When using the Auth component in CakePHP 3 you can define the findAuth() finder (or configure a different finder) to have control over what data is loaded:
// AppController
$this->loadComponent('Auth', [
//...
'authenticate' => [
'Form' => [
'finder' => 'auth'
]
],
//...
]);
// UsersTable
public function findAuth($query, array $options)
{
return $query
->...;
}
I need some functionality that cannot be done with the query builder. How can I post-process the loaded auth data before session creation?
Note that I have different ways of logging in my users, so I would prefer this be kept inside the AuthComponent logic.
(This is still for CakePHP 3, but a brief comment on how this could be done in the new CakePHP 4 Authentication plugion would also be appriciated.)
EDIT: Rough outline of what I need: data needs to re-organised in the users array based on current context, i.e. users can have an active project selected.
I'm still not really sure what exactly you need to re-organize in what way exactly, but generally you can modify the queried data using mappers/reducers and result formatters, the latter usually being the easier way.
Here's a quick example that would add an additional field named additional_data to the result in case a field named active_project_id is set:
$query->formatResults(function (\Cake\Collection\CollectionInterface $results) {
return $results->map(function ($row) {
if (isset($row['active_project_id'])) {
$row['additional_data'] = 'lorem ipsum';
}
return $row;
});
});
Such a finder query would work with the new authentication plugin too.
See also
Cookbook > Database Access & ORM > Query Builder > Adding Calculated Fields
Related
I'm trying to set up easy test data in my Acceptance tests:
public function shouldUseAFakeAccountHolder(AcceptanceTester $I) {
$I->have(AccountHolder::class);
// ...
}
I've copied the example code from the Codeception documentation and modified it with my entity names (as well as fixing the bugs).
<?php
public function _beforeSuite()
{
$factory = $this->getModule('DataFactory');
// let us get EntityManager from Doctrine
$em = $this->getModule('Doctrine2')->_getEntityManager();
$factory->_define(AccountHolder::class, [
'firstName' => Faker::firstName(),
// Comment out one of the below 'accountRole' lines before running:
// get existing data from the database
'accountRole' => $em->getRepository(AccountRole::class)->find(1),
// create a new row in the database
'accountRole' => 'entity|' . AccountRole::class,
]);
}
The relationship using existing data 'accountRole' => $em->getRepository(AccountRole::class)->find(1) always fails:
[Doctrine\ORM\ORMInvalidArgumentException] A new entity was found through the relationship 'HMRX\CoreBundle\Entity\AccountHolder#accountRole' that was not configured to cascade persist operations for entity: HMRX\CoreBundle\Entity\AccountRole#0000000062481e3f000000009cd58cbd. To solve this issue: Either explicitly call EntityManager#persist() on this unknown entity or configure cascade persist this association in the mapping for example #ManyToOne(..,cascade={"persist"}). If you cannot find out which entity causes the problem implement 'HMRX\CoreBundle\Entity\AccountRole#__toString()' to get a clue.
If I tell it to create a new entry in the related table 'accountRole' => 'entity|' . AccountRole::class, it works, but then it adds rows to the table when it should be using an existing row. All the role types are known beforehand, and a new random role type makes no sense because there's nothing in the code it could match to. Creating a duplicate role works, but again it makes so sense to have a separate role type for each user since roles should be shared by users.
I've had this error before in Unit tests, not Acceptance tests, when not using Faker / FactoryMuffin, and it's been to do with accessing each entity of the relationship with a different instance of EntityManager. As soon as I got both parts using the same instance, it worked. I don't see how to override the native behaviour here though.
It works (at least in Codeception 4.x) by using a callback for the existing relation:
<?php
public function _beforeSuite()
{
$factory = $this->getModule('DataFactory');
$em = $this->getModule('Doctrine2')->_getEntityManager();
$factory->_define(AccountHolder::class, [
'firstName' => Faker::firstName(),
'accountRole' => function($entity) use ($em) {
$em->getReference(AccountRole::class)->find(1);
},
]);
}
I've found it here: https://github.com/Codeception/Codeception/issues/5134#issuecomment-417453633
I am having an issue inserting a record into the database. I am a beginner with the Yii framework, so I may have made some stupid mistakes.
This is from the SiteController
public function actionCreatePost(){
$model = new PostForm();
$post = new Post();
if ($model->load(Yii::$app->request->post()) && $model->validate()) {
$post->body = $model->body;
$post->title = $model->title;
$post->save();
return $this->redirect('index');
}else {
return $this->render('createPost', ['model' => $model]);
}
}
This is from the Post class
public function behaviors()
{
return [
[
'class' => TimestampBehavior::className(),
'createdAtAttribute' => 'created_at',
'updatedAtAttribute' => 'updated_at',
'value' => new Expression('NOW()'),
],
[
'class' => BlameableBehavior::className(),
'createdByAttribute' => 'id_author',
]
];
}
The issue is that you have created a PostForm class for the form (which is correct) but you are then trying to load the response into the Post class - a completely different class. This won’t work without modification.
If you have a look at the response:
var_dump(Yii:$app->request->post());
You will see the form data is located within the PostForm key. Yii will therefore only load the data into the PostForm class.
The correct solution is therefore to create a savePost() function within the PostForm eg:
public function savePost(){
$model = new Post();
$model->propertyABC = $this->propertyABC
...etc...
$model->save();
So the action would appear as follows:
$model = new PostForm();
If($model->load(Yii::$app->request->post()) && $model->validate()){
$model->savePost();
The other option is to rename the key from PostForm to Post. Yii will then load the data but this is not the best approach as it is a bit obscure.
Hope that helps
I would guess the issue is with the validation.
I can see several issues I will point out. First, I cannot figure out why are you creating a new PostForm, loading the data in it and verifying it, just to dump some values in a new Post and save it. Are there some functions, you are running in the PostForm model, that are triggered by load or verify? If that is not the case, I would suggest dropping one of the models, and using only the other. Usually, that is the Form model. It serves as a link between the ActiveForm and the model handling everything. You can do everything in the createPost() function in the Form model, and then in the controller it will look like
if ($model->load(Yii::$app->request->post())) {
$model->save();
return $this->redirect('index');
}
Second of all, you can dump post->getErrors() before the save to see if there are any errors with the validation. What you can also do, is call $post->save(false) instead. If you pass false to it, it will not trigger $post->validate(), and some errors can be neglected. Please, let me know if there is anything unclear.
I am using cakephp 3.7.2 with Authentication component
$user = $this->Authentication->getIdentity();
prints:
object(Authentication\Identity) {
'config' => [
'fieldMap' => [
'id' => 'id'
]
],
'data' => object(App\Model\Entity\User) {
'id' => (int) 1,
'email' => 'aa.aaa#gmail.com',
...
}
}
I have tried $user->data but it doesn't work.
How to print user data?
Authentication Component
So I have figured it out.
In User Entity class
Add use Authentication\IdentityInterface;
and then implement the IdentityInterface.
class User extends Entity implements IdentityInterface
{
blablabla...
yale yale yale ...
Now you can print:
$user = $this->Authentication->getIdentity();
debug($user->id);
As per Authentication component documentation
The identity object is returned by the service and made available in
the request. The object provides a method getIdentifier() that can be
called to get the id of the current log in identity.
You can use this accordingly as below to get the user data:
// Service
$identity = $authenticationService
->getIdentity()
->getIdentifier()
// Component
$identity = $this->Authentication
->getIdentity()
->getIdentifier();
// Request
$identity = $this->request
->getAttribute('identity')
->getIdentifier();
The identity object provides ArrayAccess but as well a get() method to
access data. It is strongly recommended to use the get() method over
array access because the get method is aware of the field mapping.
For eg. to access email and username from the identity you can use the below code.
$identity->get('email'); // to access email
$identity->get('username'); // to access username
Reference link: Authentication -> Docs -> Identity Object
Hope this will help.
I´m using AuthComponent in CakePHP 4.xxx.
I can get User data i.e. in a view with
$user = $this->getRequest()->getAttribute('identity');
I found the Information on: http://gotchahosting.com/blog/category/cakephp/4002624
Maybe it helps someone who is looking for information about this in CakePHP4
Laravel 4: In the context of consume-your-own-api, my XyzController uses my custom InternalAPiDispatcher class to create a Request object, push it onto a stack (per this consideration), then dispatch the Route:
class InternalApiDispatcher {
// ...
public function dispatch($resource, $method)
{
$this->request = \Request::create($this->apiBaseUrl . '/' . $resource, $method);
$this->addRequestToStack($this->request);
return \Route::dispatch($this->request);
}
To start with, I'm working on a basic GET for a collection, and would like the Response content to be in the format of an Eloquent model, or whatever is ready to be passed to a View (perhaps a repository thingy later on when I get more advanced). It seems inefficient to have the framework create a json response and then I decode it back into something else to display it in a view. What is a simple/efficient/elegant way to direct the Request to return the Response in the format I desire wherever I am in my code?
Also, I've looked at this post a lot, and although I'm handling query string stuff in the BaseContorller (thanks to this answer to my previous question) it all seems to be getting far too convoluted and I feel I'm getting lost in the trees.
EDIT: could the following be relevant (from laravel.com/docs/templates)?
"By specifying the layout property on the controller, the view specified will be created for you and will be the assumed response that should be returned from actions."
Feel free to mark this as OT if you like, but I'm going to suggest that you might want to reconsider your problem in a different light.
If you are "consuming your own API", which is delivered over HTTP, then you should stick to that method of consumption.
For all that it might seem weird, the upside is that you could actually replace that part of your application with some other server altogether. You could run different parts of your app on different boxes, you could rewrite the HTTP part completely, etc, etc. All the benefits of "web scale".
The route you're going down is coupling the publisher and the subscriber. Now, since they are both you, or more accurately your single app, this is not necessarily a bad thing. But if you want the benefits of being able to access your own "stuff" without resorting to HTTP (or at least "HTTP-like") requests, then I wouldn't bother with faking it. You'd be better off defining a different internal non-web Service API, and calling that.
This Service could be the basis of your "web api", and in fact the whole HTTP part could probably be a fairly thin controller layer on top of the core service.
It's not a million miles away from where you are now, but instead of taking something that is meant to output HTTP requests and mangling it, make something that can output objects, and wrap that for HTTP.
Here is how I solved the problem so that there is no json encoding or decoding on an internal request to my API. This solution also demonstrates use of route model binding on the API layer, and use of a repository by the API layer as well. This is all working nicely for me.
Routes:
Route::get('user/{id}/thing', array(
'uses' => 'path\to\Namespace\UserController#thing',
'as' => 'user.thing'));
//...
Route::group(['prefix' => 'api/v1'], function()
{
Route::model('thing', 'Namespace\Thing');
Route::model('user', 'Namespace\User');
Route::get('user/{user}/thing', [
'uses' => 'path\to\api\Namespace\UserController#thing',
'as' => 'api.user.thing']);
//...
Controllers:
UI: UserController#thing
public function thing()
{
$data = $this->dispatcher->dispatch('GET', “api/v1/user/1/thing”)
->getOriginalContent(); // dispatcher also sets config flag...
// use $data in a view;
}
API: UserController#thing
public function thing($user)
{
$rspns = $this->repo->thing($user);
if ($this->isInternalCall()) { // refs config flag
return $rspns;
}
return Response::json([
'error' => false,
'thing' => $rspns->toArray()
], 200);
Repo:
public function thing($user)
{
return $user->thing;
}
Here is how I achieved it in Laravel 5.1. It requires some fundamental changes to the controllers to work.
Instead of outputting response with return response()->make($data), do return $data.
This allows the controller methods to be called from other controllers with App::make('apicontroller')->methodname(). The return will be object/array and not a JSON.
To do processing for the external API, your existing routing stays the same. You probably need a middleware to do some massaging to the response. Here is a basic example that camel cases key names for the JSON.
<?php
namespace App\Http\Middleware;
use Closure;
class ResponseFormer
{
public function handle($request, Closure $next)
{
$response = $next($request);
if($response->headers->get('content-type') == 'application/json')
{
if (is_array($response->original)) {
$response->setContent(camelCaseKeys($response->original));
}
else if (is_object($response->original)) {
//laravel orm returns objects, it is a huge time saver to handle the case here
$response->setContent(camelCaseKeys($response->original->toArray()));
}
}
return $response;
}
}
I'm moving some of my find code inside models.
Previously in my controller I had
$this->Book->Review->find('first', array(
'conditions' => array(
'Review.book_id' => $id,
'Review.user_id' => $this->Auth->user('id')
)
));
so in my Review model I put something like
function own($id) {
$this->contain();
$review = $this->find('first', array(
'conditions' => array(
'Review.book_id' => $id,
'Review.user_id' => AuthComponent::user('id')
)
));
return $review;
}
So I'm calling AuthComponent statically from the Model. I know I can do this for the method AuthComponent::password(), which is useful for validation. But I'm getting errors using the method AuthComponent::user(), in particular
Fatal error: Call to a member function
check() on a non-object in
/var/www/MathOnline/cake/libs/controller/components/auth.php
on line 663
Is there a way to get the info about the currently logged user from a model?
Create a new function in the "app_model.php" ("AppModel.php" in CakePHP 2.x), so it will be available at all models within our application:
function getCurrentUser() {
// for CakePHP 1.x:
App::import('Component','Session');
$Session = new SessionComponent();
// for CakePHP 2.x:
App::uses('CakeSession', 'Model/Datasource');
$Session = new CakeSession();
$user = $Session->read('Auth.User');
return $user;
}
in the model:
$user = $this->getCurrentUser();
$user_id = $user['id'];
$username = $user['username'];
The way that I use is this:
App::import('component', 'CakeSession');
$thisUserID = CakeSession::read('Auth.User.id');
It seems to work quite nicely :-)
I think the code is fine as it is and belongs in the Controller, or at the very least it needs to receive the ids from the Controller and not try to get them itself. The Model should only be concerned with fetching data from a data store and returning it. It must not be concerned with how the data is handled in the rest of the application or where the parameters to its request are coming from. Otherwise you paint yourself into a corner where the ReviewModel can only retrieve data for logged in users, which might not always be what you want.
As such, I'd use a function signature like this:
function findByBookAndUserId($book_id, $user_id) {
…
}
$this->Review->findByBookAndUserId($id, $this->Auth->user('id'));
There is a nice solution by Matt Curry. You store the data of the current logged user in the app_controller using the beforeFilter callback and access it later using static calls. A description can be found here:
http://www.pseudocoder.com/archives/2008/10/06/accessing-user-sessions-from-models-or-anywhere-in-cakephp-revealed/
EDIT: the above link is outdated: https://github.com/mcurry/cakephp_static_user
I think this is not good idea to get value from Session. Better solution to get logged user id inside any model simply try this:
AuthComponent::user('id');
This will work almost every where. View, Model and Controller
Dirtiest way would be to just access the user information in the Session. Least amount of overhead associated with that.
The "proper" way would probably be to instantiate the AuthComponent object, so that it does all the stuff it needs to be fully operational. Much like a death star, the AuthComponent doesn't really work well when not fully setup.
To get a new AC object, in the model:
App::import( 'Component', 'Auth' );
$this->Auth = new AuthComponent();
Now you can use $this->Auth in the model, same as you would in the controller.
For CakePHP 3.x this easy component is available: http://cakemanager.org/docs/utils/1.0/components/globalauth/. Direct accessing the Session is not possible because of different SessionKeys.
With the GlobalAuthComponent you can access your user-data everywhere with: Configure::read('GlobalAuth');.
Greetz
Bob
I use cake 2.2 and these both work great:
$this->Session->read('Auth.User');
//or
$this->Auth->user();
You can also get a field of currently logged in user:
$this->Session->read('Auth.User.email');
//or
$this->Auth->user()['email'];
None of these solutions work in CakePHP version 3. Anyone know of a way to do this? Right now, I'm completely stepping around the framework by accessing the $_SESSION variable directly from my model.