Toggle database value from View on button click Laravel

Question

I am using Laravel framework as a backend API and a few blade PHP files for the front end, specifically for the authentication and the admin panel from the /admin route. In /admin, I display a list of all registered users and buttons next to them. (This page is only visible for users that have their value in Admin column set as true). I want to toggle the Admin status of a user, either promoting or demoting them by clicking the button next to the user name. For this, I tried to use a form submit with get method. I have a method defined inside UserController like this:

public function setAdmin($id) {
    $user = User::find($id);
    $user->admin = !$user->admin;
    if($user->save()) {
        echo "Changed";
    }
    else {
        echo "Could not be changed";
    }
}

I want to call this method from the view on the click of a button.

I tried using a Form to send a request by specifying the action, but it gave an error saying the values passed are less than the expected number of parameters.

{!! Form::open(['action' => ['UserController@setAdmin', $user->id], 'method' => 'POST']) !!}
{{  Form::submit('Submit', ['class' => 'btn btn-primary']) }}
{!! Form::close() !!}

I have a route set up explicitly to call this action

Route::post('/admin/users/setAdmin', 'UserController@setAdmin')

Although I am not sure if I have to set an explicit route for this action or if it's possible to call a controller function directly from a view without defining the route.

I have iterated through User Model to display all users:

@if(count($users) > 0) 
    @foreach($users as $user)
        <div class="card">
            {{ $user }}
        </div>


        {!! Form::open(['action' => ['UserController@setAdmin', $user->id], 'method' => 'POST']) !!}
        {{  Form::submit('Submit', ['class' => 'btn btn-primary']) }}
        {!! Form::close() !!}
    @endforeach
@else 
    <h2>No users found!</h2>
@endif

EDIT: Added the foreach section of the blade file. Also I modified the 'action' part of the Form::open() parameters, it was a mistype, the parameters error is still there.

Can someone explain how this can be done?

{!! Form::open(['action' => ['UserController@setAdmin', $user->id], 'action' => 'POST']) !!} here parameter name should be method for post — Omi
– Omi, Commented May 27, 2019 at 7:12
The method for post is setAdmin defined in the UserController, or is it a different parameter you are talking about? — Tribunal
– Tribunal, Commented May 27, 2019 at 7:21
you have written 'action' => 'POST', it should be method not action — Omi
– Omi, Commented May 27, 2019 at 7:25
Added additional code for foreach and changed 'action' to 'method', the error still persists. Looking for an alternative better solution to this. — Tribunal
– Tribunal, Commented May 27, 2019 at 7:46

Kévin Bibollet · Accepted Answer · 2019-05-27 08:46:07Z

2

You are trying to pass a parameter to your route but there is any within its declaration. You need to add it in your route path:

Route::post('/admin/users/setAdmin/{id}', 'UserController@setAdmin')

If you don't want to have an URL like this, you should add a hidden input to your form containing your ID:

{!! Form::open(['action' => ['UserController@setAdmin'], 'method' => 'POST']) !!}

{{ Form::hidden('id', $user->id) }}
{{ Form::submit('Submit', ['class' => 'btn btn-primary']) }}

{!! Form::close() !!}

And in your controller's method:

use Request;

/* ... */

public function setAdmin(Request $request) {
    $user = User::find($request->id);

    /* ... */
}

edited May 27, 2019 at 8:46

answered May 27, 2019 at 7:59

Kévin Bibollet

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1 Comment

Tribunal Over a year ago

<input type="hidden" name="id" value="{{ $user->id }}"> translates to {{ Form::hidden('id', $user->id) }} and it changes the value in the database now.

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