I'm learning Laravel 5 from the Laracast course: Laravel 5 Fundamentals, using PhpStorm IDE.
I have issue with the static methods of model, such as where(), find(), and findOrFail().
When I use these methods PhpStorm shows:
Method 'findOrFail' not found in class App\Article.
My model is Article, and the method all() works.
How can I solve this?
Now you can use Laravel IDE helper and for properly recognition of Model methods (i.e. paginate, findOrFail) add comment for your model class
/** #mixin \Eloquent */
Thank you for interesting. But I solved, I use query() method:
public function show($id)
{
$article = Article::query()->findOrFail($id);
return view('articles.show', compact('article'));
}
It's because those methods are hitting the magic method __call which phpStorm does not know how to follow. It's really just a shortcut for the following...
$article = new \App\Article;
$query = $article->newQuery();
$query->where('column', 'value')->get();
I had the same problem, it turned at the end that I had forgotten a single quotation mark in show.balde.php when I extended the master page:
#extends('app)
when I finally realized it and fixed it everything worked perfectly:
this worked: $article = Article::findOrFail($id);
and this also worked: $article = Article::query()->findOrFail($id);
I hope this helps (I tried this in Laravel 5.3).
Related
I'm facing strange case. I face an error in production env not while in dev it's working fine.
Development:
Laravel 5.4.28
PHP 7.0.13
MYSQL 5.7.17
Production:
Laravel 5.4.28
PHP 7.2.1
MYSQL 5.7.20
In implementation code. I used:
namespace App;
use Illuminate\Support\Facades\Storage;
use Laravel\Scout\Searchable;
use Illuminate\Database\Eloquent\Model;
class Artwork extends Model
{
use Searchable;
In development it works fine. But in production it gives me this error:
count(): Parameter must be an array or an object that implements Countable
in Builder.php (line 936)
as you can see in this pic:
Any idea what is the reason behind this? and how to fix?
Put this code at the beginning of your route file, it will work fine
if(version_compare(PHP_VERSION, '7.2.0', '>=')) {
error_reporting(E_ALL ^ E_NOTICE ^ E_WARNING);
}
This is a documented change in PHP 7.2. You need to either update Laravel to 5.6 or downgrade PHP to version 7.1.
Replace
$originalWhereCount = count($query->wheres);
by
$originalWhereCount = count((array)$query->wheres);
in
\vendor\laravel\framework\src\Illuminate\Database\Eloquent\Builder.php
I was facing similar issue in Laravel 5.6. Where I was getting error for object based array. I knew that data in that particular variable will always remain object so i used to convert the object to array. Here is code sample:
$objectData = (array)$objectData;
echo "Total Elements in array are: ".count($objectData);
My server was on PHP 7.1 when I updated to PHP 7.2 I got the same issue.
After searching I found why this occurs. (This occurs because of a PHP update.).
so in my case, the error is solved by typecasting.
I just update all code where I used to count
Before
//this is before
count($adminDetails)
After updated
//after update all i typecast all the code where i used count
count((array)$adminDetails)
Goodluck
This error occurs because you are using a higher PHP version and your Laravel application is on an older PHP version.
✅ Simple solution:
Open: app/Providers/AppServiceProvider.php
And in: public function register() { ... } function add following code:
if(version_compare(PHP_VERSION, '7.2.0', '>=')) {
error_reporting(E_ALL ^ E_NOTICE ^ E_WARNING);
}
In php 7.2+ count does not work on relation objects, you need to use:
$model->relation()->exists()
Not this (less than php 7.2):
count($model->relation)
i ran into the same problem (PHP 7.2 + Laravel 5.3) but i don't see any "good" answers here. For me, the problem occurs when i tried to start a Builder from a scope method on the model: SomeModel::forUser() calls scopeForUser(). Trying to build a new Query, it trips on a count($this->wheres) that gets no initial value (null). Because the magic static call to the scope starts the builder, no other conditions have been placed in the object so the property is still null at that point.
i thought it's worth sharing my solution first, then perspective on why i consider it better than Ben's answer. It's not personal, i just disagree.
Solution
i took a cue from this answer about overriding some of the core Illuminate\Database classes...
Extend Illuminate\Database\Eloquent\Model
Mine is App\Overrides\Database\Eloquent\Model
Extend Illuminate\Database\Eloquent\Builder
Mine is App\Overrides\Database\Eloquent\Builder
Extend Illuminate\Database\Query\Builder
Can you guess? App\Overrides\Database\Query\Builder
Tell Laravel to use YOUR Eloquent\Model:
config/app.php 'aliases' array, replace the 'Eloquent' value
with your Eloquent\Model FQN
My Model:
namespace App\Overrides\Database\Eloquent;
/*
* Notes:
* * Using replacement Query\Builder with ALIAS
* * Use of Builder in this class is MY Eloquent\Builder
*/
use App\Overrides\Database\Query\Builder as QueryBuilder;
use Illuminate\Database\Eloquent\Model as EloquentModel;
class Model extends EloquentModel
{
public function newEloquentBuilder($query)
{
return new Builder($query);
}
protected function newBaseQueryBuilder()
{
$conn = $this->getConnection();
$grammar = $conn->getQueryGrammar();
return new QueryBuilder($conn, $grammar, $conn->getPostProcessor());
}
}
My Eloquent\Builder:
namespace App\Overrides\Database\Eloquent;
use Illuminate\Database\Eloquent\Builder as EloquentBuilder;
class Builder extends EloquentBuilder
{
public function __construct($query)
{
parent::__construct($query);
/*
* FIX #1: Set properties treated AS arrays
* to empty arrays on construct.
*/
$this->wheres = [];
// Any other properties treated as arrays should also be initialized.
}
}
My Query\Builder:
namespace App\Overrides\Database\Query;
use Illuminate\Database\Query\Builder as QueryBuilder;
class Builder extends QueryBuilder
{
public function __construct()
{
parent::__construct(...func_get_args());
/*
* FIX #2: Set properties treated AS arrays
* to empty arrays on construct.
*/
$this->wheres = [];
// Any other properties treated as arrays should also be initialized.
}
}
This safely preserves the framework's functionality, since the only actual change you're making is initializing properties that should have been in the first place. Everything else will pass instanceof checks used for dynamic loading and dependency injection.
Opinion
While i agree with #ben-harold about every comment he made saying "NEVER edit vendor code," i disagree with the "solution." It's an oversimplification to a much more complex problem.
Upgrade Laravel: to ensure support for PHP 7.2, jumping up several minor versions - if not major releases - is impractical for a lot of teams. As a long term objective, yes of course. As something i can do to get rid of the bug for my deadline? Nope. Upgrading takes a lot of planning and frequently a lot of rewrites as structures, names, and functionality change. It's something to prioritize, but not a need-it-now answer.
Downgrade PHP: same problem. Downgrading into PHP 5.x means A) PHP is EOL, which may be a deal breaker for a lot of customers who have security policies, and B) any usage of PHP 7.x language features have to be scrapped. As with upgrading the framework this is very likely to cause a lot of headaches. It's also an even less useful solution, since walking backward in the language just puts you farther behind and will require more long-term effort.
place the below line ob code before the class name in your controllers
if (version_compare(PHP_VERSION, '7.2.0', '>=')) {
// Ignores notices and reports all other kinds... and warnings
error_reporting(E_ALL ^ E_NOTICE ^ E_WARNING);
// error_reporting(E_ALL ^ E_WARNING); // Maybe this is enough
}
I was facing the same issue with an external created table (Not using migration or command),
After creating the model, I just assigned a table name, but the problem was in my model protected $fillable where I assign string instead of array and error occurred.
There is 2 possible solution for that.
Assign an array to your protected $fillable = ['filed1', 'filed2'];
Remove protected $fillable completely (Not Recommended)
class Abc extends Model
{
protected $table = 'cities';
protected $fillable = ['field1','field2', ...];
}
Model looking for countable parameter:
class ClassName extend Model {
protected $fillable=['column_name']; // column in DB of Model is in array
}
Before
count($collection['colors'])
Error:Expected type 'Countable|array'. Found 'string'
After
count((array)$collection['colors'])
It works for me!
'vendor\laravel\framework\src\Illuminate\Database\Eloquent\Builder.php' to:
$originalWhereCount = is_array($query->wheres) ? count($query->wheres) : 0;
I;m using laravel 6.x
for this case you can use this way:
$id = \DB::table('xxxx')->where('id', $id)->count();
I Solve this in Laravel 5.6
// in controller
public function index()
{
$todos = Todo::all();
return view('todos.index')->with(['todos' => $todos]);
}
// in view page
#if(count($todos) > 0)
#foreach($todos as $todo)
<div class="well">
<h3>{{$todo->text}}</h3>
<span class="label label-danger">{{$todo->due}}</span>
</div>
#endforeach
#endif
I try to build a path for a model on laravel
I created a function in my model:
public function path()
{
return App\Helper\GeneralController::getURL($this);
}
with dd(App\Helper\GeneralController::getURL($this)) test I got the right answer. (output is a URL)
but in view with the call: $article->path I get this error:
App\Article:: path must return a relationship instance.
What is wrong?
You need to call it:
$article->path()
When you do $article->path, you're trying to use Eloquent relationship which you don't have.
I know this has already been answered and accepted. However, if the OP did want to use a property accessor rather than a method use the "get{property name}Attribute" syntax of Laravel to create a custom attribute.
Here is what it would look like for this specific case:
public function getPathAttribute()
{
return App\Helper\GeneralController::getURL($this);
}
using this approach "path" can now be called as an attribute and will not be resolved to a relationship using the syntax:
$article->path;
You're calling a relationship.
$article->path
To call the method, use '()', like so,
$article->path()
I faced that error when I forgot to write return before relation in the model!
check it out now!
path() is method not object element you need to call as method
$article->path();
Laravel 9 introduced a new way to define accessors/mutators within a model using Illuminate\Database\Eloquent\Casts\Attribute.
https://laravel.com/docs/9.x/eloquent-mutators#defining-an-accessor
public function path(): Attribute
{
return new Attribute(fn () => GeneralController::getURL($this));
}
For future visitors from Google, all the other answers can be applicable in certain scenarios, but you might want to also look if your method access modifier, if your method is protected and you try to call it you will be welcome with this error. You need change your method to public.
I am working with laravel 4.2 and have table in db with property is_active.
When I try to access this model property:
$model->is_active
I am getting following error:
Relationship method must return an object of type Illuminate\Database\Eloquent\Relations\Relation
So question is how to access this property?
Please do not recommend to rename this field in the database if possible because this is already existing database in production.
Here is my model class:
class Position extends \Eloquent {
protected $table = "hr_positions";
protected $fillable = ['slug', 'info_small', 'info_full', 'is_active', 'start_date', 'end_date', 'tags', 'user_create_id', 'user_update_id'];
use \MyApp\Core\StartEndDateTrait;
public function postulations(){
return $this->hasMany('Postulation', 'position_id', 'id');
}
}
Latest notice:
All this error ocurrs on a page where I am creating my entity. In the controller before forwarding to the page I am doing:
$position = new \Position();
and then, for example, following code produce error as well:
dd(($position->getAttribute('is_active')));
but if I replace $position = new \Position(); with
$position = \Position::first();
error is gone?
What is going on here?????
Laravel does a lot of magic behind the scenes, as in, calls a lot of php magic methods.
If a called property is not defined, __call is invoked which in Eloquent calls getAttribute().
Steps taken by getAttribute($key) are
Is there a database field by this key? If so, return it.
Is there a loaded relationship by this key? If so, return it.
Is there a camelCase method of this key? If so, return it. (is_active looks for isActive method)
Returns null.
The only time that exception is thrown is in step 3.
When you create a new instance, eloquent has no idea what kind of fields it has, so if you have a method by the same name, it will always throw a relation error, this seems to be the case in both Laravel4 and Laravel5.
How to avoid it? Use the getAttributeValue($key) method. It has no relation checks and returns null by default.
Alternatively you can also add a get mutator for your field.
I have found a hack for this. Still not ideal but at least I have some solution. Better any than none.
So This code produce problem:
$position = new \Position();
if($position->is_active){
//
}
and this one works fine, this is solution even hacky but solution:
$position = new \Position(['is_active' => 0]);
if($position->is_active){
//
}
I will wait if someone give better, cleaner solution. If no one comes in next few days I will accept mine.
Current Paginator is using ?page=N, but I want to use something else. How can I change so it's ?sida=N instead?
I've looked at Illuminate\Pagination\Environment and there is a method (setPageName()) there to change it (I assume), but how do you use it?
In the Paginator class there is a method the change the base url (setBaseUrl()) in the Environment class, but there is no method for setting a page name. Do I really need to extend the Paginator class just to be able to change the page name?
Just came across this same issue for 5.1 and you can pass the page name like this:
Post::paginate(1, ['*'], 'new-page-name');
Just like you said you can use the setPageName method:
Paginator::setPageName('sida');
You can place that in app/start/global.php.
Well that didnt work for me in laravel 5 , in laravel 5 you will need to do more extra work by overriding the PaginationServiceProvider because the queryName "page" was hardcoded in there , so first create your new PaginationServiceProvider in /app/providers ,This was mine
<?php
namespace App\Providers;
use Illuminate\Pagination\Paginator;
use Illuminate\Support\ServiceProvider as ServiceProvider;
class PaginationServiceProvider extends ServiceProvider {
//boot
public function boot()
{
Paginator::currentPageResolver(function()
{
return $this->app['request']->input('p');
});
}//end boot
public function register()
{
//
}
}
Then in your controllers you can do this
$users = User::where("status","=",1)
->paginate(5)
->setPageName("p");
I'm quite new to ZF, and right now, i try to write a tiny app, based on ZF. It works more or less fine until now. I wanna access my db- data. For starters, i just want to use query-string, before I start messing araound with zend_db. So to keep a nice mvc-structure, I created application/models/IndexMapper.php
class Application_Models_IndexMapper{...}
it just contains one function by now to see if it works
public function test(){
return ('yay');
}
In my IndexController, which is working, i try to access my model by
$indexMapper = new Application_Models_IndexMapper();
$x = $indexMapper->test();
but the first line throws an
Fatal error: Class 'Application_Models_IndexMapper' not found in /path/to/application/controllers/IndexController.php on line 31
As I'm new, I don't understand the more complex tutorials and they don't help me fix my problem. What am I doing wrong? Do I have to include it somehow?
Thanks
edit: my application/bootstrap.php
<?php
defined('APPLICATION_PATH')
or define('APPLICATION_PATH' , dirname(__FILE__));
defined('APPLICATION_ENVIRONMENT')
or define('APPLICATION_ENVIRONMENT' , 'development');
require_once 'Zend/Loader.php';
Zend_Loader::registerAutoload();
$frontController = Zend_Controller_Front::getInstance();
$frontController->setControllerDirectory(APPLICATION_PATH . '/controllers');
$frontController->setParam('env', APPLICATION_ENVIRONMENT);
Zend_Layout::startMvc(APPLICATION_PATH . '/layouts/scripts');
//Doctype
$view = Zend_Layout::getMvcInstance()->getView();
$view->doctype('HTML5');
$view->addHelperPath('App/View/Helper', 'App_View_Helper');
unset($frontController);
The structure for a model would be found under ./application/models/IndexMapper.php. In that file you would have the class As you named it and then the autoloading will work.
A great beginner tutorial would be found on www.akrabat.com
You have your class in the wrong place and have named it incorrectly.
Your class should be in application/models/Indexmapper.php and should look like this:-
class Application_Model_Indexmapper
{
public function test(){
return ('yay');
}
}
Then you call it thus:-
$indexMapper = new Application_Model_Indexmapper();
$x = $indexMapper->test();
Notice I dropped the 's' from the end of Models, it is not required and will cause an error as you found. Also the class is in the models folder, not modules. If you want to use modules then you need to read this and this from the manual.
Your bootstrap.php should look like this for a first, basic project:-
class Bootstrap extends Zend_Application_Bootstrap_Bootstrap
{
//Yes, it's empty!
}
Well, I guess my tutorial wasn't very helpful. I'll do as recommended, and start over from scratch. Thanks though