I am using the nested set for the first time with arcanedev/laravel-nestedset.
I am following a tutorial and setup my model accordingly. But when I call the create method it gives me following error.
Symfony\Component\Debug\Exception\FatalThrowableError thrown with message "Argument 1 passed to App\Models\AmazonCategoryIndices::appendToNode() must be an instance of Arcanedev\LaravelNestedSet\Contracts\Nodeable, instance of App\Models\AmazonCategoryIndices given, called in E:\TestProjects\pld-v4-admin\vendor\arcanedev\laravel-nestedset\src\NodeTrait.php on line 1074".
The data I am using is the same as it is in tutorial.
Here is my Model
==================================
namespace App\Models;
use Arcanedev\LaravelNestedSet\NodeTrait;
use Illuminate\Database\Eloquent\Model;
class AmazonCategoryIndices extends Model
{
protected $table = 'amazon_category_indices';
protected $guarded = [];
use NodeTrait;
}
==================================
foreach ($shops as $key => $value) {
AmazonCategoryIndices::create($value);
}
Based on the error message, the parameter passed to App\Models\AmazonCategoryIndices::appendToNode() is an instance of App\Models\AmazonCategoryIndices while it expects an instance of Arcanedev\LaravelNestedSet\Contracts\Nodeable
So, adding the contract Arcanedev\LaravelNestedSet\Contracts\Nodeable in your model should resolve your error.
namespace App\Models;
use Arcanedev\LaravelNestedSet\NodeTrait;
use Illuminate\Database\Eloquent\Model;
use Arcanedev\LaravelNestedSet\Contracts\Nodeable;
class AmazonCategoryIndices extends Model implements Nodeable
{
protected $table = 'amazon_category_indices';
protected $guarded = [];
use NodeTrait;
}
When you said: "ARCANEDEV/LaravelNestedSet" package uses "lazychaser/laravel-nestedset " at its backend.
...It's not. They are two different packages and doesn't work the same. Arcanedev credited lazychaser for his work on which he probably used a lot of resources.
Now that's been said, I wouldn't recommend using arcanedev's package since it doesn't seems to be maintenained.
Related
I have a navigation model that can have many items associated with it:
<?php
namespace App\Models;
use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Relations\HasMany;
use JetBrains\PhpStorm\ArrayShape;
use Laravel\Scout\Searchable;
class Navigation extends Model
{
use HasFactory;
use Searchable;
protected $guarded = [];
public function navigation_items(): HasMany
{
return $this->hasMany(NavigationItem::class);
}
}
The navigation item model looks like this
<?php
namespace App\Models;
use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Relations\BelongsTo;
use Illuminate\Database\Eloquent\Relations\MorphTo;
class NavigationItem extends Model
{
use HasFactory;
protected $guarded = [];
public function navigation(): BelongsTo
{
return $this->belongsTo(Navigation::class);
}
public function navigatable(): MorphTo
{
return $this->morphTo();
}
}
Now an item can either be of type Page or Blog, in this case the Page model looks like this:
<?php
namespace App\Models;
use Illuminate\Database\Eloquent\Collection;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Relations\MorphOne;
use JetBrains\PhpStorm\ArrayShape;
use Laravel\Scout\Searchable;
class Page extends Model
{
protected $guarded = [];
public function navigatable(): MorphOne
{
return $this->morphOne(NavigationItem::class, 'navigatable');
}
}
When I try to save a navigation model and associate it with a item, the following error appears:
SQLSTATE[HY000]: General error: 1364 Field 'navigatable_type' doesn't have a default value
I save the model like this:
foreach ($this->selected as $id) {
$this->navigation->navigation_items()->create([
'navigation_id' => $this->navigation->id,
]);
Where $this->selected is the navigation id, it should automatically get the correct navigatable_type and navigatable_id, but this doesn't seem to be working.
passing in the type and id manually works, but this kinda defeats the point of a polymorphic relationship.
any ideas?
On NavigationItem model, since you defined polymorphic relation as 'navigatable' it is expected that NavigationItem model's table contains navigatable_type and navigatable_id. First please ensure this checks out.
Creating records through relation's base function is not a valid method. It is not clear what you are trying to achieve there but when you want to set relation there is two standard way of achieving it:
1- Associate
When a relation is defined as belongsTo, you may use associate() function. Like so:
$account = Account::find(10);
$user->account()->associate($account);
2- Attach
Attach is used when relation is defined belongsToMany (pivot). It allows you to attach multiple records to a model instance/record.
$user = User::find(1);
$user->roles()->attach($roleId);
So if you want to set a 'navigatable' to a Navigation instance, you may:
$somePageInstance=Page::find(55);
$nagivation->navigatable()->associate($somePageInstance)
$nagivation->save();//remember to save, otherwise it won't be
We created three tables called 'users', 'corso', 'iscrizione' which is a middle table between users and corso, it has two attributes that are PK/FK. We successfully built the methods to create/update/read/delete rows inside 'corso' and we populated manually 'users'. We are trying to realize a create method for iscrizione but when we try to test it on Postman it gives us an error: 🧨 Illegal offset type
We learned that Eloquent doesn't support composite keys (which is a shame) and the only way to bypass this limit is to override the method setKeysForSaveQuery() from Eloquent inside the model of our table.
Sadly it gives us this error:
Method 'App\Models\Iscrizione::setKeysForSaveQuery()' is not compatible with method 'Illuminate\Database\Eloquent\Model::setKeysForSaveQuery()'.intelephense(1038)
Iscrizione Model
<?php
namespace App\Models;
use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Builder;
class Iscrizione extends Model
{
use HasFactory;
protected $hidden = ['idCorso', 'idUtente'];
protected $table = 'iscrizione';
protected function setKeysForSaveQuery(Builder $query){
return $query->where('idCorso', $this->getAttribute('idCorso'))
->where('idUtente', $this->getAttribute('idUtente'));
}
}
You need to make it match change the below code
protected function setKeysForSaveQuery(Builder $query){
to
protected function setKeysForSaveQuery($query){
Im new to laravel, i am trying to query a specific table in my DB. I only have 1 data table and the standard user auth tables. I am getting a error: BadMethodCallException
Call to undefined method App\Figures::table().
Model
namespace App;
use Illuminate\Database\Eloquent\Model;
class Figures extends Model
{
}
controller
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use App\Figures;
class figuresController extends Controller
public function figurespag2() {
$dummyDetails = Figures::table('figures')->where('name', 'batman');
return view ( 'pagination2.index' )->withUsers($dummyDetails);
}
route
Route::get ( '/pagination2', 'figuresController#figurespag2' );
I know it's going to be something obvious, but I am new to this.
this is wrong
$dummyDetails = Figures::table('figures')->where('name', 'batman');
Method 1---------- laravel eloquent
Model
namespace App;
use Illuminate\Database\Eloquent\Model;
class Figures extends Model
{
protected $table = 'figures';
}
Controller
$dummyDetails = Figures::where('name', 'batman')->get();
and
Method 2 ---------- laravel Query Builder
$dummyDetails = \DB::table('figures')->where('name', 'batman')->get();
Use this you not need to define table name
public function figurespag2() {
$dummyDetails = Figures::where('name', 'batman')->get();
return view ( 'pagination2.index' )->withUsers($dummyDetails);
}
First you may need to know laravel model rules.
If you create a table name like "figures" (plural) you need to create its model by Figure (singular).
if you create a table other then this rule then you have to mentioned table name in model like this.
protected $table = "table_name";
you can access table with where condition in controller like this.
public function figurespag2() {
$dummyDetails = Figure::where('name', 'batman')->get();
return view ( 'pagination2.index' )->withUsers($dummyDetails);
}
Hope this may help you.
I'm trying to map an existing one to one relationship in my database using Eloquent following the Laravel tutorial and I'm receiving a string conversion error.
I have to use an existing database that is used by other program, so I can't change the database structure.
This are my models:
Pessoa.php
namespace App\Models;
use Illuminate\Support\Facades\Log; use
Illuminate\Database\Eloquent\Model;
class Pessoa extends Model {
protected $table = 'pessoas';
public function estadoCivil(){
//return EstadoCivil::find($this->estado_civil_id);
return $this->hasOne('App\Models\EstadoCivil', 'estado_civil_id');
}
}
EstadoCivil.php
namespace App\Models;
use Illuminate\Database\Eloquent\Model;
class EstadoCivil extends Model
{
protected $table = 'estados_civis';
}
I don't know what to do, but I am getting this error:
Object of class Illuminate\Database\Eloquent\Relations\HasOne could not be converted to string
I searched for typos and other mistakes and couldn't find anything. I'm not even trying to convert anything to string to get this error, the types in the database for the id and the foreign key are the same too.
Here is the controller that is using it:
namespace App\Http\Controllers;
use App\Http\Controllers\Controller;
use Illuminate\Http\Request;
use App\Models\Pessoa as Pessoa;
class Eu extends Controller
{
public function __construct()
{
}
public function dadosCompletos(Request $request){
return Pessoa::find($request->user()->pessoa_id)->estadoCivil();
}
}
If I just use return Pessoa::find($request->user()->pessoa_id) it works fine and returns the data correctly from pessoas table.
you are returning the relation not the actual EstadoCivil collection.
use return Pessoa::find($request->user()->pessoa_id)->estadoCivil without parenthesis
You have to use ->estadoCivil instead of ->estadoCivil()
My repository is like this :
<?php
namespace App\Repositories;
use App\Models\UsersBank;
use Illuminate\Contracts\Container\Container;
use Rinvex\Repository\Repositories\EloquentRepository;
class UserBankRepository extends EloquentRepository
{
public function __construct(Container $container)
{
$this->setContainer($container)
->setModel(UsersBank::class)
->setRepositoryId('rinvex.repository.uniqueid');
}
}
My model is like this :
<?php
namespace App\Models;
use Illuminate\Database\Eloquent\Model;
class UsersBank extends Model
{
protected $fillable = ['user_id','bank_id','status','account_name','account_number'];
}
My service is like this :
public function setMain($id)
{
$param['status'] = 1;
$this->user_bank_repository->update($id, $param);
}
When setMain function executed, It will update field status = 1 by the id
I want to update status = 1 to all record. So, Not by id
How can I do it?
Note :
Update which I mean here is the update via the repository. Not update through a model
Look into the source code of update() method of the package and you'll see it's impossible, so you'll need to use foreach() and create a bunch of queries.
My opinion is this and similar packages are useless since they are still using Eloquent and do not provide any handful functionality. I'd recommend you to use Eloquent directly to update all rows with just one query:
Model::query()->update(['status' => 1]);