Hello i'm trying to display the authenticated user but i'm getting this error :
Undefined variable: users (View:
C:\wamp64\www\Blan\resources\views\users\index.blade.php)
this is my code :
usercontroller :
public function index()
{
return view('users.index')->with('user', Auth::user());
}
the view :
#extends('layouts.app')
#section('content')
<div class="card">
<div class="card-header">
Update Your Profile
</div>
<div class="card-body">
<table class="table table-hover">
<thead>
<th>Image</th>
<th>Name</th>
<th>Update</th>
</thead>
<tbody>
#if( $users -> count() > 0 )
#foreach ($users as $user)
<tr>
<td>
#if(isset($user->profile->avatar))
<img src="{{ asset($user->profile->avatar)}}" width="60px" height="60px" style="border-radius: 50%" alt="">
#else
No image
#endif
</td>
<td>
{{$user->name}}
</td>
<td>
Update
</td>
</tr>
#endforeach
#else
<tr>
<th colspan="5" class="text-center">No Users </th>
</tr>
#endif
</tbody>
</table>
</div>
</div>
#stop
But if i return all the users using this code in the controller :
return view('users.index')->with('users', User::all());
its work fine.
so how i can return only the current authenticated user ?
You don't need to send the auth user via your controller.Auth is the global facade in laravel.Also you don't need foreach because there is only 1 authenticated user. Just type Auth::user()->name in the blade
where you want to display the user
Yea and btw the error ur getting is because you are doing foreach for $users but sending $user to blade but im pretty sure it wont work even if you correct that typo
In your example
public function index()
{
return view('users.index')->with('user', Auth::user());
}
You have used "user"
But tried to access variable as "users"
Unless you just made a typo in your example?
Related
I am trying to get all users and their associated (1:n) journals. However I want to add pagination to the associated journals, not the users
My Controller:
public function index()
{
$users = User::with(['journal'])->orderBy('name', 'asc')->get();
return view('journals/journals', ['users' => $users]);
}
My Blade:
#foreach($users as $user)
<a class="list-group-item list-group-item-action" data-toggle="collapse" href="#collapse{{$user->id}}" role="button" aria-expanded="false" aria-controls="collapse{{$user->id}}">
{{$user->name}}
</a>
<div class="collapse mt-1" id="collapse{{$user->id}}">
<div class="card card-body">
<div class="list-group">
<table class="table table-sm table-hover">
<thead>
<tr>
<th scope="col">Kunde</th>
<th scope="col">Betreff</th>
<th scope="col">Leistungsart</th>
<th scope="col">Beginn</th>
<th scope="col">Ende</th>
<th scope="col">Dauer</th>
<th scope="col">Arbeitszeit</th>
</tr>
</thead>
<tbody>
#foreach($user->journal as $journal)
<tr onclick="window.location='{{route('journal.show', [$journal->id])}}'">
<th scope="row">{{$journal->customer->name}}</th>
<td>{{$journal->title}}</td>
<td>{{$journal->type}}</td>
<td>{{$journal->started}}</td>
<td>{{$journal->ended}}</td>
<td>{{$journal->duration}}</td>
<td>{{$journal->worked}}</td>
</tr>
#endforeach
</tbody>
</table>
</div>
</div>
</div>
#endforeach
My Journal Model:
public function user(){
return $this->belongsTo('App\User', 'user_id', 'id');
}
My User Model:
public function journal(){
return $this->hasMany('App\Journal', 'user_id', 'id');
}
Again, what I am trying to achieve is that I can print out every user and paginate their journals, not paginate the users
I really hope someone can help me there
Render in blade
Above your journals for-each loop in your blade, try and put something like this:
#php
$paginated_journals = $user->journal()->paginate(10);
#endphp
And then, your for-each loop should look like this:
#foreach($paginated_journals as $journal)
...
#endforeach
After the for-each loop you can just put:
$paginated_journals->links()
to get the links for pagination
Pre-load data in controller
You can do the same thing server-side. Create a custom array that is empty, go through each user, and add sub array to that custom array:
array_push($custom_array, ['user' => $user, 'journals' => $user->journal()->paginate(10)])
This way you can send the custom array to your blade, loop through it, and render user data and paginated journals.
home blade
<div class="card-header">Total Users: {{ $users->count() }} </div>
<div class="card-body">
#if (session('status'))
<div class="alert alert-success" role="alert">
{{ session('status') }}
</div>
#endif
<table class="table table-dark table-striped">
<tr>
<th>SL</th>
<th>Name</th>
<th>Email</th>
<th>Created at</th>
</tr>
#foreach($users as $user)
<tr>
<td>{{$loop->index +1}}</td>
<td>{{ $user-> name}}</td>
<td>{{$user->email}}</td>
<td>{{$user->created_at->diffForHumans()}}</td>
</tr>
#endforeach
</table>
</div>
HomeController
public function index()
{
$users = User::all();
return view('home', compact('users'));
}
When I am logged in I want to see all the other users information in a table other than the logged-in user in Laravel. How can I accomplish that?
While the accepted answer works, a nicer approach would be to make use of Laravel's except() Collection method:
The except method returns all items in the collection except for those with the specified keys
Your query then just returns all users except the currently logged in user to your view. No logic in your view, no other changes required:
public function index()
{
$users = User::all()->except(Auth::id);
return view('home', compact('users'));
}
If you pass the current user's ID along with the user list you can simply test against that ID.
public function index()
{
$users = User::all();
return View::make('home', array(
'users' => $users,
'user_id' => Auth::id()
));
}
#foreach($users as $user)
#if ($user->id != $user_id)
<tr>
<td>{{$loop->index+1}}</td>
<td>{{$user->name}}</td>
<td>{{$user->email}}</td>
<td>{{$user->created_at->diffForHumans()}}</td>
</tr>
#endif
#endforeach
I'm Doing a "take home assignment" for job interview. While I have some experience in web development, its not my forte. I am trying to delete a row in a SQLite table using a HTML DELETE button. I am using Laravel-php framework.
I've tried different solutions on google and stack Overflow, but none seem to solve the problem. I modeled my approach after this Laracasts video
Link to my code
The blade seems to be passing the correct info ($id from $contact->id) & the controller seems to be receiving. But the given contact associated with the id isn't being deleted.
FROM BLADE:
<div class="col-md-6">
<table class="table table-striped table-hover">
<tr>
<th>First Name</th>
<th>Last Name</th>
<th>Address</th>
</tr>
#foreach($contacts as $contact)
<tr>
<td> {{$contact->f_name}} </td>
<td> {{$contact->l_name}} </td>
<td> {{$contact->address}} </td>
<td>
<form method="POST" action="/delete/{{ $contact->id }}">
#method('DELETE')
#csrf
<div class="field">
<div class="control">
<button type="submit" class="button">Delete Contact</button>
</div>
</div>
</form>
</td>
</tr>
#endforeach
<tr>
<td></td>
<td></td>
<td></td>
</tr>
</table>
</div>
FROM CONTROLLER:
public function delete($id) {
Contact::find($id)->delete();
return view('index');
}
FROM ROUTE:
Route::delete('/delete', [
'uses'=>'ContactController#delete',
'as'=>'contacts.delete'
]);
you're not getting the id on your delete method, you need to include it on the url like
Route::delete('/delete/{id}', [
'uses'=>'ContactController#delete',
'as'=>'contacts.delete'
]);
in this case, you don't need to change the delete method.
you can also retrieve the id from the request object without changing the delete route, int this case you need to include the id as a hidden input on your view and your delete method will look like this
public function delete(Request $request) {
Contact::find($request->id)->delete();
return view('index');
}
I am using Laravel 5.2.
My question is:
Can variable be classified in view?
For example:
There is a controller ,it can get all of the articles belonging to the current user, like this:
public function index()
{
$user=\Auth::user();
$articles = $user->articles;
return view('user.dashboard.index', compact('articles'));
}
In view,these codes below can show the articles,like this:
<div class="card">
<div class="card-header">
Articles
</div>
#if ($articles!=null)
<table class="table table-sm">
<thead>
<tr>
<th>title</th>
<th>created time</th>
<th>status</th>
</tr>
</thead>
<tbody>
#foreach ($articles as $article)
<tr>
<td>{{$article->title}}</td>
<td>{{$article->created_at}}</td>
<td>{{$article->status}}</td>
</tr>
#endforeach
</tbody>
</table>
#else
<p>Nothing</p>
#endif
</div>
These articles can be classified into two types:
1、“published”，status is 1.
2、“unpublished”，status is 0.
Question：
I would want the published articles to be shown in a card(div class="card"), and, the unpublished articles to be shown in another card.
Can they be classified in view? Thus,it doesn't need to query twice.
Try to use collection method where()
#foreach ($articles->where('published', 1)->all() as $article)
Just change published and 1 to real ones.
I am having some trouble implementing server side pagination in Laravel 5.
Do you have any idea please tell
try with this one
//in controller
$users = \App\User::paginate(15)
return view('your desired view file name', compact('users'));
// in view
<div class="text-center text-muted" role="status" aria-live="polite">Showing {{$users->firstItem()}} to {{$users->lastItem()}} of {{$users->total()}} entries</div>
<div style="display:flex;justify-content:center;align-items:center;" >
<p></p>
{!! with(new Illuminate\Pagination\BootstrapThreePresenter($users))->render()!!}
</div>
In controller you do somrthing like following
$users = \App\User::paginate(15)
return view('template.file', compact('users'));
and in View file you put following
<table>
<thead>
<tr>
<th>Name</th>
</tr>
</thead>
<tbody>
#foreach ($users as $user)
<tr>
<td> {{ $user->name }} </td>
</tr>
#endforeach
</tbody>
</table>
{!! $users->appends(\Input::except('page'))->render() !!}
Last line renders pagination links.