I use this package for follow management, but I'm having a hard time showing all posts from all users that are being followed.
https://github.com/overtrue/laravel-follow
In particular, the problem is that only posts from only ONE user are shown. So even that 10 users are being followed, it only shows one.
My code is this, I'll do something wrong in the query (first()), but putting get() generates error.
$items = Auth::user()->followings()->first()->items();
Related
Is it possible to use cypress for testing relationship between models?
For example, I have this kind of relationship: a teacher has many students, each student belongs to a teacher. After the teacher A logged in, at url "/my-students", he or she will see a list of all his or her students.
What I want to test is to make sure none of the students listed on "/my-students" page belong to teacher B than teacher A.
Can I test this case with cypress? Is it possible and how to do it if it's possible?
The short answer is "yes" you can absolutely do this kind of testing. There are dozens of ways, but I'm going to suggest what I consider to be the simplest approach.
Make sure the data being used by your website doesn't change. You want your tests to be deterministic... none of this run a database query to determine the expected results stuff.
The first time through, verify the page contents manually
Use cy.snapshot() to record the current page state for future comparison. This is an additional npm package from Gleb Bahmutov (a Cypress developer). Full instructions, including installation, can be found here.
Your hypothetical test would look something like this:
describe('student directory page', () => {
beforeEach(() => {
// Log in
cy.logIn('lizstewart#example.com') // This is usually a custom command; up to you
})
it('displays the correct students', () => {
// Go to the page
cy.visit('my-students')
// Check for the correct students
cy.get('#studentList').snapshot()
})
})
The first time the test runs it will pass no matter what, and will write out a file titled snapshots.js, which you can commit to your repo. All subsequent test runs will fail if the HTML output doesn't match the previous content exactly.
It's a blunt approach, but it's quick and effective.
Need help with Facebook SDK Php Facebook Like.
My website fetches Logged in FB User's Timeline and user can Like posts in his timeline.
Posting Likes works for me successfully. But my question is how should I know he had already liked the post.
I want to show Like button if he has not liked that object and unlike button if he has already liked it.
PS: I tried collecting all the list of users who liked and then compared it with logged in user's fb id, but this method takes long time if likes are more than 10k.
Is there any other methods to accomplish this task.
I am using facebook-php-sdk-v4-4.0-dev
PS: My code for showing his feed is:
(new FacebookRequest($session, 'POST',"/me/feed", $params))->execute()->getGraphObject()->asArray();
The only solution available now is looping through the list of likes like you do , as of now .
there is another solution using FQL
select user_id from like where object_id=your object_id AND user_id=me()
but the problem is that he LIKE table only considers videos, notes, links, photos and albums, not posts.
check facebook fql like documentation . also note that fql is on track for full deprecation .
May be this will help you
$fql= "select user_id from like where object_id=REQUIRE_OBJECT_ID_HERE AND user_id=me()";
$request = (new FacebookRequest($session, 'GET', "/fql?q=$fql"))->execute()->getGraphObject()->asArray();
Change REQUIRE_OBJECT_ID_HERE variable to the id that you need.
I'm trying to find all users w/ a specific permissions list in Sentry with laravel. The problem is that Sentry::findAllUsersWithAccess() returns an array().
as stated in their github repository i pinpointed their code to be
public function findAllWithAccess($permissions)
{
return array_filter($this->findAll(), function($user) use ($permissions)
{
return $user->hasAccess($permissions);
});
}
right now, it gets all users and filter it out with users with permission list. the big problem would be when I as a developer would get the set of users, it'll show ALL users, i'm developing an app which may hold thousands of users and i only need to get users with sepcific permission lists.
With regards to that would love to use one with a ->paginate() capability.
Any thoughts how to get it without getting all the users.
Why dont you override the findAllWithAccess() method and write your own implementation, which uses mysql where instead of array_filter().
I dont know your project structure and the underlying db schema, so all i can give you atm is the link to the eloquent documentation Querying Relations (whereHas).
In case you dont know where to start: its always a good idea to look at the ServiceProvider (SentryServiceProvider, where the UserProvider, which holds the findAllWidthAccess() method, is registered). Override the registerUserProvider method and return your own implementation of the UserProvider (with the edited findAllWithAccess() method).
Hope that will point you in the right direction.
In Laravel you can do pagination manually on arrays:
$paginator = Paginator::make($items, $totalItems, $perPage);
Check the docs: http://laravel.com/docs/pagination
I am making a site for a client and decided i would use code igniter.
The site essentially has two backends, one for designers, and one for a sales team. So after logging in, the user will be redirected to either
mysite.com/sales/
mysite.com/design/
The sales team for example can view orders, containers, products, therefore i need a controller for each of these.
mysite.com/sales/orders/
The sales need to be able to view, edit, delete certain orders...
mysite.com/sales/orders/vieworder/235433
Basically my problem is i dont have enough url segments to play with.
My thoughts on solving my problem
removing the orders, containers, products classes and making ALL of their methods as methods of the sales class, although this would mean a large number of methods and loading all models so it seemed kind of pointless.
removing the sales/designer classes and controlling what each kind of user has access to based on a user type stored in session data.
a way of having an extra url segment?
I appreciate any advice, I just dont want to get 3 weeks into the project and realise i started wrong from the beginning!
Use folders.
If you make a subfolder in /application/ called sales you can put different controllers in there:
/application/
/sales/
orders.php /* Controller */
/design/
Then in orders.php you will put your vieworders($id) method and so on, and you will be able to acces it with domain.com/sales/orders/vieworders/id.
You can also make subfolders in the /models/ and /views/ to organize your files.
Now, Access Control is something apart and it depends more in the auth system you are using.
Give the user/designer a privilege, a column in the user table for example, check the permission of the user at the beginning of the function, then prevent or execute it.
This would be the exact way i would do it.
Seems like you should have a look into the routing class. Might be a dirty solution but rerouting the sales/(:any) to something like sales_$1 would mean you'd make controllers with names like sales_orders.
Same for the design part.
(FYI: $routing['sales/(:any)'] = 'sales_$1'; should do the trick; see application/config/routing.php).
So, I'm not quite sure how I should structure this in CakePHP to work correctly in the proper MVC form.
Let's, for argument sake, say I have the following data structure which are related in various ways:
Team
Task
Equipment
This is generally how sites are and is quite easy to structure and make in Cake. For example, I would have the a model, controller and view for each item set.
My problem (and I'm sure countless others have had it and already solved it) is that I have a level above the item sets. So, for example:
Department
Team
Task
Equipment
Department
Team
Task
Equipment
Department
Team
Task
Equipment
In my site, I need the ability for someone to view the site at an individual group level as well as move to view it all together (ie, ignore the groups).
So, I have models, views and controls for Depart, Team, Task and Equipment.
How do I structure my site so that from the Department view, someone can select a Department then move around the site to the different views for Team/Task/Equipment showing only those that belong to that particular Department.
In this same format, is there a way to also move around ignoring the department associations?
Hopefully the following example URLs clarifies anything that was unclear:
// View items while disregarding which group-set record they belong to
http://www.example.com/Team/action/id
http://www.example.com/Task/action/id
http://www.example.com/Equipment/action/id
http://www.example.com/Departments
// View items as if only those associated with the selected group-set record exist
http://www.example.com/Department/HR/Team/action/id
http://www.example.com/Department/HR/Task/action/id
http://www.example.com/Department/HR/Equipment/action/id
Can I get the controllers to function in this manner? Is there someone to read so I can figure this out?
Thanks to those that read all this :)
I think I know what you're trying to do. Correct me if I'm wrong:
I built a project manager for myself in which I wanted the URLs to be more logical, so instead of using something like
http://domain.com/project/milestones/add/MyProjectName I could use
http://domain.com/project/MyProjectName/milestones/add
I added a custom route to the end (!important) of my routes so that it catches anything that's not already a route and treats it as a "variable route".
Router::connect('/project/:project/:controller/:action/*', array(), array('project' => '[a-zA-Z0-9\-]+'));
Whatever route you put means that you can't already (or ever) have a controller by that name, for that reason I consider it a good practice to use a singular word instead of a plural. (I have a Projects Controller, so I use "project" to avoid conflicting with it.)
Now, to access the :project parameter anywhere in my app, I use this function in my AppController:
function __currentProject(){
// Finding the current Project's Info
if(isset($this->params['project'])){
App::import('Model', 'Project');
$projectNames = new Project;
$projectNames->contain();
$projectInfo = $projectNames->find('first', array('conditions' => array('Project.slug' => $this->params['project'])));
$project_id = $projectInfo['Project']['id'];
$this->set('project_name_for_layout', $projectInfo['Project']['name']);
return $project_id;
}
}
And I utilize it in my other controllers:
function overview(){
$this->layout = 'project';
// Getting currentProject id from App Controller
$project_id = parent::__currentProject();
// Finding out what time it is and performing queries based on time.
$nowStamp = time();
$nowDate = date('Y-m-d H:i:s' , $nowStamp);
$twoWeeksFromNow = $nowDate + 1209600;
$lateMilestones = $this->Project->Milestone->find('all', array('conditions'=>array('Milestone.project_id' => $project_id, 'Milestone.complete'=> 0, 'Milestone.duedate <'=> $nowDate)));
$this->set(compact('lateMilestones'));
$currentProject = $this->Project->find('all', array('conditions'=>array('Project.slug' => $this->params['project'])));
$this->set(compact('currentProject'));
}
For your project you can try using a route like this at the end of your routes.php file:
Router::connect('/:groupname/:controller/:action/*', array(), array('groupname' => '[a-zA-Z0-9\-]+'));
// Notice I removed "/project" from the beginning. If you put the :groupname first, as I've done in the last example, then you only have one option for these custom url routes.
Then modify the other code to your needs.
If this is a public site, you may want to consider using named variables. This will allow you to define the group on the URL still, but without additional functionality requirements.
http://example.com/team/group:hr
http://example.com/team/action/group:hr/other:var
It may require custom routes too... but it should do the job.
http://book.cakephp.org/view/541/Named-parameters
http://book.cakephp.org/view/542/Defining-Routes
SESSIONS
Since web is stateless, you will need to use sessions (or cookies). The question you will need to ask yourself is how to reflect the selection (or not) of a specific department. It could be as simple as putting a drop down selection in the upper right that reflects ALL, HR, Sales, etc. When the drop down changes, it will set (or clear) the Group session variable.
As for the functionality in the controllers, you just check for the Session. If it is there, you limit the data by the select group. So you would use the same URLs, but the controller or model would manage how the data gets displayed.
// for all functionality use:
http://www.example.com/Team/action/id
http://www.example.com/Task/action/id
http://www.example.com/Equipment/action/id
You don't change the URL to accommodate for the functionality. That would be like using a different URL for every USER wanting to see their ADDRESS, PHONE NUMBER, or BILLING INFO. Where USER would be the group and ADDRESS, PHONE NUMBER< and BILLING INFO would be the item sets.
WITHOUT SESSIONS
The other option would be to put the Group filter on each page. So for example on Team/index view you would have a group drop down to filter the data. It would accomplish the same thing without having to set and clear session variables.
The conclusion is and the key thing to remember is that the functionality does not change nor does the URLs. The only thing that changes is that you will be working with filtered data sets.
Does that make sense?