Laravel password reset "No Sender" - email

When I do a password reset in Laravel 5.1, I get the email, but it says "No Sender" in the title.
Is there a way to specify the sender from somewhere for password reset emails? Apart from the sender, I assume Laravel should automatically uses the email settings as specified in the config files? It's strange because when I set the Laravel mail config to use the 'mail' driver, I get bounced emails saying I can't send from a dynamic address (which is to be expected on dev), but still the password reset emails go through. Shouldn't email reset use the same config settings?

Jack, you can set in the From attributes for email id and name in config/mail.php. I too had the same problem and just got it sorted, as i mentioned above.

#SeriousJelly Answer Update for Laravel 5.2
in your Auth PasswordController Override resetEmailBuilder Method
class PasswordController extends Controller
{
protected function resetEmailBuilder()
{
return function (Message $message) {
$message->subject($this->getEmailSubject());
$message->from('you#email.com', 'you');
};
}
}
This might help someone

So, Alexey Mezenin's answer is almost there, however, one big no no is overwriting the core files as any future updates can break functionality.
As your PasswordController should be using the ResetsPassword trait you should be able to just overwrite any methods from ResetsPassword trait in your PasswordController.
For example, adding your own from and subject line to emails is a simple case of finding the relevant function in your trait, copy and pasting into your PasswordController and amending it.
Here is an example PasswordController with a function that over writes the sendResetLinkEmail() function.
<?php
namespace App\Http\Controllers\Auth;
use Illuminate\Http\Request;
use Illuminate\Mail\Message;
use Illuminate\Support\Facades\Password;
use App\Http\Controllers\Controller;
use Illuminate\Foundation\Auth\ResetsPasswords;
class PasswordController extends Controller
{
/*
|--------------------------------------------------------------------------
| Password Reset Controller
|--------------------------------------------------------------------------
|
| This controller is responsible for handling password reset requests
| and uses a simple trait to include this behavior. You're free to
| explore this trait and override any methods you wish to tweak.
|
*/
use ResetsPasswords;
/**
* Create a new password controller instance.
*
* #return void
*/
public function __construct()
{
$this->middleware('guest');
}
/**
* Send a reset link to the given user.
*
* #param \Illuminate\Http\Request $request
* #return \Illuminate\Http\Response
*/
public function sendResetLinkEmail(Request $request)
{
$this->validate($request, ['email' => 'required|email']);
$broker = $this->getBroker();
$response = Password::broker($broker)->sendResetLink($request->only('email'), function (Message $message) {
$message->subject($this->getEmailSubject());
$message->from(env('MAIL_FROM'), env('APP_NAME'));
});
switch ($response) {
case Password::RESET_LINK_SENT:
return $this->getSendResetLinkEmailSuccessResponse($response);
case Password::INVALID_USER:
default:
return $this->getSendResetLinkEmailFailureResponse($response);
}
}

Maybe there is a better solution, but you could manually add code into \vendor\laravel\framework\src\Illuminate\Foundation\Auth\ResetPasswords.php, after this line:
$message->subject($this->getEmailSubject()); // this is line 66
Add something like this:
$message->from('my#email.com', 'My Site');
https://laravel.com/docs/5.1/mail#sending-mail

Related

Laravel: Sending Mail with Queue ignores Locale

My Email template looks like this:
#component('mail::message')
# {{ $helloUser }}
#lang('welcome.message')
This
\App::setLocale('de);
$activeMail = new \App\Mail\Register\Activate($user);
\Mail::to($user)->send($activeMail);
will send an mail with German text.
However, when I use a queue
\App::setLocale('de);
$activeMail = new \App\Mail\Register\Activate($user);
\Mail::to($user)->queue($activeMail);
The mail is send in English, which is the default language of my app.
How can I send a message in German with the queue without changing the default language?
In Laravel 5.6. the Mailable class has gotten a locale method to care for this:
$activeMail = new \App\Mail\Register\Activate($user);
$locale = $user->lang; // de
\Mail::to($user)->locale($locale)->queue($activeMail);
For Laravel < 5.6 one could save the text in the mail object
class Activate extends Mailable
{
public $mainText
public function __construct()
{
$this->mainText = __('welcome.message');
}
}
and change the template to
#component('mail::message')
# {{ $helloUser }}
{{$mainText}}
The difference is that $mainText is the string from the language when the mail object was created, while #lang('welcome.message') would be the string of the default language from your app.
Since Laravel 5.7, there's something that can help you with that. Take a look at Localizing Mailables in the documentation.
use Illuminate\Contracts\Translation\HasLocalePreference;
class User extends Model implements HasLocalePreference
{
/**
* Get the user's preferred locale.
*
* #return string
*/
public function preferredLocale()
{
return $this->locale;
}
}

ManyToOne with FOSUSerBundle ignoring exclusion policy

Building a JSON response for an API type thing, to retrieve a specific set of data that includes a ManyToOne relationship in the entity for my entity that extends FOSUSerBundle's User entity (called Account in my case).
The problem is, the Account entity thats included as a field in the response, is wanted, but I dont want to include all of the password and role type stuff.
I've been browing the internet for a couple hours now, and I've followed many guides on this, and I've cleared my cache every single time, and to no avail; So here's where I ended up:
// app/config/config.yml
jms_serializer:
metadata:
auto_detection: true
directories:
FOSUserBundle:
namespace_prefix: "FOS\\UserBundle"
path: "%kernel.root_dir%/Resources/serializer/FOS"
I've for below I've tried User.Model.yml and Model.User.yml and User.Entity.yml as well in a vain thought that the file name actually matters
// app/Resources/serializer/FOS/Entity.User.yml
FOS\UserBundle\Model\User:
exclusion_policy: ALL
properties:
id:
expose: true
and what I get still looks like this:
{
"status":"ok",
"api_version":"1.0",
"code":200,
"data":{
"video":{
"id":1,
"published":true,
"visibility":true,
"title":"Megaman 2",
"slug":"megaman-2",
"summary":"A rap song about Megaman",
"description":"A rap song\r\nAbout megaman",
"youtube_id":"R6L9bUouDr8",
"date_published":"2014-07-02T14:09:26-0700",
"date_created":"2014-07-02T14:09:26-0700",
"date_updated":"2014-07-02T14:09:26-0700",
"author_id":3,
"author":{
"id":3,
"username":"kharrison",
"username_canonical":"kharrison",
"email":"(sorry private)",
"email_canonical":"(sorry, private)",
"enabled":true,
"salt":"(sorry, private)",
"password":"(sorry, private)",
"last_login":"2014-07-04T15:17:34-0700",
"locked":false,
"expired":false,
"roles":[
"ROLE_SUPER_ADMIN"
],
"credentials_expired":false,
"display_name":"Kyle Harrison",
"slug":"kyle-harrison",
"bio":"Test"
}
}
}
}
The "author" field, is my Account entity thats being run through the JMSSerializer
I want to exclude ALL of that, except the user ID, Display name, and slug.
And finally this is how the API works:
// My/Bundle/Controller/BaseAPIController.php
//......... other code
/**
* #param string $status
* #param integer $code
* #return Response
*/
public function render_api($status, $code)
{
$this->apiResponse->setStatus($status);
$this->apiResponse->setCode($code);
return new Response($this->apiResponse->serialize($this->get('jms_serializer')), $this->apiResponse->getCode(), ["Content-type"=>"application/json"]);
}
//............. other code
and finally, that calls this:
// My/Bundle/Models
class APIResponse {
protected $status;
protected $apiVersion;
protected $code;
protected $data;
public function __construct($apiVersion, $status = "OK", $code = 500)
{
$this->status = $status;
$this->code = $code;
$this->apiVersion = $apiVersion;
$this->data = [];
}
// ... getters and setters
/**
* #return mixed
*/
public function serialize($serializer) {
return $serializer->serialize($this, "json");
}
}
I've for below I've tried User.Model.yml and Model.User.yml and
User.Entity.yml as well in a vain thought that the file name actually
matters.
It does matter, actually. It's a concatenation of the namespace and class name. In this case, you're trying to configure the FOS\UserBundle\Model\User class, so the file name should be Model.User.yml. (FOS\UserBundle\ should be excluded from the file name, since you configured it as namespace_prefix in your config.yml)
Also make sure that your Account class doesn't re-declare (overwrite) the properties, as the serializer config only works if you configure it for the class that actually declares the properties.
Ok So, the actual answer, couldn't have been arrived to via the information I provided. But Nic's Answer did lead me towards the solution. The description of how the the serializer looks at and deciphers the config file lead me to the real problem at hand.
This is what I failed to show:
<?php
namespace [PRIVATE]\[PRIVATE]Bundle\Entity;
use Doctrine\ORM\Mapping as ORM;
use FOS\UserBundle\Model\User as BaseUser;
use JMS\Serializer\Annotation\ExclusionPolicy;
use JMS\Serializer\Annotation\Expose;
use JMS\Serializer\Annotation\Groups;
use JMS\Serializer\Annotation\VirtualProperty;
/**
* Account
*
* #ORM\Table()
* #ORM\Entity(repositoryClass="[PRIVATE]\[PRIVATE]Bundle\Entity\AccountRepository")
*/
class Account extends BaseUser
{
The problem lays with the Alias I provided the FOS\UserBundle\Model\User namespace. I no longer remember why I wrote that that way. However, the moment I remove the Alias and rewrote the extends to resemble this instead:
<?php
namespace [PRIVATE]\[PRIVATE]Bundle\Entity;
use Doctrine\ORM\Mapping as ORM;
use FOS\UserBundle\Model\User;
use JMS\Serializer\Annotation\ExclusionPolicy;
use JMS\Serializer\Annotation\Expose;
use JMS\Serializer\Annotation\Groups;
use JMS\Serializer\Annotation\VirtualProperty;
/**
* Account
*
* #ORM\Table()
* #ORM\Entity(repositoryClass="[PRIVATE]\[PRIVATE]Bundle\Entity\AccountRepository")
*/
class Account extends User
{
combined with the new correct filename from Nic's answer, the config based Exclusion policy for JMSSerializerBundle totally kicks in, and every instance of FOSUserBundle's items are now completely hidden, except for the fields I've now explicitly told it to expose.
This is exactly what I wanted :)
Thanks everyone for your help! Cheers
~k
I'm not sure it's the exact way you want it, more a way around:
way around 1: Select only the properties you want (via the entity manager) and then serialize the array obtained.
It's what I do with what I call my API (which is not a class as you but controllers)

Symfony2 FOSUserBundle – Validate against "user active" flag on login

I have a flag on my users for 'active' and if set to zero or null, I will not allow login.
I have tried a couple of approaches and come up short.
If I do the logout route the flash message is not preserved, so the user sees nothing.
I looked into adding a validation on the login form so that it would throw a normal form error if the flag was not set to true, but in that folder (vendor/Bundles/FOS/UserBundle/Form/Type) I find nothing for login form, only registration and such, so I wouldn't know where to put it or where to inherit from in order to override.
I also tried as suggested here to manually log out, but that left me with a white screen of death...
Any suggestions how to easily accomplish this?
************** UPDATE ************
I realized that I probably want to go about it adding a validator on the login form. I currently have it coded into the controller of the first route a user gets sent to, but that won't provide much security if a user types a route before logging in, because on a successful login attempt, my default "landing page" after login will not be the route that the user is taken to, but he will be landing on the route of his choice...
***UPDATE AGAIN ****
So the service config file has this...
<service id="security.user_checker" class="%security.user_checker.class%" public="false" />
And that parameter is defined here...
<parameter key="security.user_checker.class">Symfony\Component\Security\Core\User\UserChecker</parameter>
So in order to modify the login logics I need to override
Symfony\Component\Security\Core\User\UserChecker
Now I have done that by overriding that parameter above in my own parameters.ini in the symfony app/config like this
security.user_checker.class = BizTV\UserBundle\Controller\UserChecker
.. and added this check to my userChecker overrider...
//Test for companylock...
if ( !$user->getCompany()->getActive() ) {
throw new LockedException('The company of this user is locked.', $user);
}
Here's the entire file:
<?php
/*
* This file is part of the Symfony package.
*
* (c) Fabien Potencier <fabien#symfony.com>
*
* For the full copyright and license information, please view the LICENSE
* file that was distributed with this source code.
*/
//Override by Mattias
namespace BizTV\UserBundle\Controller;
//namespace Symfony\Component\Security\Core\User;
use Symfony\Component\Security\Core\Exception\CredentialsExpiredException;
use Symfony\Component\Security\Core\Exception\LockedException;
use Symfony\Component\Security\Core\Exception\DisabledException;
use Symfony\Component\Security\Core\Exception\AccountExpiredException;
use Symfony\Component\Security\Core\User\UserInterface;
use Symfony\Component\Security\Core\User\UserChecker as OriginalUserChecker;
/**
* UserChecker checks the user account flags.
*
* #author Fabien Potencier <fabien#symfony.com>
*/
class UserChecker extends OriginalUserChecker
{
/**
* {#inheritdoc}
*/
public function checkPreAuth(UserInterface $user)
{
//Test for companylock...
if ( !$user->getCompany()->getActive() ) {
throw new LockedException('The company of this user is locked.', $user);
}
if (!$user instanceof AdvancedUserInterface) {
return;
}
if (!$user->isCredentialsNonExpired()) {
throw new CredentialsExpiredException('User credentials have expired.', $user);
}
}
/**
* {#inheritdoc}
*/
public function checkPostAuth(UserInterface $user)
{
//Test for companylock...
if ( !$user->getCompany()->getActive() ) {
throw new LockedException('The company of this user is locked.', $user);
}
if (!$user instanceof AdvancedUserInterface) {
return;
}
if (!$user->isAccountNonLocked()) {
throw new LockedException('User account is locked.', $user);
}
if (!$user->isEnabled()) {
throw new DisabledException('User account is disabled.', $user);
}
if (!$user->isAccountNonExpired()) {
throw new AccountExpiredException('User account has expired.', $user);
}
}
}
* Update nb 3 ********
Now I only have left to make it actually check for the standard user lock which surprisingly it doesn't do out of the box. (Thanks nifr for getting me this far!)
My user entity starts off like this, and like Nifr said, I need to implement the AdvancedUserInterface, but this is probably not the way to do it since it still doesn't check for this lock... but it throws me no error message either (if I change them up and put implememts AdvancedUserInterface and then EXTENDs baseUser it throws an error so...)
<?php
// src/BizTV/UserBundle/Entity/User.php
namespace BizTV\UserBundle\Entity;
use BizTV\UserBundle\Validator\Constraints as BizTVAssert;
use Symfony\Component\Security\Core\User\AdvancedUserInterface;
use FOS\UserBundle\Entity\User as BaseUser;
use Doctrine\ORM\Mapping as ORM;
use BizTV\BackendBundle\Entity\company as company;
/**
* #ORM\Entity
* #ORM\Table(name="fos_user")
*/
class User extends BaseUser implements AdvancedUserInterface
{
Not sure if that's how you do it when you both extend base user and try and implement AdvancedUserInterface, when done as above I still can't use the features it's supposed to add (but it throws me no error message either), but If I switch places of the EXTENDS and IMPLEMENTS like this (line 18)...
class User implements AdvancedUserInterface extends BaseUser
...I get this error:
Parse error: syntax error, unexpected T_EXTENDS, expecting '{' in /var/www/cloudsign/src/BizTV/UserBundle/Entity/User.php on line 18
FOSUserBundle / Symfony already has some kind of "active" flag integrated.
FOS\UserBundle\Model\User already provides the properties "locked" and "enabled" which are intended basically for this purpose. The difference between those two properties is the following ( quoting #stof's comment here)
From the Security component point of view, there is no real
difference: both are forbidden to log in. The difference is a
semantic one: disabled users are generally users that need to
activate their account (for instance, when you activate the need to
confirm the email in FOSUserBundle, the user is disabled on creation
and enabled on confirmation). On the other hand, locking a user is
generally an action done by the admin of the site to ban a user.
Using the same field in the database does not make sense as it would
allow banned user to have access again by simply going through the
confirmation process.
The check for locked/disabled users is being performed by a UserChecker ( symfony provides this one as #security.user_checker ) in FOSUserBundle's AuthenticationListener which implements Symfony\Component\Security\Core\User\UserCheckerInterface.
Now in order to redirect inactive user's to a different route you would:
Catch the Symfony\Component\Security\Core\Exception\DisabledException in the try/catch block in an extended AuthenticationListener
Redirect the user to a certain route if the caught exception is of type InactiveUserException
Optionally move the redirect to a newly created EventListener/-Subscriber which is being dispatched in the extended AuthenticationListener. This way you could later create additional Listeners i.e. for logging purposes and just subscribe them to the inactive-user login-attempt event.

Laravel 4 Auth with Facebook (no password authentication)

I'm trying to set up an authentication system with Laravel 4 with a Facebook login. I am using the madewithlove/laravel-oauth2 package for Laravel 4.
Of course, there is no password to add to my database upon a user loggin in with Facebook. I am, however, trying to check to see if a user id is in the database already to determine if I should create a new entity, or just log in the current one. I would like to use the Auth commands to do this. I have a table called "fans".
This is what I'm working with:
$fan = Fan::where('fbid', '=', $user['uid']);
if(is_null($fan)) {
$fan = new Fan;
$fan->fbid = $user['uid'];
$fan->email = $user['email'];
$fan->first_name = $user['first_name'];
$fan->last_name = $user['last_name'];
$fan->gender = $user['gender'];
$fan->birthday = $user['birthday'];
$fan->age = $age;
$fan->city = $city;
$fan->state = $state;
$fan->image = $user['image'];
$fan->save();
return Redirect::to('fans/home');
}
else {
Auth::login($fan);
return Redirect::to('fans/home');
}
Fan Model:
<?php
class Fan extends Eloquent {
protected $guarded = array();
public static $rules = array();
}
When I run this, I get the error:
Argument 1 passed to Illuminate\Auth\Guard::login() must be an instance of Illuminate\Auth\UserInterface, instance of Illuminate\Database\Eloquent\Builder given
EDIT: When I use: $fan = Fan::where('fbid', '=', $user['uid'])->first();
I get the error:
Argument 1 passed to Illuminate\Auth\Guard::login() must be an instance of Illuminate\Auth\UserInterface, null given, called in /Applications/MAMP/htdocs/crowdsets/laravel-master/vendor/laravel/framework/src/Illuminate/Auth/Guard.php on line 368 and defined
I do not know why it is giving me this error. Do you have suggestions on how I can make this work? Thank you for your help.
You have to implement UserInterface to your model for Auth to work properly
use Illuminate\Auth\UserInterface;
class Fan extends Eloquent implements UserInterface{
...
public function getAuthIdentifier()
{
return $this->getKey();
}
/**
* Get the password for the user.
*
* #return string
*/
public function getAuthPassword()
{
return $this->password;
}
}
getAuthIdentifier and getAuthPassword are abstract method and must be implemented in you class implementing UserInterface
To login any user into the system, you need to use the User model, and I bet inherited classes will do the trick as well but I'm not sure.
Anyway, your Fan model does not associate with the User model/table in any way and that's a problem. If your model had a belong_to or has_one relationship and a user_id field then you could replace Auth::login($user) with Auth::loginUsingId(<some id>).
Original answer:
You are missing an extra method call: ->get() or ->first() to actually retrieve the results:
$fan = Fan::where('fbid', '=', $user['uid'])->first();
Alternatively, you can throw an exception to see what's going on:
$fan = Fan::where('fbid', '=', $user['uid'])->firstOrFail();
If you see different errors, update your question with those errors.

Can (and should?) Zend_Auth return class as the Identity?

I have a class R00_Model_User, which, curiously enough, represents user as he is. Can $result->getIdentity() return me an object of this class? (Or maybe it's stupid?)
(There is a factory method in R00_Model_User which prevents from duplicating objects. I'd like Zend_Auth to use it instead of creating a new object, if it can)
Two options:
write your own authentication adapter subclassing the out-of-the-box-adapter that matches your scenario best
class R00_Auth_Adapter extends Zend_Auth_Adapter_*
{
/**
* authenticate() - defined by Zend_Auth_Adapter_Interface. This method is called to
* attempt an authentication. Previous to this call, this adapter would have already
* been configured with all necessary information to successfully connect to a database
* table and attempt to find a record matching the provided identity.
*
* #throws Zend_Auth_Adapter_Exception if answering the authentication query is impossible
* #return Zend_Auth_Result
*/
public function authenticate()
{
$result = parent::authenticate();
if ($result->isValid() {
return new Zend_Auth_Result(
$result->getCode(),
R00_Model_User::load($result->getIdentity()),
$result->getMessages()
);
} else {
return $result;
}
}
}
This will allow you to code
$adapter = new R00_Auth_Adapter();
//... adapter initialisation (username, password, etc.)
$result = Zend_Auth::getInstance()->authenticate($adapter);
and on successfull authentication your user-object is automatically stored in the authentication storage (session by default).
or use your login-action to update the stored user identity
$adapter = new Zend_Auth_Adapter_*();
$result = $adapter->authenticate();
if ($result->isValid()) {
$user = R00_Model_User::load($result->getIdentity());
Zend_Auth::getInstance()->getStorage()->write($user);
}
In one of my applications, I have getIdentity() return a user object, and it works pretty well for me. To use your factory method, do like this:
$auth = Zend_Auth::getInstance();
$user = R00_Model_User::getInstance(...);
$auth->getStorage()->write($user);
Then when you call getIdentity(), you will have your user object.