderBy($column, $order)->toSql();
$query->remember(1)->from(DB::raw("({$sub}) as sub"));
}
public function getRowNumber($column = 'created_at', $order = 'asc')
{
$order = ($order == 'asc') ? 'asc' : 'desc';
$key = "userRow.{$this->id}.{$column}.{$order}";
if (Cache::get($key)) return Cache::get($key);
$row = $this->withRowNumber($column, $order)
->where($column, '<=',$this->$column)
->whereId($this->id)->pluck('row');
Cache::put($key, $row);
return $row;
}
This needs to select all the rows from the table till the one you are looking for is found, then selects only that particular row number.
It will let you do this:
$user = User::find(15);
$user->getRowNumber(); // as default ordered by created_at ascending
$user->getRowNumber('username'); // check order for another column
$user->getRowNumber('updated_at', 'desc'); // different combination of column and order
// and utilizing the scope:
User::withRowNumber()->take(20)->get(); // returns collection with additional property 'row' for each user
As this scope requires raw statement setting #row to 0 everytime, we use caching for 1 minute to avoid unnecessary queries.
$query = \DB::table(\DB::raw('Products, (SELECT #row := 0) r'));
$query = $query->select(
\DB::raw('#row := #row + 1 AS SrNo'),
'ProductID',
'ProductName',
'Description',
\DB::raw('IFNULL(ProductImage,"") AS ProductImage')
);
// where clauses
if(...){
$query = $query->where('ProductID', ...));
}
// orderby clauses
// ...
// $query = $query->orderBy('..','DESC');
// count clause
$TotalRecordCount = $query->count();
$results = $query
->take(...)
->skip(...)
->get();
I believe you could use Raw Expresssions to achieve this:
$users = DB::table('users')
->select(DB::raw('ROW_NUMBER() OVER(ORDER BY ID DESC) AS Row, status'))
->where('status', '<>', 1)
->groupBy('status')
->get();
However, looking trough the source code looks like you could achieve the same when using SQLServer and offset. The sources indicates that if you something like the following:
$users = DB::table('users')->skip(10)->take(5)->get();
The generated SQL query will include the row_number over statement.
[For Postgres]
In your model
public function scopeWithRowNumber($query, $column = 'id', $order = 'asc'){
$sub = static::selectRaw('*, row_number() OVER () as row_number')
->orderBy($column, $order)
->toSql();
$query->from(DB::raw("({$sub}) as sub"));
}
In your controller
$user = User::withRowNumber()->get();