As the title suggests, I need to call a class.
Example config file within Laravel:
return [
'user' => App\User::class
];
The problem I have is that I need to do run the non-static method and not sure how to do that.
i.e. (new User())->getTable()
If you have this in your config file example_config_file.php:
return [
'user' => App\User::class,
];
You can call it like this using the new command to create an instance of the User model:
$userClass = config('example_config_file.user');
$user = new $userClass(); // blank instance of User model
Or like this if you want to use Eloquent methods:
$userClass = config('example_config_file.user');
$user = $userClass::find(1); // User model with id = 1
You can create a helper class and get any type of model like this:
class ModelHelper
{
public static function getModel($type, $id = null)
{
$modelClass = config("example_config_file.$type");
if($id) {
return $modelClass::find($id);
}
return new $modelClass();
}
}
Then in your code, whenever you need a user instance, call the helper class:
$user = ModelHelper::getModel('user'); // get a blank User model
$post = ModelHelper::getModel('post', 1); // get the post with id = 1
Related
In My project there is a PackageRequest model and this model have a tracking_code.
I want to set this attribute automatically with a mutator.
this is my mutator:
public function setTrackingCodeAttribute()
{
$code = mt_rand(1000000, 9999999);
$this->attributes['tracking_code'] = "$code";
}
but when I save new request into database there is no tracking_code
this is my part of code that create new PackageRequest:
public function store(Request $request)
{
$input = $request->only(['requested_time', 'address_id']);
$valid = validator($input, [
'requested_time' => 'required|date',
'address_id' => 'required|numeric|exists:addresses,id'
]);
if ($valid->fails())
return Response::fail($valid->errors());
$package_request = new PackageRequest($input);
$package_request->user_id = \Auth::id();
$package_request->status = 'waiting';
if(!$package_request->save())
return Response::error(__('errors.save_data', ['attr' => __('errors.attribures.package_request')]));
return Response::success(PackageRequest::find($package_request->id), 201);
}
The mutator will receive the value that is being set on the attribute, allowing you to manipulate the value and set the manipulated value on the Eloquent model's internal $attributes property.
The mutator will be automatically called when you attempt to set the value to attribute tracking_code, and you are not setting anything to it. What you need is setting default value to tracking_code before creating, so you can add these codes in model PackageRequest:
protected static function boot()
{
parent::boot();
static::creating(function($package_request)
{
$package_request->tracking_code = mt_rand(1000000, 9999999);
});
}
I have a function to add new property. But i want to check for duplicate data at column "code" before add new data into database. If data exists will appear a message error.
function addPro(Request $req)
{
$id = $req->type_id;
$type = AssetType::find($id);
if($req->save == 'save'){
$pro = new TypeProperties;
$pro->name = $req->name;
$pro->code = $req->code;
$pro->type = $req->type;
$pro->assettype_id = $req->type_id;
$pro->save();
Schema::table($type->code, function ($table) use ($pro) {
if ($pro->type == "textbox")
$table->string($pro->code )->nullable();
if ($pro->type == "textarea")
$table->text($pro->code )->nullable();
});
return redirect(url($type->id.'/add/property'))->with('message','Save successful');
}
return redirect(url('asset/type/'.$type->id));
}
You can use laravel Request Validation
function addPro(Request $req)
{
$id = $req->type_id;
$type = AssetType::find($id);
if($req->save == 'save'){
$req->validate([
'code' => 'required|unique:tablename'
]);
$pro = new TypeProperties;
$pro->name = $req->name;
$pro->code = $req->code;
$pro->type = $req->type;
$pro->assettype_id = $req->type_id;
$pro->save();
Schema::table($type->code, function ($table) use ($pro) {
if ($pro->type == "textbox")
$table->string($pro->code )->nullable();
if ($pro->type == "textarea")
$table->text($pro->code )->nullable();
});
return redirect(url($type->id.'/add/property'))->with('message','Save successful');
}
return redirect(url('asset/type/'.$type->id));
}
The most simple way to do this is by checking if code is_null :
if (is_null($pro->code)) {
// It does not exist
} else {
// It exists
}
The other way is to make a validation using Laravel's built in ValidateRequest class. The most simple use-case for this validation, is to call it directly in your store() method like this:
$this->validate($req, [
'code' => 'required|unique,
//... and so on
], $this->messages);
With this, you're validating users $req by saying that specified columns are required and that they need to be unique, in order for validation to pass. In your controller, you can also create messages function to display error messages, if the condition isn't met:
private $messages = [
'code.required' => 'Code is required',
'code.unique' => 'Code already exists',
//... and so on
];
You can also achieve this by creating a new custom validation class:
php artisan make:request StorePro
The generated class will be placed in the app/Http/Requests directory. Now, you can add a few validation rules to the rules method:
public function rules()
{
return [
'code' => 'required|unique,
//... and so on
];
}
All you need to do now is type-hint the request on your controller method. The incoming form request is validated before the controller method is called, meaning you do not need to clutter your controller with any validation logic:
public function store(StorePro $req)
{
// The incoming request is valid...
// Retrieve the validated input data...
$validated = $req->validated();
}
If you have any additional question about this, feel free to ask. Source: Laravel official documentation.
What does your migration look like for AssetType?
I ask because you can do this in the schema with ->unique() added to the column on the creation or make a migration to add the constraint.
You can also check with something like this:
// Search database table for entry
$entry = AssetType::where('code', '=', $pro->code)->first();
// If not found
if ($entry === null) {
// Save method here.
}
Otherwise, you can use the manual validator or create a Request with validation
References:
https://laravel.com/docs/5.8/queries#where-clauses
https://laravel.com/docs/5.8/validation#creating-form-requests
https://laravel.com/docs/5.8/validation#manually-creating-validators
I have a variable which holds the model name like so
$fooTableName = 'foo_defs';
$fooModel = 'FooDefs';
Now I would like to insert in the DB using that model like so
$fooModel::insert([..foo..array...]);
Throws an error
"message": "Parse error: syntax error, unexpected '$fooModel' (T_VARIABLE), expecting identifier (T_STRING)",
Is it possible to do something like that? or will I be forced to use
DB::table('fooTableName')->insert([...foo...array...]);
If I do it in the latter way, the timestamps in the table are wrong. The created_at column is null and the updated_at has the value
EDIT 1
$model = CustomHelper::getNameSpace($this->tableNames[$i]);
// $model => /var/www/html/erp/app/Models/sales/InvoiceDefs
$model::insert($this->tableCollections[$this->tableNames[$i]]);
Most of them said that, it was namespace issue, so I have corrected it, but still it is throw error like
"message": "Class '/var/www/html/erp/app/Models/sales/InvoiceDefs' not
found",
What you are doing wrong is using model name as string, you need to refactor your code as like below :
$fooModel = 'App\Models\FooDefs';
I have a same situation before and i have created the function to do this
function convertVariableToModelName($modelName='',$nameSpace='')
{
//if the given name space iin array the implode to string with \\
if (is_array($nameSpace))
{
$nameSpace = implode('\\', $nameSpace);
}
//by default laravel ships with name space App so while is $nameSpace is not passed considering the
// model namespace as App
if (empty($nameSpace) || is_null($nameSpace) || $nameSpace === "")
{
$modelNameWithNameSpace = "App".'\\'.$modelName;
}
//if you are using custom name space such as App\Models\Base\Country.php
//$namespce must be ['App','Models','Base']
if (is_array($nameSpace))
{
$modelNameWithNameSpace = $nameSpace.'\\'.$modelName;
}
//if you are passing Such as App in name space
elseif (!is_array($nameSpace) && !empty($nameSpace) && !is_null($nameSpace) && $nameSpace !== "")
{
$modelNameWithNameSpace = $nameSpace.'\\'.$modelName;
}
//if the class exist with current namespace convert to container instance.
if (class_exists($modelNameWithNameSpace))
{
// $currentModelWithNameSpace = Container::getInstance()->make($modelNameWithNameSpace);
// use Illuminate\Container\Container;
$currentModelWithNameSpace = app($modelNameWithNameSpace);
}
//else throw the class not found exception
else
{
throw new \Exception("Unable to find Model : $modelName With NameSpace $nameSpace", E_USER_ERROR);
}
return $currentModelWithNameSpace;
}
How To user it:
Arguments
First Argument => Name of the Model
Second Argument => Namespcce of the Model
For Example we have the model name as Post
$postModel = convertVariableToModelName('Post');
dd($postModel::all());
Will returns all the values in the posts table
But in Some Situation You Model Will in the
Custom Namespace such as App\Models\Admin\User
So this function is created to overcome that
$userModel = convertVariableToModelName('User',['App','Models','Admin']);
dd($userModel::all());
You are feel free to customize the function
Hope it helps
Try the below one,
$fooModel = new FooDefs();
and then you can do the following also,
$fooModel->column1 = $value1;
$fooModel->column2 = $value2;
$fooModel->column2 = $value2;
$fooModel->save();
or
$fooModel->save([
'column1' => $value1,
'column2' => $value2,
'column3' => $value3,
])
Edited answer
$path = 'my\project\path\to\Models';
$fooModel = app($path.'\FooDefs');
$fooModel::save([
'column1' => $value1,
'column2' => $value2,
'column3' => $value3,
])
dd($fooModel ::all());
Try my edited answer.
When you have the name of class stored as string you can call method on that class by using the php method call_user_func_array like this
$class_name = "FooDefs";
call_user_func_array(array($class_name, "insert"), $data);
$data is an Array of data which will be past to the called function as arguments.
Just for simple advice It's will be Good if you save in the $class_name variable the FQN Fully Qualified Name of the class which is the __NAMESPACE__ follow by the name of the class. For sample purpose FQN look like Illuminate\Support\Facades\DB which you can get when you save you use User::class I presume you have some User model. That will return the Fully Qualified Name of the User model will will be App\User in case of Laravel.
$requests = $post['request'] // posting the data from view page
$models = "app\models".'\\'.$requests //geting the model
$model = $models::findOne($referenceId) //fetching value from database
Alright , I have used this way to save the users info and It works perfect,
static public function memberSave($request) {
$signup = false;
$member = new Members();
$member->name = $request['name'];
$member->email = $request['email'];
$member->password = bcrypt($request['password']);
$member->save();
if (!empty($member->id)) {
$new_id = $member->id;
DB::insert("INSERT INTO roles VALUES ($new_id, 5613)");
$signup = true;
Session::flash('sm', 'Thank you! You have signed up successfully!');
}
return $signup;
}
but when making this for editing the profile(by user) It doesn't work
becuase I use new(); (making object)
I also didn't succeed to use find(); so I tried to use this
static public function saveProfile($id,$name,$email,$password) {
$sql = "UPDATE members SET name=?,email=?,password=? WHERE id=?";
$member = DB::select($sql, [$name,$email,$password,$id]);
but when I want to bcrypt the password in laravel doesnt work .
this is the code also in the second page
public function postProfile(ProfileValidation $request) {
if (Members::saveProfile($request['id'], $request['name'], $request['email'], $request['password'])) {
return redirect('');
}
}
I hope getting helped for editing the users profile by laravel , thanks.
Your Members class must extend Eloquent\Model for following this code to work.
class Members extends Model {
// optional
protected $table = 'members';
...
To find and update the member using email,
// find the single member
$member = Members::where('email', request['email'])->first();
// update the member
$member->name = $request['name'];
$member->password = $request['password'];
// now save the updated member
$member->save();
In order to to encrypt Password, Laravel provides Hash Facade,
// import this
use Hash;
...
// encrypt Password
$encrypted = Hash::make($request['password']);
...
if you want your user automatically hash the password at your model put:
public function setPasswordAttribute($value)
{
$this->attributes['password'] = Hash::make($value);
}
and you can directly check for the user if exist create new or update it:
public function saveMember($request)
{
$member = Member::findOrNew($request->email);
//All your input you want to save
$member->save();
}
I'm planning to use ZF2 in a future project, so I'm trying Zend Framework 2 RC1 now. I started with authentication step, and noticed that when i chose a different name than 'Zend_Auth' for session storage namespace, i can't access to object stored in session (AuthenticationService class' hasIdentity method returned false, despite User object data set in session).
<?php
namespace Application\Controller;
use Zend\Authentication\Adapter\DbTable as AuthAdapter;
use Zend\Authentication\AuthenticationService;
use Zend\Authentication\Storage\Session as SessionStorage;
use Zend\Mvc\Controller\AbstractActionController;
use Zend\View\Model\ViewModel;
use Application\Model\User;
use Application\Form\LoginForm;
class LoginController extends AbstractActionController
{
public function indexAction()
{
$auth = new AuthenticationService();
if ($auth->hasIdentity()) {
return $this->redirect()->toRoute('application');
}
$form = new LoginForm();
return array('form' => $form);
}
public function loginAction()
{
$auth = new AuthenticationService();
$form = new LoginForm();
$form->get('submit')->setAttribute('value', 'Add');
$request = $this->getRequest();
if ($request->isPost()) {
$user = new User();
$form->setInputFilter($user->getInputFilter('login'));
$form->setData($request->getPost());
if ($form->isValid()) {
$data = $form->getData();
// Configure the instance with constructor parameters...
$sm = $this->getServiceLocator();
$dbAdapter = $sm->get('db-adapter');
$authAdapter = new AuthAdapter($dbAdapter, 'users', 'username', 'password');
$authAdapter
->setIdentity($data['username'])
->setCredential(sha1($data['password']));
// Use 'users' instead of 'Zend_Auth'
$auth->setStorage(new SessionStorage('users'));
$result = $auth->authenticate($authAdapter);
if ($result->isValid()) {
// store the identity as an object where only the username and
// real_name have been returned
$storage = $auth->getStorage();
// store the identity as an object where the password column has
// been omitted
$storage->write($authAdapter->getResultRowObject(
null,
'password'
));
// Redirect to list of application
return $this->redirect()->toRoute('application');
}
}
}
// processed if form is not valid
return array('form' => $form);
}
}
In this code, when i changed the below line,
$auth->setStorage(new SessionStorage('users'));
like this:
$auth->setStorage(new SessionStorage());
hasIdentity method returned true.
I checked two classes Zend\Authentication\AuthenticationService and Zend\Authentication\Storage\Session, and didn't see a way to access session data which has different session namespace other than default.
What i need to understand is how can i access session data which has a different namespace and if there is no way to do it for now, should we define this as a bug?
I can update the question if any other information needed.
We are kinda missing one part of your code, the one where you try and receive the user identity. im guessing that you have forgotten to pass the the SessionStorage Object with the same namespace.
Also the configuration of the Authentication object should be moved to a factory so these kind of issues to not arrise.
Thats my five cents atleast :)