Laravel Form::Select - forms

Question

I am using dropdown menu to select the site as:
{{ Form::select('site', $sites, null, ['class' => 'form-control', 'placeholder' => 'Select a Site']) }}
{{ Form::label('site','site*') }}
I am storing the selected site in the database as site_id.
As this returns the index of the selected item from the list, so if the first item is selected the site_id would be stored as 0 in the database.
The problem here is, site_id is a foreign_key in my database and this is causing errors while matching it with the column:id of the sites table. As the id column generated by laravel migration scaffolding starts from 1 and the site_id returned by Form::select starts from 0.
Is there anyway that the index returned from Form::select would start from 1?
Or is there any other way to solve the problem?

Answer 1

When you're getting the sites from DB, keep original IDs to solve the problem:
$sites = Site::pluck('name', 'id');
This code will generate a list of sites' names with real IDs:
[1 => 'Site name 1', 2 => 'Site name 2']

Answer 2

You have many solutions
First, just verify with the validation, site must not be 0 since your controller
use Illuminate\Validation\Rule;
/** ... */
$this->validate($request, [
'site'=>['required',Rule::notIn([0])]
])
You can also check where you save Site model :
DB::transaction(function() use($request){
/** mapping of site attribute */
if($site->save())
{
}
}
You can also check since client side via JavaScript
{{ Form::select('site', $sites, null, ['id'=>'sites', 'class' => 'form-control', 'placeholder' => 'Select a Site']) }}
{{ Form::label('site','site*') }}
$("form input['type=submit']").click(function(e){
e.preventDefault();
if($('#sites').val() == 0)
{
alert('Please, select a site');
return;
}else{
$(this).submit();
}
})

Answer 3

if you want to inset name without id you can use $sites = Site::pluck('name', 'name'); I hope this will work.