I have this question about Laravel:
I have a my model and my RestfulAPI controller.
Into the store() method I would check if I have an element that already has the field 'myField' (myField id different from 'id') equal to what I have to create. If it already exist then I would like to update, otherwise I would simply create (save())..
Have I to use find() method?
From my experience, you'll have to traverse table and check for uniqueness.
You can create your helper function and use something like array_unique function. Maybe it is worth checking how Validator class is checking that users entry is unique.
Currently we have firstOrCreate or firstOrNew, but I don't think they really fit your needs. For instance, firstOrCreate will try to locate a row by all attributes, not just some, so an update in this case wouldn't make sense. So I think you really would have to find it, but you can create a BaseModel and create a createOrUpdate method that could look like this:
This is untested code
class BaseModel extends Eloquent {
public function createOrUpdate($attributes, $keysToCheck = null)
{
// If no attributes are passed, find using all
$keysToCheck = $keysToCheck ?: $attributes;
if ($model = static::firstByAttributes(array_only($keysToCheck, $attributes))
{
$model->attributes = $attributes;
$model->save();
}
else
{
$model = static::create($attributes);
}
return $model;
}
}
This is an implementation of it:
class Post extends BaseModel {
public function store()
{
$model = $this->createOrUpdate(Input::all(), ['full_name']);
return View::make('post.created', ['model' => $model]);
}
}
Related
Thought I'd ask this as Laravel is the most elegant Framework I've come across and wondered if there was a "prettier way" of doing this.
I have a system which records books such that:
class Chapter extends Model
{
public function book()
{
return $this->belongsTo('\App\Book');
}
}
In the system there are number of other models which extend from "Book" such as "Novel", "Biography" etc. Is there a way for Eloquent to provide me with a correctly cast object given the right info (i.e. a namespaced class)? Currently, I am obtaining the book and the casting it using the function at https://gist.github.com/borzilleri/960035 which works but doesn't feel very "tidy".
I can see a few different options here. One would be to write your class like this:
class Chapter extends Model
{
public function book()
{
return $this->belongsTo('\App\Book');
}
public function biography()
{
return $this->belongsTo('\App\Biography')->where('type', 'biography');
}
public function novel()
{
return $this->belongsTo('\App\Novel')->where('type', 'novel');
}
}
You'd then need to know ahead of time which type of book it is though. Another would be to do something like this:
class Chapter extends Model
{
protected function parent_book()
{
return $this->belongsTo('\App\Book');
}
public function getBookAttribute()
{
$book = $this->parent_book;
if (!$book) return $book; // No related book.
if ($book->type == 'novel') return (Novel)$book;
if ($book->type == 'biography') return (Biography)$book;
return $book;
}
}
You still have to do all of the casting yourself, but at least it's all in one place and transparent to the rest of the app, as it can still just reference $chapter->book For this second solution, if you ever set $chapter->book = new Book(), you'd also need to make sure to make a setBookAttribute() function.
One more complicated possibility would be to create your own custom relationship type by extending the BelongsTo class and overriding getResults() to to the casting before returning the result. This would be pretty transparent from the outside and would let you still call $chapter->book() and treat it as a relationship.
This should be attributed to Joshua Dwire as he set me on the path to this solution. I was intrigued by his reference to extending the standard BelongsTo class and make it work for me. Ideally I want to be able to call a custom relationship:
$this->belongsToBook('\App\Book');
And for that function to return a correctly cast object.
Routing through the code I found that it was the trait HasRelationship used by Model which was responsible for returning the relationship. By changing that relationship we can change the implementation and therefore the returned object.
I also wanted to replicate the same methodology that Laravel employs so have mimiced it in my own app.
With all that in mind the first step is to create a new trait HasBookRelationship which can be used in a model to handle the call to $this->belongsToBook('\App\Book'):
trait HasBookRelationship
{
public function belongsToBook($related, $foreignKey = null, $ownerKey = null, $relation = null)
{
if (is_null($relation)) {
$relation = $this->guessBelongsToRelation();
}
$instance = $this->newRelatedInstance($related);
if (is_null($foreignKey)) {
$foreignKey = \Str::snake($relation).'_'.$instance->getKeyName();
}
$ownerKey = $ownerKey ?: $instance->getKeyName();
//We change the return relationship here
**return new BelongsToBook(
$instance->newQuery(), $this, $foreignKey, $ownerKey, $relation
);**
}
}
This is simply copied from the existing belongsTo method in the HasRelationships trait. The key thing here is that we are going to return a custom relationship BelongsToBook and use that to override what is returned. The last line of the method is changed to return our desired relationship class.
The class we use is extended from BelongsTo but we change the get method to cast the object before returning it.
class BelongsToBook extends BelongsTo
{
public function __construct(Builder $query, Model $child, $foreignKey, $ownerKey, $relationName)
{
parent::__construct($query, $child, $foreignKey, $ownerKey, $relationName);
}
public function get($columns = ['*'])
{
$objs = $this->query->get($columns);
//iterate over the collated objects...
$objs->transform(function($item)
{
//..and return a cast object with whatever method you want
return castTheCorrectObject($item);
});
return $objs;
}
}
castTheCorrectObject can be any casting function you like perhaps set up as a helper or another method in the relationship.
Once these are set up, we can empoy it in our own Model:
class Author extends Model
{
use HasBookRelationship;
public function books()
{
return $this->belongsToBook('\App\Book');
}
}
This will return a collection of correctly cast objects and maintains the relationship.
One thing did puzzle me though. The method I overrode in my BelongsToBook class was get() and not getResults() as suggested by Joshua. get() is defined in Relation and is inherited by BelongsTo where as getResults() is defined in BelongsTo. I'm not sure what the difference between getResults() and get() is nor why I had to override get() rather than getResults(). If anyone can shed any light , it would be appreciated.
I'm currently rebuilding my vanilla-PHP-App with Laravel and I have the following problem.
I have multiple database-tables, that represent word categories (noun, verb, adverb, ...). For each table I created a separate Model, a route::resource and a separate resource-Controller. For example:
NomenController.php
public function show($id)
{
$vocab = Nomen::find($id);
return view('glossarium.vocab_update', compact('vocab'));
}
and
VerbController.php
public function show($id)
{
$vocab = Verb::find($id);
return view('glossarium.vocab_update', compact('vocab'));
}
...which are essentially the same except the Model class.
I don't want to create a separate Controller for each model, that does exactly the same. What would be the most simple and elegant way to solve this?
Should I just create a VocabController.php and add a parameter for the Model-name like:
Route::resource('/vocab/{category}', 'VocabController');
and then add a constructor method in this controller like
public function __construct ($category) {
if ($category == 'nomen') {
$this->vocab = App\Nomen;
}
else if ($category == 'verb') {
$this->vocab = App\Verb;
}
}
I wonder if there is a simpler method to do that. Can I somehow do this with Route Model Binding?
Thanks in advance
Simply create a trait like this in App\Traits, (you can name it anything... Don't go with mine though... I feel its pretty lame... :P)
namespace App\Traits;
trait CommonControllerFunctions {
public function show($id) {
$modelObject = $this->model;
$model = $modelObject::find($id);
return view('glossarium.vocab_update', compact('model'));
}
}
and in your NomenController and VerbController, do this:
use App\Traits\CommonControllerFunctions;
class NomenController {
use CommonControllerFunctions;
protected $model = Nomen::class;
}
and
use App\Traits\CommonControllerFunctions;
class VerbController {
use CommonControllerFunctions;
protected $model = Verb::class;
}
Note: Please note that this example is just a work-around for your particular situation only... Everyone practices code differently, so this method might not be approved by all...
I think the simpliest way it to create only one controller, eg VocabController with methods nomen, verb and whatever you want.
Routes:
Route::get('/vocab/nomen/{nomen}', 'VocabController#item');
Route::get('/vocab/verb/{verb}', 'VocabController#item');
And the model binding:
Route::model('nomen', 'App\Nomen');
Route::model('verb', 'App\Varb');
Then your method shoud look like that:
public function item($item)
{
return view('glossarium.vocab_update', $item);
}
Keep in mind, that $item is already fetched model from the database.
Laravel documentation suggests the following way to set up an eloquent model:
$user = user::with($conditions)->first();
What if I want to set up my eloquent model inside the model itself:
$user = new user();
$user->setup($conditions);
// class definition
class user extends Eloquent{
public function setup($conditions){
// load current object with table data
// something like
$this->where($conditions)->first();
// previous line output is dangling, is ok to assign it to $this variable?
}
}
If you're extending from Eloquent model, you may try the following approach. I assume you have a unique id column.
public function setup($conditions)
{
$model = self::with($conditions)->first();
if (! is_null($model)) {
$this->exists = true;
$this->forceFill(self::find($model->id)->toArray());
}
return $this;
}
Hope this solve your issue.
Before anyone asks, I've looked into CRUD generators and I know all about the Laravel Resource routes, but that's not exactly what I'm pulling for here.
What I'm looking to do is create one Route with a couple parameters, and one global class that (uses/extends?) the Model controller for simple CRUD operations. We have 20 or so Models and creating a Resource Controller for each table would be more time consuming than finding a way to create a global CRUD class to handle all "api" type calls and any ajax json request like a create / update / destroy statement.
So my question is what is the cleanest and best way to structure a class to handle all CRUD requests for every Model we have without having to have a resource controller for every model? I've tried researching this and can't seem to find any links except ones to CRUD generators and links describing the laravel Resource route.
The easiest way would be to do the following:
Add a route for your resource controller:
Route::resource('crud', 'CrudController', array('except' => array('create', 'edit')));
Create your crud controller
<?php namespace App\Http\Controllers;
use Illuminate\Routing\Controller;
use App\Models\User;
use App\Models\Product;
use Input;
class CrudController extends Controller
{
const MODEL_KEY = 'model';
protected $modelsMapping = [
'user' => User::class,
'product' => Product::class
];
protected function getModel() {
$modelKey = Input::get(static::MODEL_KEY);
if (array_key_exists($modelKey, $this->modelsMapping)) {
return $this->modelsMapping[$modelKey];
}
throw new \InvalidArgumentException('Invalid model');
}
public function index()
{
$model = $this->getModel();
return $model::all();
}
public function store()
{
$model = $this->getModel();
return $model::create(array_except(Input::all(), static::MODEL_KEY));
}
public function show($id)
{
$model = $this->getModel();
return $model::findOrFail($id);
}
public function update($id)
{
$model = $this->getModel();
$object = $model::findOrFail($id);
return $object->update(array_except(Input::all(), static::MODEL_KEY));
}
public function destroy($id)
{
$model = $this->getModel();
return $model::remove($id);
}
}
Use your new controller :) You have to pass the model parameter that will contain the model key - it must be one of the allowed models in the whitelist. E.g. if you want to get a User with id=5 do
GET /crud/5?model=user
Please keep in mind that it's as simple as possible, you might need to make the code more sophisticated to match your needs.
Please also keep in mind that this code has not been tested - let me know if you see any typos or have some other issues. I'll be more than happy to get it running for you.
Unless you want to implement CRUD manually, consider to integrate a ready-made datagrid such as phpGrid.
Check out integration walkthrough: http://phpgrid.com/example/phpgrid-laravel-5-twitter-bootstrap-3-integration/ No models are required and the code is minimum. It can almost do anything.
A basic working CRUD:
// in a controller
public function index()
{
$dg = new \C_DataGrid("SELECT * FROM orders", "orderNumber", "orders");
$dg->enable_edit("FORM", "CRUD");
$dg->display(false);
$grid = $dg -> get_display(true);
return view('dashboard', ['grid' => $grid]);
}
You need one generic class for all CRUD operations and there are many ways to achieve that and one rule for all may not fit but you may try the approach that I'm going to describe now. This is an abstract idea, you need to implement it, so at first, think the URI for all CRUD operations. In this case you must follow a convention and it could be something like this:
example.com/user/{id?} // get all or one by id (if id is available in the URI)
example.com/user/create // Show an empty form
example.com/user/edit/10 // Show a form populated with User model
example.com/user/save // Create a new User
example.com/user/save/10 // Update an existing User
example.com/user/delete/10 // Delete an existing User
In ths case the user could be something else to specify the name of the model for example, example.com/product/create and keeping that on mind, you need to declare routes as given below:
Route::get('/{model}/{id?}', 'CrudController#read');
Route::get('/{model}/create', 'CrudController#create');
Route::get('/{model}/edit/{id}', 'CrudController#edit');
Route::post('/{model}/save/{id?}', 'CrudController#save');
Route::post('/{model}/delete/{id}', 'CrudController#delete');
Now, in your app\Providers\RouteServiceProvider.php file modify the boot method and make it look like this:
public function boot(Router $router)
{
$model = null;
$router->bind('model', function($modelName) use (&$model, &$router)
{
$model = app('\App\User\\'.ucfirst($modelName));
if($model)
{
if($id = $router->input('id'))
{
$model = $model->find($id);
}
return $model ?: abort(404);
}
});
parent::boot($router);
}
Then declare your CrudController as given below:
class CrudController extends Controller
{
protected $request = null;
public function __construct(Request $request)
{
$this->request = $request;
}
public function read($model)
{
return $model->exists ? $model : $model->all();
}
// Show either an empty form or a form
// populated with the given model atts
public function createOrEdit($model)
{
$classNameArray = explode('\\', get_class($model));
$className = strtolower(array_pop($classNameArray));
$view = view($className . '.form');
$view->formAction = "$className/save";
if(is_object($model) && $model->exists)
{
$view->model = $model;
$view->formAction .= "/{$model->id}";
}
return $view;
}
public function save($model)
{
// Validation required so do it
// Make sure each Model has $fillable specified
return $this->model->fill($this->request)->save();
}
public function delete($model)
{
return $this->model->delete();
}
}
Since same form is used to creating and updating a model, use something like this to create a form:
<form action="{{url($formAction)}}" method="POST">
<input
type="text"
class="form-control"
name="first_name" value="{{old('first_name', #$model->first_name)}}"
/>
<input type="Submit" value="Submit" />
{!!csrf_field()!!}
</form>
Remember that, each form should be in a directory corresponding to the model, for user add/edit, form should be in views/user/form.blade.php and for product model use views/product/form.blade.php and so on.
This will work and don't forget to add validation before saving a model and validation could be done inside the model using model events or however you want. This is just an idea but probably not the best way to it.
I'd like to be able to add a custom attribute/property to an Laravel/Eloquent model when it is loaded, similar to how that might be achieved with RedBean's $model->open() method.
For instance, at the moment, in my controller I have:
public function index()
{
$sessions = EventSession::all();
foreach ($sessions as $i => $session) {
$sessions[$i]->available = $session->getAvailability();
}
return $sessions;
}
It would be nice to be able to omit the loop and have the 'available' attribute already set and populated.
I've tried using some of the model events described in the documentation to attach this property when the object loads, but without success so far.
Notes:
'available' is not a field in the underlying table.
$sessions is being returned as a JSON object as part of an API, and therefore calling something like $session->available() in a template isn't an option
The problem is caused by the fact that the Model's toArray() method ignores any accessors which do not directly relate to a column in the underlying table.
As Taylor Otwell mentioned here, "This is intentional and for performance reasons." However there is an easy way to achieve this:
class EventSession extends Eloquent {
protected $table = 'sessions';
protected $appends = array('availability');
public function getAvailabilityAttribute()
{
return $this->calculateAvailability();
}
}
Any attributes listed in the $appends property will automatically be included in the array or JSON form of the model, provided that you've added the appropriate accessor.
Old answer (for Laravel versions < 4.08):
The best solution that I've found is to override the toArray() method and either explicity set the attribute:
class Book extends Eloquent {
protected $table = 'books';
public function toArray()
{
$array = parent::toArray();
$array['upper'] = $this->upper;
return $array;
}
public function getUpperAttribute()
{
return strtoupper($this->title);
}
}
or, if you have lots of custom accessors, loop through them all and apply them:
class Book extends Eloquent {
protected $table = 'books';
public function toArray()
{
$array = parent::toArray();
foreach ($this->getMutatedAttributes() as $key)
{
if ( ! array_key_exists($key, $array)) {
$array[$key] = $this->{$key};
}
}
return $array;
}
public function getUpperAttribute()
{
return strtoupper($this->title);
}
}
The last thing on the Laravel Eloquent doc page is:
protected $appends = array('is_admin');
That can be used automatically to add new accessors to the model without any additional work like modifying methods like ::toArray().
Just create getFooBarAttribute(...) accessor and add the foo_bar to $appends array.
If you rename your getAvailability() method to getAvailableAttribute() your method becomes an accessor and you'll be able to read it using ->available straight on your model.
Docs: https://laravel.com/docs/5.4/eloquent-mutators#accessors-and-mutators
EDIT: Since your attribute is "virtual", it is not included by default in the JSON representation of your object.
But I found this: Custom model accessors not processed when ->toJson() called?
In order to force your attribute to be returned in the array, add it as a key to the $attributes array.
class User extends Eloquent {
protected $attributes = array(
'ZipCode' => '',
);
public function getZipCodeAttribute()
{
return ....
}
}
I didn't test it, but should be pretty trivial for you to try in your current setup.
I had something simular:
I have an attribute picture in my model, this contains the location of the file in the Storage folder.
The image must be returned base64 encoded
//Add extra attribute
protected $attributes = ['picture_data'];
//Make it available in the json response
protected $appends = ['picture_data'];
//implement the attribute
public function getPictureDataAttribute()
{
$file = Storage::get($this->picture);
$type = Storage::mimeType($this->picture);
return "data:" . $type . ";base64," . base64_encode($file);
}
Step 1: Define attributes in $appends
Step 2: Define accessor for that attributes.
Example:
<?php
...
class Movie extends Model{
protected $appends = ['cover'];
//define accessor
public function getCoverAttribute()
{
return json_decode($this->InJson)->cover;
}
you can use setAttribute function in Model to add a custom attribute
Let say you have 2 columns named first_name and last_name in your users table and you want to retrieve full name. you can achieve with the following code :
class User extends Eloquent {
public function getFullNameAttribute()
{
return $this->first_name.' '.$this->last_name;
}
}
now you can get full name as:
$user = User::find(1);
$user->full_name;
In my subscription model, I need to know the subscription is paused or not.
here is how I did it
public function getIsPausedAttribute() {
$isPaused = false;
if (!$this->is_active) {
$isPaused = true;
}
}
then in the view template,I can use
$subscription->is_paused to get the result.
The getIsPausedAttribute is the format to set a custom attribute,
and uses is_paused to get or use the attribute in your view.
in my case, creating an empty column and setting its accessor worked fine.
my accessor filling user's age from dob column. toArray() function worked too.
public function getAgeAttribute()
{
return Carbon::createFromFormat('Y-m-d', $this->attributes['dateofbirth'])->age;
}