Laravel 5.6 dynamic pages from a database best practice - laravel

I am using/learning Laravel 5.6 and wondered if this is the best approach in trying to accomplish dynamic pages from a database.
The approach I have taken works but I feel it could be improved especially with having to retrieve the pages for the navigation bar with every request.
I have a route in web.php
Route::get('/', 'PageController#index')->name('index');
Route::get('/{page}', 'PageController#show');
I then have my page controller with index and show functions.
public function index()
{
$pages = Page::all();
$posts = Post::latest('created_at')->paginate(2);
return view('index', compact('posts','pages'));
}
public function show($uri)
{
$pages = Page::all();
$page = Page::where('uri', $uri)->first();
return view('templates.page', compact('page','pages'));
}
Now in my header.blade.php I display the list of pages like this:
#foreach($pages as $page)
<li class="nav-item">
<a class="nav-link" href="/{{ $page->uri }}">{{ $page->title }}</a>
</li>
#endforeach
Now my problem is with all the other controllers I have to get the page's information from the database everytime which seems inefficient.
Any advice would be appreciated. Thank you in advance.

You can share your pages data with every view automatically by adding View::share('key', 'value');in the boot method of a service provider. Alternatively you can create a view composer.

one layout blade.
yield title by
<title>#yield('page-title')</title>
then on view,
//process title using php from DB
then display using
#section('page-title')
{{$page_title}}
#endsection

Related

how to display data based on specific id in laravel?

I am new to laravel. I want to send data as id of existing data. Id comes from products.blade I send via href tag as shown below to gallery page. I have tried to find a way through other sites but it still doesn't work
<a class="btn btn-success" href="/dashboard/galleries/{{ $product->id }}"><i class="ri-image-add-line text-white"></i></a>
then i create a route like this
Route::resource('/dashboard/galleries', DashboardGalleryController::class)->middleware('admin')->shallow();
in the controller gallery, I made like this
public function index($id)
{
$gallery = Gallery::where('products_id', $id)->get();
return view('dashboard.galleries.index', compact('gallery'));
}
then inside the gallery page, I want to display a table to add images based on that id.
dashboard/galleries/index.blade.php
<h1>{{ $gallery->id }}</h1>
should i add data inside foreach or outside?
A restful resource controller sets up some default routes for you and even names them.
and ur case the logic should be inside show() function not index()
check this issue it will help u same issue solved here

How to prevent loops in laravel global variable

I've created file GlobalVariable.php inside app\Composers
public function compose($view)
{
$categories = \App\Models\Category::all();
$view->with('globCategory', $categories);
}
then register to AppServiceProvider the code view()->composer('*', GlobalVariable::class);
I use global $globCategory for creating dynamic navbar
<ul class="nav nav-tabs border-0 flex-column flex-lg-row">
#foreach ($globCategory as $item)
<li class="nav-item">
{{$item->name}}
</li>
#endforeach
</ul>
the only problem here when I see laravel debuggar it show repetition of categories query.
here is the result
How to avoid this looping query? is there correct way?
The way you're registering your view composer (using '*' instead of a particular view name), it's going to call the compose method for every single rendered view + subview.
What you can do is instead of this:
view()->composer('*', GlobalVariable::class);
Have this:
\View::share('globCategory', \App\Models\Category::all());
This will globally share your categories (within views), and run the query only once.
View composers, as described from the laravel documentation, bind data to a view every time it is rendered. They clean our code by getting fetching data once and passing it to the view.
While it is possible to get the data in every controller method and pass it to the single view, this approach may be undesirable.
Replace the view name with an asterisk wildcard.

Laravel 5.7 Pagination Customization

$users = Student::paginate(2);
I was trying to paginate some data with paginate function() but am want to show only data no links
You can customize the way pagination is displayed, first step:
php artisan vendor:publish --tag=laravel-pagination
This command publishes the pagination views under the resources/views/vendor directory.
Here you can modify files as you want, the bootstrap-4.blade.php file is the default view for pagination.
You can find more info in the laravel official documentation
Using the $students->links() is optional if yo have a custom blade file showing pagination. However, if you do not need detailed pagination, you can use simplePaginate
$students = Student::simplePaginate(2);
And then in blade file just loop through it :
#foreach($students as $student)
Show student details
#endforeach
If you want to show just next or prev page links then you can use $students->links(), otherwise do not use links at all.
it's simple you must create a function to display your students in your template, add this code in your controller :
public function index()
{
$students = Student::paginate(10);
return view('student.index', ['students ' => $students ]);
}
And on the bottom of your template for example views\student\index.blade.php add :
<div class="row">
<div class="col-md-12">
{{ $students->links() }}
</div>
</div>
if you want display additional pagination information use the following methods: :
Paginator Instance Methods

Making object accessible throughout application

I am adding user notifications to my system. To access these notifications for a user, I call an API I have created in another system. So, my IndexController looks something like the following
public function index()
{
if ($user = Sentinel::getUser()) {
$notifications = MyAPI::returnNotifications($user->id);
return view('centaur.dashboard', compact('notifications'));
}
}
Now to problem with the above is that notifications is now only available on the dashboard view. Within my header view I have something like this
#if($notifications)
#foreach($notifications as $notification)
<a class="content" href="#">
<div class="notification-item">
<h4 class="item-title">{{ $notification->subject }}</h4>
<p class="item-info">{{ $notification->body }}</p>
</div>
</a>
#endforeach
#endif
But if I now visit another page besides the dashboard page I get a Undefined variable: notifications error. This is because header is on every page, but I am only passing my notification object to the dashboard page.
Is there any way to make this notification object universally available?
Thanks
UPDATE
if($user = Sentinel::getUser()) {
view()->composer('*', function ($view) {
$view->with('notifications', MyAPI::returnNotifications($user->id));
});
}
You can use a view composer. In your App\Providers\AppServiceProvider#boot method add:
view()->composer('*', function ($view) {
$view->with('notifications', MyAPI::returnNotifications($user->id););
});
Now you'll have the variable $notifications in all of your views. If you want it for specific ones just replace the * with the view name.

Laravel Sub-menu Within View

Hi I am very new to Laravel and MVC frameworks in general and am looking to create a list of links (in a view within a template) that links to some content. I am using this to display a list of nine people and to display their profile description when the link is clicked on. I have created a model of what the page looks like at http://i.imgur.com/8XhI2Ba.png. The portion that I am concerned with is in blue. Is there a way to route these links to something like /about/link2 or /about?link2 while maintaining the same exact page structure but modifying the ‘link content’ section (on the right of the link menu) to show the specific link's content? I would greatly appreciate it if someone could point me in the right direction, as I have literally no clue where to go with this!
There are a couple ways you can go about doing this.
Templates
Create your route.
Im assuming a lot about your app here but hopefully you get the picture. If you need help with anything in particular, be sure to update your question with the code youve tried so it will be easier to help you.
Route::get('about/{page}', function($page)
{
$profile = Profile::where('name', $page)->first();
return View::make('about')->with('profile', $profile);
});
Modify Template.blade.php
Put this line where you wish for About.blade.php to appear.
#yield('content')
Create your view which will extend your template
#extends('Template')
#section('content')
<h2>User Profile</h2>
<ul>
<li>Name: {{ $profile->name }}</li>
<li>Last Updated: {{ $profile->updated_at }}</li>
</ul>
#stop
AJAX
This solution will utilize AJAX to grab the data from the server and output it on the page.
Route for initial page view
Route::get('about', function($page)
{
$profiles = Profile::all();
return View::make('about')->with('profiles', $profiles);
});
Feel free to follow the same templating structure as before but this time we need to add some javascript into the template to handle the AJAX. Will also need to id everything which needs to be dynamically set so we can easily set it with jquery.
#extends('Template')
#section('content')
<h2>Links</h2>
#foreach($profiles as $profile)
{{ $profile->name }}
#endforeach
<h2>User Profile</h2>
<ul>
<li>Name: <span id="profile_name">{{ $profile->name }}</span></li>
<li>Last Updated: <span id="profile_updated_at">{{ $profile->updated_at }}</span></li>
</ul>
<script>
function setProfile(a)
{
$.ajax({
method: 'get',
url: 'getProfile',
dataType: 'json',
data: {
profile: $(a).data('id')
},
success: function(profile) {
$('#profile_name').html(profile.name);
$('#profile_updated_at').html(profile.updated_at);
},
error: function() {
alert('Error loading data.');
}
});
}
</script>
#stop
Route to handle the AJAX request
Route::get('getProfile', function()
{
$profile_id = Input::get('profile');
$profile = Profile::find($profile_id);
return $profile->toJson();
});
Now, the page should not have to reload and only the profile information should be updated.
Making some assumptions here as no code posted and assuming you're using the latest version of Laravel, Laravel 5.
Lets say you have a table in your database named users and you have a Model named Users (Laravel 5 comes with the Users model as default, see app/Users.php). The users will be the base of our data for the links.
Firstly, you want to register a route so you can access the page to view some information. You can do this in the routes file. The routes file can be found here: app/Http/routes.php.
To register a route add the following code:
Route::get('users', ['uses' => 'UserController#index']);
Now what this route does is whenever we hit the URL http://your-app-name/public/users (URL might be different depending on how you have your app set up, i.e. you may not have to include public) in our web browser it will respond by running the index method on the UserController.
To respond to that route you can set up your UserController as so:
<?php namespace App\Http\Controllers;
class UserController extends Controller {
public function index()
{
}
}
Controllers should be stored in app/Http/Controllers/.
Now lets flesh out the index method:
public function index()
{
// grab our users
$users = App\Users::all();
// return a view with the users data
return view('users.index')->with('users');
}
This grabs the users from the database and loads up a view passing the users data.
Here's what your view could look like:
<!DOCTYPE html>
<html>
<head>
<title>Users Page</title>
</head>
<body>
#foreach($users as $user)
<a href="{{ URL::route('user', ['id' => $user->id]) }}">
{{ $user->username }}
</a>
#endforeach
</body>
</html>
The view code will loop through each user from the $users data we passed to the view and create a link to their user page which is different for each user based on their id (their unique identifier in the DB)
Due to the way I've named it, this would be found in app/views/users/index.blade.php - if you save files ending in blade.php you can use Laravel's templating language, blade.
Now you need to finally set up another route to respond to a user page, for example http://your-app-name/public/user/22.
Route::get('user/{id}', ['uses' => 'UserController#show']);
Then add the show method to UserController
public function show($id)
{
// this will dump out the user information
return \App\User::find($id);
}
Hope that helps a little! Wrote most of it off the top of my head so let me know if you get any errors via comment.
This question is very bare, and it is difficult to actually help your situation without you showing any code. Just to point you in the right direction though, here is what you would need.
A Model called People, this is how you will access your data.
A controller. In this controller you will do the following
Get the ID of the profile you want from the functions parameters
Find that persons information e.g. People::find($person_id);
return the profile view with that persons data e.g. return view('profile')->with('person', $person);
In your view you can then use that data on that page e.g. {{ $person->name }}
For the page that needs to display the links to the people you would have a method in your controller which..
Get all the people data e.g. People::all();
Return a view with that data return view('all-people')->with('people', $people);
You will then need a route to access an individual person. The route will need to pass the persons ID into a controller method e.g.
Route::get('get-person/{id}',
[ 'as' => 'get-person',
'uses' => 'PeopleController#getPerson' ]);
You can then use this route in your view to get the links to each person
#foreach($people as $person)
{{$person->name}}
#endforeach
This would produce the list of links you want.

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