I need help with Laravel 5 (using Eloquent)
I have 2 tables...
Model Driver
drivers
id
company
Model DriverOnline
drivers_online
id
name
driver_id
I need to search for a result on (company=1 and driver_id=driver.id).
Please help me!
If you want to only fetch Driver based on a condition, you can just do this:
Driver::with('online')->where('company', 1)->get();
If the clause is on the relationship, use with and specify a query on that.
$company = 1;
$drivers = Driver::with(['online' => function($query) use ($company)
{
$query->where('company', $company);
}]);
See "Eager Load Constraints":
https://laravel.com/docs/5.0/eloquent
Take note of my use. This allows you to include variables from the scope into your Closure instance.
And be aware, if you use either solution, you must set up a relationship. Consult the link I shared with more information on that.
Edit: As per our conversation.
$drivers = Driver::where('company_id','=',1)
->with('driversOnline')
->whereHas('driversOnline', function($query) {
$query->where('online','=',1);
})
->get();
Related
hello I am new to laravel and maybe I am a bit confused between eloquent and query builder way for writing a query but anyway can you please tell me what could be the best eloquent way to retrieve info like this in laravel 6 or 7
User > hasMany > Recipes
Recipe > belongsTo > User
I want to check if user id 2 present in users table then get only one post which id is 3
Query builder is for explicitly building SQL queries, and does not return instances of your models. Eloquent query builder, is similar but the result will contain the model(s) loaded with all their attributes, and has some handy functions for querying the relations you define in your models.
Given the limited information in your post, I am assuming when you say a post, you mean a recipe:
Query Builder:
DB::table('users')
->join('recipes', 'recipes.user_id', '=', 'users.id')
->select(['users.some_col', ... 'recipes.some_col'])
->where('users.id', 2)
->get();
If you have your models setup with the relations. You can use Eloquent like so:
User::where('id', 2)->with('recipes')->get();
If I understand you correctly it would be like this:
User::whereId($userId) //asuming it is 2
->with(['recipes' => function($q) use($recipeId) {
$q->where('id', $recipeId); //assuming it is 3
}])->first();
you can do this if i understand correctly:
$user = User::findOrFail(2); //auto 404 if user not found
$recipe = $user->Recipes()->where('id',3)->first();
You may use conditional eager loading for better performance.
$userId = 2;
$receiptId = 3;
$user = User::with(['receipts'=> function ($query) use($receiptId){
$query->where('id', $receiptId);
}
])->find($userId)
I came accros a problem with laravel's ORM, eloquent and found no solution yet.
I have some tables as follows
Team
- id
- name
User
- id
- name
- role
- team_id
Student_Info
- id
- user_id
- data1
- data2
- etc ...
Project
- id
- student_id
- name
Now, I want to query all projects a certain team, where team = 'some team'
Now the thing here is, without an ORM, it's simple, I would have done multiple joins in raw SQL.
However, because all these tables have a common column "name" I will have to alias all this stuff, which is really boring
With eloquent I can't find a way to do this query using "has many through" because it only allows on intermediate and I can't do a raw SQL as the alis thing is really a pain in the ass and as it would be very difficult to map the result to laravel's Models
There is no native relationship for this case.
I created a HasManyThrough relationship with unlimited levels: Repository on GitHub
After the installation, you can use it like this:
class Team extends Model {
use \Staudenmeir\EloquentHasManyDeep\HasRelationships;
public function projects() {
return $this->hasManyDeep(Project::class, [User::class, StudentInfo::class],
[null, null, 'student_id']);
}
}
$projects = Team::where('name', 'some team')->first()->projects;
Try with relationship existence. This assumes you have all relationships properly defined
$projects = Project::whereHas('students.user.team', function ($query) {
$query->where('name', '=', 'some team');
})->get();
That's 3 levels of nesting. Never tested. However, if you already define a Project-User relationship via hasManyThrough() you can shorten it to 2 levels only.
$projects = Project::whereHas('user.team', function ($query) {
$query->where('name', '=', 'some team');
})->get();
Those will give you the data for projects only. If you also want the the intermediate data, use eager loading instead with with(). Just replace whereHas() by with().
$projects = Project::with('user.team', function ($query) {
$query->where('name', '=', 'some team');
})->get();
Using Laravel 4 I have the following models and relations: Event which hasMany Record which hasMany Item. What I would like to do is something like this
Item::where('events.event_type_id', 2)->paginate(50);
This of cause doesn't work as Eloquent doesn't JOIN the models together when retrieving the records. So how do I go about this without just writing the SQL myself (which I would like to avoid as I want to use pagination).
What you want is eager loading.
It works like this if you want to specify additional constraints:
Item::with(array('events' => function($query) {
return $query->where('event_type_id', 2);
}))->paginate(50);
There is a pull request pending here https://github.com/laravel/framework/pull/1951.
This will allow you to use a constraint on the has() method, something like this:
$results = Foo::has(array('bars' => function($query)
{
$query->where('title', 'LIKE', '%baz%');
}))
->with('bars')
->get();
The idea being you only return Foos that have related Bars that contain the string 'baz' in its title column.
It's also discussed here: https://github.com/laravel/framework/issues/1166. Hopefully it will be merged in soon. Works fine for me when I update my local copy of the Builder class with the updated code in the pull request.
I have got 2 joined tables in Eloquent namely themes and users.
theme model:
public function user() {
return $this->belongs_to('User');
}
user model:
public function themes() {
return $this->has_many('Theme');
}
My Eloquent api call looks as below:
return Response::eloquent(Theme::with('user')->get());
Which returns all columns from theme (that's fine), and all columns from user (not fine). I only need the 'username' column from the user model, how can I limit the query to that?
Change your model to specify what columns you want selected:
public function user() {
return $this->belongs_to('User')->select(array('id', 'username'));
}
And don't forget to include the column you're joining on.
For Laravel >= 5.2
Use the ->pluck() method
$roles = DB::table('roles')->pluck('title');
If you would like to retrieve an array containing the values of a single column, you may use the pluck method
For Laravel <= 5.1
Use the ->lists() method
$roles = DB::table('roles')->lists('title');
This method will return an array of role titles. You may also specify a custom key column for the returned array:
You can supply an array of fields in the get parameter like so:
return Response::eloquent(Theme::with('user')->get(array('user.username'));
UPDATE (for Laravel 5.2)
From the docs, you can do this:
$response = DB::table('themes')
->select('themes.*', 'users.username')
->join('users', 'users.id', '=', 'themes.user_id')
->get();
I know, you ask for Eloquent but you can do it with Fluent Query Builder
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->get(array('themes.*', 'users.username'));
This is how i do it
$posts = Post::with(['category' => function($query){
$query->select('id', 'name');
}])->get();
First answer by user2317976 did not work for me, i am using laravel 5.1
Using with pagination
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->select('themes.*', 'users.username')
->paginate(6);
Another option is to make use of the $hidden property on the model to hide the columns you don't want to display. You can define this property on the fly or set defaults on your model.
public static $hidden = array('password');
Now the users password will be hidden when you return the JSON response.
You can also set it on the fly in a similar manner.
User::$hidden = array('password');
user2317976 has introduced a great static way of selecting related tables' columns.
Here is a dynamic trick I've found so you can get whatever you want when using the model:
return Response::eloquent(Theme::with(array('user' => function ($q) {
$q->addSelect(array('id','username'))
}))->get();
I just found this trick also works well with load() too. This is very convenient.
$queriedTheme->load(array('user'=>function($q){$q->addSelect(..)});
Make sure you also include target table's key otherwise it won't be able to find it.
This Way:
Post::with(array('user'=>function($query){
$query->select('id','username');
}))->get();
I know that this is an old question, but if you are building an API, as the author of the question does, use output transformers to perform such tasks.
Transofrmer is a layer between your actual database query result and a controller. It allows to easily control and modify what is going to be output to a user or an API consumer.
I recommend Fractal as a solid foundation of your output transformation layer. You can read the documentation here.
In Laravel 4 you can hide certain fields from being returned by adding the following in your model.
protected $hidden = array('password','secret_field');
http://laravel.com/docs/eloquent#converting-to-arrays-or-json
On Laravel 5.5, the cleanest way to do this is:
Theme::with('user:userid,name,address')->get()
You add a colon and the fields you wish to select separated by a comma and without a space between them.
Using Model:
Model::where('column','value')->get(['column1','column2','column3',...]);
Using Query Builder:
DB::table('table_name')->where('column','value')->get(['column1','column2','column3',...]);
If I good understood this what is returned is fine except you want to see only one column. If so this below should be much simpler:
return Response::eloquent(Theme::with('user')->get(['username']));
#You can get selected columns from two or three different tables
$users= DB::Table('profiles')->select('users.name','users.status','users.avatar','users.phone','profiles.user_id','profiles.full_name','profiles.email','profiles.experience','profiles.gender','profiles.profession','profiles.dob',)->join('users','profiles.user_id','=','users.id')
->paginate(10);
Check out, http://laravel.com/docs/database/eloquent#to-array
You should be able to define which columns you do not want displayed in your api.
I'm using Laravel and having a small problem with Eloquent ORM.. I can get this working simply with SQL query using a JOIN but I can't seem to get it working with Eloquent!
This is what I want, I have two tabels. one is 'Restaurants' and other is 'Restaurant_Facilities'.
The tables are simple.. and One-To-One relations. like there is a restaurant table with id, name, slug, etc and another table called restaurant_facilities with id, restaurant_id, wifi, parking, etc
Now what I want to do is.. load all restaurants which have wifi = 1 or wifi = 0..
How can i do that with Eloquent ? I have tried eager loading, pivot tables, with(), collections() and nothing seems to work!
The same problem I have for a Many-To-Many relation for cuisines!
I have the same restaurant table and a cuisine table and a restaurant_cuisine_connection table..
but how do I load all restaurants inside a specific cuisine using it's ID ?
This works.
Cuisine::find(6)->restaurants()->get();
but I wanna load this from Restaurant:: model not from cuisines.. because I have many conditions chained together.. its for a search and filtering / browse page.
Any ideas or ways ? I've been struggling with this for 3 days and still no answer.
Example Models :
class Restaurant extends Eloquent {
protected $table = 'restaurants';
public function facilities() {
return $this->hasOne('Facilities');
}
}
class Facilities extends Eloquent {
protected $table = 'restaurants_facilities';
public function restaurant() {
return $this->belongsTo('Restaurant');
}
}
PS :
This seems to be working.. but this is not Eloquent way right ?
Restaurant::leftJoin(
'cuisine_restaurant',
'cuisine_restaurant.restaurant_id',
'=', 'restaurants.id'
)
->where('cuisine_id', 16)
->get();
Also what is the best method to find a count of restaurants which have specific column value without another query ? like.. i have to find the total of restaurants which have parking = 1 and wifi = 1 ?
Please help on this.
Thank you.
I don't see anything wrong with doing the left join here, if you have to load from the Restaurant model. I might abstract it away to a method on my Restaurant model, like so:
class Restaurant extends Eloquent {
protected $table = 'restaurants'; // will be default in latest L4 beta
public function facility()
{
return $this->hasOne('Facility');
}
// Or, better, make public, and inject instance to controller.
public static function withWifi()
{
return static::leftJoin(
'restaurant_facilities',
'restaurants.id', '=', 'restaurant_facilities.restaurant_id'
)->where('wifi', '=', 1);
}
}
And then, from your routes:
Route::get('/', function()
{
return Restaurant::withWifi()->get();
});
On the go - haven't tested that code, but I think it should work. You could instead use eager loading with a constraint, but that will only specify whether the facility object is null or not. It would still return all restaurants, unless you specify a where clause.
(P.S. I'd stick with the singular form of Facility. Notice how hasOne('Facilities') doesn't read correctly?)
I stumbled across this post while trying to improve my REST API methodology when building a new sharing paradigm. You want to use Eager Loading Constraints. Let's say you have an api route where your loading a shared item and it's collection of subitems such as this:
/api/shared/{share_id}/subitem/{subitem_id}
When hitting this route with a GET request, you want to load that specific subitem. Granted you could just load that model by that id, but what if we need to validate if the user has access to that shared item in the first place? One answer recommended loading the inversed relationship, but this could lead to a confusing and muddled controller very quickly. Using constraints on the eager load is a more 'eloquent' approach. So we'd load it like this:
$shared = Shared::where('id', $share_id)
->with([ 'subitems' => function($query) use ($subitem_id) {
$query->where('subitem_id', $subitem_id)
}]);
So where only want the subitem that has that id. Now we can check if it was found or not by doing something like this:
if ($shared->subitems->isEmpty())
Since subitems is a collection (array of subitems) we return the subitem[0] with this:
return $shared->subitems[0];
Use whereHas to filter by any relationship. It won't join the relation but it will filter the current model by a related property. Also look into local scopes to help with situations like this https://laravel.com/docs/5.3/eloquent#local-scopes
Your example would be:
Restaurant::whereHas('facilities', function($query) {
return $query->where('wifi', true);
})->get();
Restaurant::whereHas('cuisines', function($query) use ($cuisineId) {
return $query->where('id', $cuisineId);
})->get();
To achieve the same thing with local scopes:
class Restaurant extends Eloquent
{
// Relations here
public function scopeHasWifi($query)
{
return $query->whereHas('facilities', function($query) {
return $query->where('wifi', true);
});
}
public function scopeHasCuisine($query, $cuisineId)
{
return $query->whereHas('cuisines', function($query) use ($cuisineId) {
return $query->where('id', $cuisineId);
});
}
}
For local scopes you DO NOT want to define them as static methods on your model as this creates a new instance of the query builder and would prevent you from chaining the methods. Using a local scope will injects and returns the current instance of the query builder so you can chain as many scopes as you want like:
Restaurant::hasWifi()->hasCuisine(6)->get();
Local Scopes are defined with the prefix scope in the method name and called without scope in the method name as in the example abover.
Another solution starring whereHas() function:
$with_wifi = function ($query) {
$query->where('wifi', 1);
};
Facilities::whereHas('restaurant', $with_wifi)
Nice and tidy.
Do you absolutely have to load it from the Restaurant model? In order to solve the problem, I usually approach it inversely.
Facilities::with('restaurant')->where('wifi' ,'=', 0)->get();
This will get all the restaurant facilities that match your conditions, and eager load the restaurant.
You can chain more conditions and count the total like this..
Facilities::with('restaurant')
->where('wifi' ,'=', 1)
->where('parking','=', 1)
->count();
This will work with cuisine as well
Cuisine::with('restaurant')->where('id','=',1)->get();
This grabs the cuisine object with the id of 1 eager loaded with all the restaurants that have this cuisine