I am trying to add new models to RocketChat specifically to store Location information. I have extended the models._Base model to ModelLocations and assigned it to RocketChat.models.Locations as following
RocketChat.models.Locations = new ModelLocations('location', true);
But when I try to use RocketChat.models.Locations it's always undefined. I observed the migrations folder and thought I need to create a migration file but I couldn't find documentation for that. Can anybody point me on how to add a new model.
I forgot to add the new model's file path in Rocket.Chat/packages/rocketchat-lib/package.js. Once added it works.
Related
I can't figure out how to set up the links. I have created a collection. I need an creator to be automatically specified in that record when adding a record to the collection. How do I do that?
I've faced this problem...
I found the solution in the documentation:
need to add
"populateCreatorFields": true
in the file (strapi v3) /api/your-type-name/models/your-type-name.settings.json
in options object
like:
...
"options": {
"increments": true,
"timestamps": true,
"draftAndPublish": true,
"populateCreatorFields": true
},
...
More information can be found at strapi documentation
Strapi does not support it by default. As mentioned in this form you can achieve it by editing the strapi's controller. But I will not recommend you to edit the strapi controller at all. Please avoid it.
There is a simple and better solution to this. You can achieve this by creating one to one relationship. Create an author table/collection. Make one to one relationship with your other collection. You can make it a required option as well. So whenever someone creates an entry they have to select an author from your already created collection of authors.
And now you can get relation in your API and use it wherever you want to.
As stated in my comment, Strapi (tested on v3) comes with a created by field. To ascertain the claim, the following steps can be followed;
Create a new content type. Here I am setting only one field, called test
Add an entry for that content type. Notice the last update and by field on the right.
Save and open the entry. The last update and by fields have been automatically populated.
How would I override the View::make('cashier::receipt'); view so that when that particular namespace is called like that, it checks my folder first and then defaults back to the vendor path.
View::addNamespace('cashier', [
'/path/to/my/views', // check first
'/path/to/original/views' // check second
]);
I believe that's how Laravel handles custom views for packages already. https://laravel.com/docs/5.2/packages#views - see Overriding Package Views here.
Laravel registers two locations to load views so they can be easily customised, the standard vendor path and something customisable.
Laravel will first check if a custom version of the view has been provided by you, for example in
resources/views/vendor/cashier.
Let me know if there was something more specific you were trying to achieve but I believe the info there will get you going.
yo meanjs:crud-module
generates a module similar to 'articles' modules.
Is there a way to generate module from a custom model/schema ?
The generator is only a starting point for your module.
When creating the module, start from the back and edit your way to the front:
modules/articles/server/models/articles.server.model.js
change the schema to fit your requirements. The controllers and routes on the server side don't need any changing. Then you move to:
modules/articles/client/views/view-articles.client.view.html
Here you add new inputs that matches your schema. After that you should be done with your 'custom model/schema'. Hope this helps.
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 had my users profile at
www.domain.com/user/username
and moved it at
www.domain.com/username
but this required to add most classes functions into the routes.php file in the config and if i want to add new features to my app i will need to add all the functions into the routes.php file which doesnt sound good practice...
What is the best way that other deal with it on CodeIgniter ?
Perhaps, you can do it the other way round - make a whitelist of usernames that can't be taken (those would be names of your controllers, like admin, contact, etc...) and route anything except the whitelist items.
seems i got the answer
what i did is add the below code for every controller i have
$route['controller'] = "controller";
$route['controller/(:any)'] = "controller/$1";
and this code at the bottom
$route['(:any)'] = "user/$1";