I have users table and each user have a column for 'qr_url' and it's a random 170 characters length string generated with Str::random
I have a page that displays all the users and for each user there is a button named inquiry to check if the user exists in the database
I have a route name inquiry that check if this qr_url already for a user or not
Route::get('/users/inquiry/{query}', [UserController::class, 'inquiry'])->name('users.inquiry');
in the inquiry() function I receive this query that contain the qr_url
public function inquiry($query)
{
$user = User::where('qr_url', '=', $query)->first();
return view('users/inquiry', compact('user'));
}
I need to know how can I make the URL looks like that
http://example.com/Home/GetResultByQR?query=theqr_url
Hello guy you should use request()->get('query'). For example:
public function inquiry()
{
$user = User::where('qr_url', '=', request()->get('query'))->first();
return view('users/inquiry', compact('user'));
}
Related
I want to select all the users in my table "User" except the first One cause its the admin,
im using this function index in my controller but it doesn't work .
public function index()
{
// this '!=' for handling the 1 row
$user = User::where('id', '!=', auth()->id())->get();
return view('admin.payroll',compact('user'))->with(['employees' => User::all()]);
}
Better to used here whereNotIn Method
Note: first you find the admin role users and create a static array
$exceptThisUserIds = [1];
$user = User::whereNotIn('id', $exceptThisUserIds)->get();
It's not a good idea to specify the admin user with just id. A better design would be using some sort of a flag like is_admin as a property of your User model.
Still you can use the following code to get the users who have an id greater than 1:
User::where('id', '>', 1)->get()
For getting data skipping the first one you should use skip() method and follow the code like below
public function index()
{
$user = User::orderBy('id','asc')->skip(1)->get();
return view('admin.payroll',compact('user'))->with(['employees' => User::all()]);
}
I have a table named 'Feedbacks' with 5 fields:
('id', 'user_id', 'instruction', 'description', 'fk_eleve')
The admin can create several recordings, here is a screenshot.
Here, I have a recording which is Menier Jeremy it's the student
In my rubric 'Eleves' (english: Student) we can see several recordings:
Here, Menier Jeremy has like email address test#gmail.com.
Menier Jeremy wants to connect...
There are 2 rubrics for now:
- Student Profil and Feedback
the user Menier Jeremy can see his profil
However, when the user wants to see his 'feedback'.
Unfortunately, I get the following error message:
SQLSTATE [42S22]: Column not found:
1054 Field Field 'email' unknown in where (SQL: select count (*) as aggregate fromreturnswhere email= test#gmail.com)
I have a problem with the email ?
public function index(Request $request)
{
$user = $request->user();
$feedbacks = Feedback::query()
->when($user->hasRole('admin') !== true, function (Builder $query) use ($user) {
$query->where('email', $user->email);
})
->when($request->has('search'), function (Builder $query) use ($request) {
$query->join('eleves', 'feedbacks.fk_eleve', '=', 'eleves.id')
->orderBy('eleves.nom', 'asc')
->where('eleves.nom', 'like', '%'.$request->input('search').'%');
})
->paginate(5);
return view('admin.feedbacks.index', compact('feedbacks'))
->with('display_search', $user->hasRole('admin'));
}
Do know you where is the problem?
Thank you a lot
Edit code #Watercayman
When, the user Jeremy Menier is connected, I see several recordings:
I have to retrieve only ID n° 1
public function index(Request $request)
{
$user = $request->user();
$feedbacks = Feedback::query()
->when($user->hasRole('admin') !== true, function (Builder $query) use ($user) {
\Auth::user()->load('feedbacks');
$feedbacksForThisUser = \Auth::user()->feedbacks;
})
->when($request->has('search'), function (Builder $query) use ($request) {
$query->join('eleves', 'feedbacks.fk_eleve', '=', 'eleves.id')->orderBy('eleves.nom', 'asc')->where('eleves.nom','like','%'.$request->input('search').'%');
})
->paginate(5);
return view('admin.feedbacks.index', compact('feedbacks'))
->with('display_search', $user->hasRole('admin'));
}
My model User:
public function feedbacks()
{
return $this->hasMany('App\Feedback', 'user_id', 'id');
}
My model Feedback
public function user()
{
return $this->belongsTo('App\User', 'id', 'user_id');
}
Sure, you can get the email for the users that have provided feedback using the FK. But I don't think you need all of the query you are using. I think you can make this much simpler. Because the user has a relation with the Feedback model (I assume through an Eleve model), you can simply eager load the relationship using the name search if you wish:
// Note I have removed the query() method you had in your original query
$feedbacks = Feedback::with('eleves', function ($query) use($request)
$query->orderBy('eleves.nom','asc')->where('eleves.nom','like','%'.$request->input('search').'%');
})->paginate(5);
To simplify the explanation I've removed the where from the query. To demonstrate I'll just use an if check before the query - you can make this more complex if you wish, but it is a little easier to see this way.
Something like:
if($request->has('search'){ do the query above }
else { $feedbacks = Feedback::with('eleves')->paginate(5); }
Then, because you have pulled the relations, you can then just get the email from any eleves that have been loaded:
$feedback->first()->eleve->email
So if you use the $feedbacks in a loop in your blade file:
#foreach($feedbacks as $feedback)
{{ $feedback->eleve->email }}
#endforeach
OK, I think because of the edit, the question is pretty different than what was first asked, so I'll add another answer rather than try to convey this in the comments.
I think you are asking that when the user tries to see his feedbacks, you are getting the SQL error. If the user is selecting the feedback option, it may only be the single user that is looking for his own feedbacks. I suggest you change your query to the reverse (pull the feedbacks from the user instead of pulling all the feedbacks and then getting user to match):
\Auth::user()->load('feedbacks');
$feedbacksForThisUser = \Auth::user()->feedbacks;
Or if you want to get another user's feedbacks (and the user has permission to see them), just query on that user:
$user = User::with('feedbacks')->find($idOfUserYouWantToSee);
$feedbacks = $user->feedbacks;
Same thing goes for if you want to search for feedbacks from a user with a specific name:
$user = User::with('feedbacks')->where('nom','like','%'.$request->input('search').'%')->first();
$feedbacksForSearchedUser = $user->feedbacks;
The key is to forget about the emails - the relationship will link all feedbacks to the user. You don't need to join, or query on the email. You always have it on the $user object. (i.e. $user->email will always work).
If you want to see ALL feedbacks from ALL users, this isn't really any harder, just query all users and eager load the feedbacks relationship:
$users = User::with("feedbacks")->get();
Then in your blade:
#foreach($users as $user)
// If you want to display only a special set of users to the Auth::User(), then do an if-check here
#if(\Auth::user()->hasPerm("someperm") && $user->isInSomeCategory)
#foreach($user->feedbacks as $feedback)
{{ $feedback->text // or whatever the field you want to show }}
#endforeach
#endif // The if check is totally optional in this - just giving you an example
#endforeach
EDIT: to show a simple index()
public function index(Request $request)
{
if(!\Auth::user()->hasRole('admin')) {
\Auth::user()->load('feedbacks');
$feedback = \Auth::user()->feedbacks;
}
elseif($request->has('search')) {
$user = User::with('feedbacks')->where('nom','like','%'.$request->input('search').'%')->first();
$feedbacks = $user->feedbacks;
}
return view('admin.feedbacks.index', compact('feedbacks'));
}
Note: to make this as simple as possible, I also removed:
->with('display_search', $user->hasRole('admin'));
You can check this on the Blade page because \Auth::user() is already loaded automatically.
%7Bid%7D Showing this except real id
http://localhost/laravel/public/Images/%7Bid%7D/show
but when i change this %7Bid%7D in to number it also work but now shows automatically.
Routes: Route::get('Images/{id}/show','ImageController#index');
I think you are wrong when calling route.
ex: if you want to make route to show image with id 2 in your view you must do something like:
link
or
link
//simple change (->get(); into ->first();
public function userprofile($id)
{
$userdata = User::where('id', $id)->first();
// $userdata = Auth::login($user)->where('role', '=', 'admin');
return view('layouts.userdetail')->With('userdata', $userdata);
}
I want to paginate the data according to current logged in user.
this is my controller :
public function index()
{
$dosen = Dosen::paginate(5);
return view('dosen.index', compact('dosen'));
}
I'm assuming you've a user_id column in your Dosen model table. So you can say
$dosen = Dosen::where('user_id', auth()->user()->id)->paginate(5);
return view('dosen.index', compact('dosen'));
You can get current login user by Auth::user()->id
So pass it in your query.
$dosen = Dosen::where('user_id', '=', Auth::user()->id)->paginate(5);
Note:- user_id is your dosen table column name.
You can first filter the result using query and then apply paginate on the result to get the result-set of logged in user only.
Like this,
$doesn_query = Dosen::query();
$doesn_query->where('user','=',$userid); // Replace $userid with logged in user id.
$result = $doesn_query->paginate(5);
How to List all rows from a DB where $id matches the logged user id.
I'm using default Auth from Laravel.
At the moment i can list them all with this method in my controller:
public function index(){
$invoices = Invoice::all();
return view('index', compact('invoices'));
}
But i just want the ones that are from this user which is logged in:
Something like
$invoices = Invoice::where('id', '=', Auth::user()->id);
Your code Seems almost right. I would do:
$invoice = Invoice::where('id', Auth::user()->id)->get();
So basically use the get in order to fetch a collection. And maybe I would separate the user id in a varaible in case that you change the authentication in the future ;)
When using any condition then of course you must need to add the get() method. Otherwise, you can't show your data.
$invoices = Invoice::where('id', '=', Auth::user()->id)->get();
This means that you want to see data on which user is logged in now