When you create model in laravel for eloquent, the User model is treated like a users table. Why is that? Can we use an exact table name for the model?
I think more than anything it's convention. I don't see why you couldn't create a Users model for your users table, but a User is an instance of users, which is why it's done that way. You can always specify the table the model uses with:
protected $table = 'name_of_table';
which can be different than the model name. For example, a Data model can use the table userdata, as long as you specify that.
Hope that helps.
Related
Here are some models:
UserModel
SpecializationModel
UserSpecializationModel
I need to recieve authorized user's specialization. I can do that:
$specializations = UserSpecialization::where("user_id", Auth::user()->id)->get();
Also I can do this through the UserModel model using relation hasMany() specializations().
When to use first case and the second?
$specializations = Auth::user()->specializations();
Do I need a model UserSpecializationModel?
In general you don't need UserSpecilizatonModel, in most situations you wont access data directly from that table, you'll either do it through user or specialization model.
Check also https://laravel.com/docs/9.x/eloquent-relationships#retrieving-intermediate-table-columns for accessing data from pivot table.
I am changing my File-model to go from one to many to be a one to many polymorphic relation. Before only my User-model could have a file but now my model Guest will also be able to have a file.
So before I had a column in the file table named user_id I have renamed that one following the pattern here: https://laravel.com/docs/7.x/eloquent-relationships#many-to-many-polymorphic-relations
So it is now named fileable_id.
In addition to the renaming I have added a column named fileable_type. I now want to backfill that column so everything existing points on the User-model. What should I write in that column?
I have tried App\User and User without success.
App\User should be correct if that is how you have your User model namespaced. The polymorphich relationships use the getMorphClass() method on the Eloquent model. You can verify this by doing:
$u = new User();
dd($u->getMorphClass());
If you have not created a custom polymorphic type for the User model, then getMorphClass() should be equivalent to User::class.
Which relation to use in Laravel to bind two table through third?
When Doctors can be assigned to some Centers. The intermediate table will be as:
doctor_id | center_id
How to create model in Laravel for this case?
You don't need a model for the intermediate table, simply use attach
Example:
$center = Center::create();
$doctor = Doctor::find(1);
$doctor->centers()->attach($doctor->id);
This is a very simple example but should give you the idea, of how to approach it.
All of it of course requires you have set up your Center and Doctor model with the correct many to many relations
Doctor.php model:
public function centers()
{
return $this->belongsToMany(Doctor::class);
}
See the documentation, for more information.
You could obviously create a model called DoctorsCenter and create it manually by doing this, whenever you want to attach a relation.
DoctorsCenter::create(['center_id' => $center->id, 'doctor_id' => $doctor->id]);
I don't see any good reason for doing this, and would not recommend it.
You can use hasMany or belongsTo relationship of Laravel.
See the laravel documentation, for more information
I have this schema
All the relations here must be one-to-zero/one.
A user can be either an employee or a customer. The user_type ENUM gives me the type so I know where to go from there.
Then an employee can be either basic or a manager. The employee_type discriminator let's me know that.
How am I supposed to build the Eloquent Model relations?
Let's say I have a user that is an employee. I need to get it's common fields from the users table but also need to get common fields from employees table. Do I need to hard code, and know that when user_type=emp I need to select from the employees table? What if I need to add another user type later?
UPDATE
Would it make sense to change my schema into something simpler?
My problem is that by using, as suggested, polymorphic relations I would end up to something like this:
$user = new User::userable()->employable()->...
Would a schema in which I drop the employees table and have employee_managers and employee_basics linked straight to the users table?
this is an polymorphic relationship. but if you want to be easy, you need to fix some things.
in the table employees
- user_id
- employable_id
- employable_type enum(Manager, Basic) # References to the target model
.... this last two are for the polymorphic relation, this is the nomenclature
in the basics and managers table you could delete the user_id field, but you need an id field as increments type
and now in the model Employee you need to make this function
public function employable(){
return $this->morphTo();
}
I hope this works :)
There's a missing link I fail to understand.
I use migrations to create database tables and I define the relationships there. meaning.. if I have a person table and a job table and I need a one to many relationship between the person and jobs, I'd have the job table contain a "person_id".
When I seed data or add it in my app, I do all the work of adding the records setting the *_id = values etc.
but somehow I feel Laravel has a better way of doing this.
if I define that one to many relationship with the oneToMany Laravel Eloquent suports:
in my Person model.....
public function jobs()
{
return $this->hasMany('Jobs);
}
what's done on the database level? how do I create the migration for such table? Is Laravel automagically doing the "expected" thing here? like looking for a Jobs table, and having a "person_id" there?
Yep, Laravel is doing what you guess in your last paragraph.
From the Laravel documentation for Eloquent Relationships (with the relevant paragraph in bold):
For example, a User model might have one Phone. We can define this
relation in Eloquent:
class User extends Model {
public function phone()
{
return $this->hasOne('App\Phone');
}
}
The first argument passed to the hasOne method is the name of the
related model. Once the relationship is defined, we may retrieve it
using Eloquent's dynamic properties:
$phone = User::find(1)->phone;
The SQL performed by this statement
will be as follows:
select * from users where id = 1
select * from phones where user_id = 1
Take note that Eloquent assumes the foreign key of the relationship based on the model name. In this case, Phone model is assumed to use a user_id foreign key.
Also note that you don't actually have to explicitly set the foreign key indexes in your database (just having those "foreign key" columns with the same data type as the parent key columns is enough for Laravel to accept the relationship), although you should probably have those indexes for the sake of database integrity.
There is indeed support to create foreign key relationships inside migration blueprints and it's very simple too.
Here is a simple example migration where we define a jobs table that has a user_id column that references the id column on users table.
Schema::create('jobs', function($table)
{
$table->increments('id');
$table->integer('user_id')->unsigned();
$table->foreign('user_id')->references('id')->on('users');
});
You can also use some other methods that laravel provides such as onDelete() or onUpdate
Of course to understand better the options that are available to you please read the documentation here.
Edit:
Keep in mind that Eloquent is just using fluent SQL wrapper and behind the scenes there are just raw sql queries, nothing magical is happening, fluent just makes your life a lot easier and helpers you write maintainable code.
Take a look here about the Query Builder and how it works and also, as #Martin Charchar stated , here about Eloquent and relationships.