I have project in laravel 5.3 and i am using Observer for activity log of user, for that i have created one obeserver with saved() and deleted() method.
The saved() method is working fine for new record, while update the record saved() is not getting call nor updated() method.
I also try with deleted() method, that is also not getting call, here below is my code, please help.
public function __construct()
{
// echo "dsd"; die();
}
public function saved($user)
{
if ($user->wasRecentlyCreated == true) {
// Data was just created
$action = 'created';
} else {
// Data was updated
$action = 'updated';
}
UserAction::create([
'user_id' => Auth::user()->id,
'action' => $action,
'action_model' => $user->getTable(),
'action_id' => $user->id
]);
}
public function deleting($user)
{
dd($user);
}
}
public static function boot() {
parent::boot();
parent::observe(new \App\Observers\UserObserver);
}
Everything seems ok, so i guess something bigger is at fault here. Normally the best practice for registering observers is to do it in a provider class boot() method.
public function boot()
{
User::observe(UserObserver::class);
}
EDIT
For model events to trigger you have to use the model and not update the data through a query.
$discount = Discounts::find($request->edit_id);
$discount->fill($data);
$discount->save();
I have a page with a some content on it and a comments section. Comments can only be left by users who are signed in so I have added a login form to the page for users to sign in with (this only shows if they are not already logged in).
The problem I have is that when the user signs in they get redirected back to the home page and not the page they were previously on.
I have not changed the login method from the out of the box set-up.
Can anyone suggest a simple way to set the redirect url. My thoughts are that it would be good to be able to set it in the form.
Solution for laravel 5.3:
In loginController overwrite the showLoginForm() function as this one:
public function showLoginForm()
{
if(!session()->has('url.intended'))
{
session(['url.intended' => url()->previous()]);
}
return view('auth.login');
}
It will set the "url.intended" session variable, that is the one that laravel uses to look for the page which you want to be redirected after the login, with the previous url.
It also checks if the variable has been set, in order to avoid the variable to be set with the login url if the user submit the form with an error.
For Laravel 5.5, following code worked for me by just updating LoginController.php
public function showLoginForm()
{
session(['link' => url()->previous()]);
return view('auth.login');
}
protected function authenticated(Request $request, $user)
{
return redirect(session('link'));
}
Please use redirect()->intended() instead in Laravel 5.1
You can also see more about it here: http://laravel.com/docs/5.1/authentication
For Laravel 5.3
inside App/Http/Controllers/Auth/LoginController
add this line to the __construct() function
$this->redirectTo = url()->previous();
So the full code will be
public function __construct()
{
$this->redirectTo = url()->previous();
$this->middleware('guest', ['except' => 'logout']);
}
It works like a charm for me i'm using laravel 5.3.30
For Laravel 5.4, following code worked for me by just updating LoginController.php
use Illuminate\Support\Facades\Session;
use Illuminate\Support\Facades\URL;
public function __construct()
{
$this->middleware('guest', ['except' => 'logout']);
Session::put('backUrl', URL::previous());
}
public function redirectTo()
{
return Session::get('backUrl') ? Session::get('backUrl') : $this->redirectTo;
}
The Laravel 5.6, When user insert wrong credentials then login page will reload and session(['link' => url()->previous()]); will take login URL in link variable. So the user will redirect to a login page again or redirect to /home if login success. So to avoid these below code working for me! After that no matter how much time user insert wrong credentials he will redirect after login to exactly where he was before login page.
Update or overwrite public function showLoginForm() in LoginController.
public function showLoginForm()
{
if (session('link')) {
$myPath = session('link');
$loginPath = url('/login');
$previous = url()->previous();
if ($previous = $loginPath) {
session(['link' => $myPath]);
}
else{
session(['link' => $previous]);
}
}
else{
session(['link' => url()->previous()]);
}
return view('auth.login');
}
Also, Update or Overwrite protected function authenticated(Request $request, $user) in LoginController.
protected function authenticated(Request $request, $user)
{
return redirect(session('link'));
}
If you want to redirect always to /home except for those pages with comments, then you should overwrite your redirectTo method in your LoginController:
public function redirectTo()
{
return session('url.intended') ?? $this->redirectTo;
}
On all pages where you want to remain on the site, you should store the url for one request in the session:
public function show(Category $category, Project $project){
// ...
session()->flash('url.intended' , '/' . request()->path());
}
Redirect to login with the current's page url as a query string:
login
In your LoginController check if exists and save the query string in session then redirect to the url after login
public function __construct() {
parent::__construct();
if ( \request()->get( 'redirect_to' ) ) {
session()->put( 'redirect.url', \request()->get( 'redirect_to' ) );
}
$this->middleware( 'guest' )->except( 'logout' );
}
protected function authenticated(Request $request, $user) {
if(session()->has('redirect.url') {
return redirect( session()->get( 'redirect.url' ) );
}
}
Look into laravel cheat sheet
and use:
URL::previous();
to go to the previous page.
Laravel 5
(maybe 6 also, not tested, if someone knows it please update the answer)
add this to LoginController:
protected function redirectTo(){
return url()->previous();
}
Note: if present the field $redirectTo , remove it
in your RedirectIfAuthenticated.php change this code
public function handle($request, Closure $next, $guard = null)
{
if (Auth::guard($guard)->check()) {
return redirect()->intended('/contactus');
}
return $next($request);
}
please notice to :
return redirect()->intended('/contactus');
Inside your template file you can just use:
{{ url()->previous() }}
To redirect from the controller you should use
return redirect()->back();
or Just
return back();
use Illuminate\Support\Facades\Redirect;
public function Show_Login_Form()
{
$back = Session::put('url_back',url()->previous());
$current = url()->current();
if(Session::get('user_id'))
{
if ($back == $current) { // don't back Login Form
return Redirect::to('home');
}
elseif (Session::has('url_back')) {
return Redirect::to('home');
}
else{
return redirect()->back();
}
}
else{
if ($back == $current) {
return Redirect::to('home');
}
else{
Session::put('url_back',url()->previous());
}
return view('account.customer-account.login');
}
}
public function signin_user(Request $request) // Login post
{
$username = $request->input_username_login;
$password = md5($request->input_password_login);
$result = DB::table('tbl_user')
->where([['user_email',$username],['user_password',$password]])
->orWhere([['user_phone',$username],['user_password',$password]])
->first();
if($result){
Session::put('user_id', $result->user_id );
Session::put('user_name', $result->user_name);
Session::put('user_username', $result->user_username);
Session::put('user_avatar', $result->user_avatar);
return Redirect::to(Session::get('url_back')); // Back page after login
} else {
Session::put('message_box', 'Error !!!');
return redirect()->back();
}
}
You can use redirect back with Laravel 5:
<?php namespace App\Http\Controllers;
use Redirect;
class SomeController extends Controller {
public function some_method() {
return Redirect::back()
}
}
Use Thss
return Redirect::back('back-url')
I'm currently playing around with Exception Handler, and creating my own custom exceptions.
I've been using PHPUnit to run tests on my Controller Resource, but when I throw my custom exceptions, Laravel thinks it's coming from a regular HTTP request rathen than AJAX.
Exceptions return different response based on wether it's an AJAX request or not, like the following:
<?php namespace Actuame\Exceptions\Castings;
use Illuminate\Http\Request;
use Exception;
use Actuame\Exceptions\ExceptionTrait;
class Already_Applied extends Exception
{
use ExceptionTrait;
var $redirect = '/castings';
var $message = 'castings.errors.already_applied';
}
And the ExceptionTrait goes as follows:
<?php
namespace Actuame\Exceptions;
trait ExceptionTrait
{
public function response(Request $request)
{
$type = $request->ajax() ? 'ajax' : 'redirect';
return $this->$type($request);
}
private function ajax(Request $request)
{
return response()->json(array('message' => $this->message), 404);
}
private function redirect(Request $request)
{
return redirect($this->redirect)->with('error', $this->message);
}
}
Finally, my test goes like this (excerpt of the test that's failing)
public function testApplyToCasting()
{
$faker = Factory::create();
$user = factory(User::class)->create();
$this->be($user);
$casting = factory(Casting::class)->create();
$this->json('post', '/castings/apply/' . $casting->id, array('message' => $faker->text(200)))
->seeJsonStructure(array('message'));
}
My logic is like this although I don't think the error is coming from here
public function apply(Request $request, User $user)
{
if($this->hasApplicant($user))
throw new Already_Applied;
$this->get()->applicants()->attach($user, array('message' => $request->message));
event(new User_Applied_To_Casting($this->get(), $user));
return $this;
}
When running PHPUnit, the error I get returned is
1) CastingsTest::testApplyToCasting PHPUnit_Framework_Exception:
Argument #2 (No Value) of PHPUnit_Framework_Assert:
:assertArrayHasKey() must be a array or ArrayAccess
/home/vagrant/Code/actuame2/vendor/laravel/framework/src/Illuminate/Foundation/T
esting/Concerns/MakesHttpRequests.php:304
/home/vagrant/Code/actuame2/tests/CastingsTest.php:105
And my laravel.log is over here http://pastebin.com/ZuaRaxkL (Too large to paste)
I have actually discovered that PHPUnit is not actually sending an AJAX response, because my ExceptionTrait actually changes the response on this. When running the test it takes the request as a regular POST request, and runs the redirect() response rather than ajax(), hence it's not returning the correspond.
Thanks a bunch!
I have finally found the solution!
As I said, response wasn't the right one as it was trying to redirect rathen than return a valid JSON response.
And after going through the Request code, I found out that I need to use also wantsJson(), as ajax() may not be the case always, so I have modified my trait to this:
<?php
namespace Actuame\Exceptions;
trait ExceptionTrait
{
public function response(Request $request)
{
// Below here, I added $request->wantsJson()
$type = $request->ajax() || $request->wantsJson() ? 'ajax' : 'redirect';
return $this->$type($request);
}
private function ajax(Request $request)
{
return response()->json(array('message' => $this->message), 404);
}
private function redirect(Request $request)
{
return redirect($this->redirect)->with('error', $this->message);
}
}
I am new to CodeIgniter framework. I am using 2.1.4 version. I designed a simple login form, with a javascript validation, and the home page of a site. Can you please help me to understand how to declare session , and how to destroy the session on clicking signout link.
controller file of login page ( to load the view page login.php ):-
class Login extends CI_Controller {
function __construct() {
parent::__construct();
$this->load->helper('url');
}
function index(){
$this->load->view('login');
}
function success() {
redirect ('home');
}
}
The controller file home.php for the view home.php
class Home extends CI_Controller {
// local constructor will be overriding the one in the parent controller class
// for using a constructor in any of my Controllers
function __construct() {
parent::__construct();
}
public function index()
{
$this->load->view('home');
}
}
I have designed the view page home.php, and gave the signout link:-
<div class="logout">Signout</div>
For initializing the session, i need to know, what all constructor changes/ config changes need, and the method of session destoy.
To start session library, Go to application/config/config.php and change the below line:
$autoload['libraries'] = array('session');
It would be better if you start your session in the autoload.php. To destroy session you would use :
$this->session->sess_destroy();
To set session :
$newdata = array(
'username' => 'johndoe',
'email' => 'johndoe#some-site.com',
'logged_in' => TRUE
);
$this->session->set_userdata($newdata);
here is an controller... first of all u need to declare a session so that you have two choice to declare one is Go to application/config/config.php change the code as
$autoload['libraries'] = array('session');
and follow this following method (controller)
class Login extends CI_Controller{
function __construct(){
parent::__construct();
$this->load->library('session');
}
function index(){
$this->load->view('login');
}
function success() {
$user=$this->input->post('user');
$psw=$this->input->post('pswd');
$this->load->model('validation');
$result=$this->validation->useraccess($user,$psw);
if($result)
{
$this->session->set_userdata('username', $user); //setting session
redirect ('home');
}
else
{
$this->index();
}
}
function logout()
{
$this->session->unset_userdata('username');
redirect('login','refresh');
}
}
this is model where validation done
Class Validation extends CI_Model{
function __construct(){
parent::__construct();
}
function useraccess($user,$pswd)
{
$query = $this->db->query("select * from user where username='$user' AND password='$pswd'");
foreach ($query->result_array() as $row)
{
if($row['username']==$user AND $row['password']==$pswd)
{
return true;
}
else
{
return false;
}
}
}
}
here is a view
login page
create 2 text box and 1 submit button and declare form action as
localhost/index.php/login/success
for logut
localhost/index.php/login/logout
I'm not CI programmer, just trying to learn it. Maybe this is wrong approach, please advice.
my controller(not in sub directory) :
class Users extends CI_Controller {
function __construct() {
parent::__construct();
}
public function index($msg = NULL) {
$this->load->helper(array('form'));
$data['msg'] = $msg;
$this->load->view('user/login' , $data);
}
public function process_logout() {
$this->session->sess_destroy();
redirect(base_url());
}
}
And a route for login :
$route['user/login'] = 'users/index';
Problem is when I wanna logout, it shows me 404 because I do not have it in my route :
$route['user/process_logout'] = 'users/process_logout';
and in my view I put logout
When I add that, it works, and that is stuppid to add a route for everything. What I'm I doing wrong, please advice.
Thank you
Don't know why you are trying to implement login feature in index() function. However since you said you are learning CI I'm telling something about _remap() function.
Before that. You can try the following routing:
$route['user/:any'] = 'users/$1';
$route['user/login'] = 'users/index';
If you want to take value immediately after controller segment you need to use _remap() function and this function may be solve your routing problem, i mean you don't need to set routing. Lets implement your code controller 'users' using _remap() function.
class Users extends CI_Controller {
private $sections = array('login', 'logout');
function __construct() {
parent::__construct();
}
public function _remap($method)
{
$section = $this->uri->segment(2);
if(in_array($section, $this->sections))
call_user_func_array(array($this, '_'.$section), array());
else show_404(); // Showing 404 error
}
private function _login()
{
$msg = $this->uri->segment(3);
$this->load->helper(array('form'));
$data['msg'] = $msg;
$this->load->view('user/login' , $data);
}
public function _logout() {
$this->session->sess_destroy();
redirect(base_url());
}
}