For cakephp 3.6, cakephp.org tell how to customise user finder query for auth component at following link:
link
But I am not getting how to implement it?
I have 'department_id' column in users table which belongs to departments table. I want to change the following query:
public function findAuth(){
$query
->select(['id', 'username', 'password'])->contain(['Departments'])
->where(['Users.active' => 1]);
return $query;
}
Will the above code work?
Please tell me in which file I have to write the function?
And what are other necessary steps to get it done, so that I get all user related info in $this->Auth->User ?
Firstly, you will need to load the Auth component and pass in custom configuration in any controller that you want to use the custom finder like so:
public function initialize()
{
parent::initialize();
$this->loadComponent('Auth', [
'authenticate' => [
'Form' => [
'finder' => 'auth'
]
],
]);
}
then, in your UsersTable you would have the custom finder:
public function findAuth(\Cake\ORM\Query $query, array $options)
{
$query
->select(['id', 'username', 'password'])->contain(['Departments'])
->where(['Users.active' => 1]);
return $query;
}
This answer may also help Containing tables in custom Auth finder
Related
I am having trouble creating multi-role application in laravel5 since in laravel 5 the authentication is pre defined so I am not willing to mess around with predefined codes of laravel 5 authentication. I have a constructor that authenticates every controller in my project but I am unable to check user roles for the following roles:-
1. Admin
2. Agent
3. User
I can check manually for every functions but that is not the right process of doing so and if I have a total of around 500 functions I cant go in every function and define manually. please any help
Thank you
Personally I would use middleware and route groups to accomplish the task, which would be similar the way Laravel checks for user authentication.
You just have to determine when you need to run the middleware, which can be done by nesting Route::group's or injecting the middleware from your controller.
So, for an example of nesting you can have something like this in your routes file:
Route::group(['middleware' => ['auth']], function () {
Route::get('dashboard', ['as' => 'dashboard', function () {
return view('dashboard');
}]);
Route::group(['prefix' => 'company', 'namespace' => 'Company', 'middleware' => ['App\Http\Middleware\HasRole'], function () {
Route::get('dashboard', ['as'=>'dashboard', function () {
return view('company.dashboard');
}]);
Route::resource('employees', 'EmployeesController');
...
...
});
});
or you can inject the middleware to your controllers like so:
use Illuminate\Routing\Controller;
class AwesomeController extends Controller {
public function __construct()
{
$this->middleware('hasRole', ['only' => 'update'])
}
}
And then add a one or more Middleware files using something like php artisan make:middleware HasRole which will give you the middleware boiler plate which you could then add your role checking logic:
<?php namespace App\Http\Middleware;
use Closure;
class HasRole {
/**
* Handle an incoming request.
*
* #param \Illuminate\Http\Request $request
* #param \Closure $next
* #return mixed
*/
public function handle($request, Closure $next)
{
if($request->is('admin/*')){
[******ADD YOUR LOGIC HERE TO DETERMINE THE ROLE ******]
[******YOU CAN ALSO INCLUDE ANY REDIRECTS IF NECESSARY******]
}
return $next($request);
}
}
Notice I used the $route->is('admin/*') to filter any routes as an example of further filtering requests, which you would probably not include if you are injecting the middleware from the controller.
But if the user passes the required role check you do not need to do anything and they will be allowed to continue to the view. If they fail the role check, you can handle that accordingly, but beware of getting them caught in a failed permission loop.
I assume you get the gist of it, feel free to look into the Laravel middleware docs for more info.
How can I use middleware with resources?
Route::resource('myitem', ['middleware' => 'auth', 'uses' => 'App\\Controllers\\MyitemsController']);
Just followed https://laracasts.com/discuss/channels/general-discussion/struggling-with-routeresource-and-auth-middleware but unfortunately could not solve.
Getting error:
ErrorException (E_UNKNOWN)
Array to string conversion
Open: /vendor/laravel/framework/src/Illuminate/Routing/Router.php
protected function getResourceAction($resource, $controller, $method, $options)
{
$name = $this->getResourceName($resource, $method, $options);
return array('as' => $name, 'uses' => $controller.'#'.$method);
}
Using filter with resource was not working that why had to use Route::group
Route::group(array('before' => 'auth'), function()
{
Route::resource('myitem', 'App\\Controllers\\MyitemsController');
});
https://stackoverflow.com/a/17512478/540144
Middleware is a new feature of Laravel 5. In Laravel 4, filters where something similar. So instead of using the key middleware you should use before or after. Also, and that's where the error comes from, the second argument of Route::resource should be the controller name as string and the third one is an array of options:
Route::resource('myitem', 'App\\Controllers\\MyitemsController', ['before' => 'auth']);
Edit
Apparently before filters only work with resource routes when you wrap a group around it. See the OPs answer for an example...
I just came up against this and found the easiest way is to add the middleware straight to the controller.
I found my answer here:
http://laravel.com/docs/master/controllers
class MyitemsController extends Controller {
/**
* Instantiate a new MyitemsController instance.
*/
public function __construct()
{
$this->middleware('auth');
}
}
How to do this in Laravel 5. The Answer you have been waiting for.
Use middleware instead of before
Route::group(array('middleware' => 'auth'), function()
{
Route::resource('user', 'UserController',
['only' => ['edit']]);
}
To check if the route is setup, run:
php artisan route:list
which should show the following:
GET|HEAD | user/{user}/edit | user.edit | App\Http\Controllers\UserController#edit | auth
Note auth instead of guest
Better solution
Use middleware instead of before
Route::group(['middleware' => 'auth'], function(){
Route::resource('myitem', 'MyitemsController');
});
You can check if it's ok with:
php artisan route:list
i bulit api using yii restful extension http://www.yiiframework.com/extension/restfullyii/
i want create api for login which accept username and password then return all user
information i think the correct way to do that is to create custom router after checking
extension documentation section Defining Custom Routes the author say i must create http verb combination (event name = 'req..\.render')
[POST] api/login/login
to create url i do this
main.php
array('<controller>/req.post.login', 'pattern'=>'api/<controller:\w+>', 'verb'=>'POST'),
login Controller :
public function restEvents()
{
$this->onRest('req.post.login.render', function() {
echo "sss";
//Custom logic for this route.
//Should output results.
$this->emitRest('req.render.json', [
[
'type'=>'raw',
'data'=>['active'=>true]
]
])
});
}
restEvents didn't fire any help ??
to create custom routing add this to controler and url will be http://css.local.com/index.php/api/controlername/test
$this->onRest('req.post.test.render',function() {
$this->emitRest('req.render.json', [
[
'type'=>'raw',
'data'=>data
]
]);
});
Im new to zf2.
zfcuser is setup as in the installation guide. Im able to register, login and logout.
I had created modules for frontend and backend. Im trying to check if user is login for the backend - admin and all the child modules.
I tried to include
$sm = $app->getServiceManager();
$auth = $sm->get('zfcuser_auth_service');
if (!$auth->hasIdentity()) {
//redirect to login page
}
in my admin/module.php function onBootStrap
it did check for login, BUT not only for the admin, is for the entire modules including the frontend.
I just need to check login for the admin modules, and all the child modules of admin.
Couldn't figure out how. Please help
In you module.config.php
<?php
namespace AdminModule;
use Zend\Authentication\AuthenticationService;
return array(
__NAMESPACE__ => array(
'params' => array(),
'service_manager' => array(
'invokables' => array(
'Zend\Authentication\AuthenticationService' => 'Zend\Authentication\AuthenticationService',
),
),
'services' => array(
// Keys are the service names
// Values are objects
'Auth' => new AuthenticationService(),
),
),
);
Example in your controller/Manager
$this->auth = new AuthenticationService();
if($this->auth->hasIdentity()){
$this->userid = $this->auth->getIdentity();
}
You can use a factory too.
Example in your controller
using tools from ZfcUser module
$this->zfcUserAuthentication()->hasIdentity()
I use custom routes which include namespace besides controller and action. So for ACL purposes I use MVC route name as ACL resource name. Now I need to obtain currently DISPATCHING route name. The only solution I've come up with is to get namespace/controller/action from Dispatcher and iterating over all the routes find an appropriate one.
Is there any easiest way to obtain currently dispatching (not just matched) route name?
Pretty easy
\Phalcon\DI::getDefault()->get('router')->getMatchedRoute()->getName();
You can use your router, dispatcher and base controller to get what you need. Consider this:
$router = new \Phalcon\Mvc\Router(false);
$routes = array(
'/{namespace:"[a-zA-Z]+}/:controller' => array(
'controller' => 2,
),
'/{namespace:"[a-zA-Z]+}/:controller/:action/:params' => array(
'controller' => 2,
'action' => 3,
'params' => 4,
),
);
foreach($routes as $route => $params) {
$router->add($route, $params);
}
Now in your base controller you can do this:
public function getNamespace()
{
return $this->dispatcher->getParam('namespace');
}
This way you can have the namespace currently being served in your controllers (so long as they extend your base controller).
If you need to get the namespace in a model you can always use the DI like so (base model):
public function getNamespace()
{
$di = \Phalcon\DI::getDefault();
return $di->dispatcher->getParam('namespace');
}