Laravel static Model is not returning error

Question

I have this Model UserSetting.php:

class UserSetting extends Model
{
    protected $fillable = ['user_id', 'name', 'setting_value'];

    public static function set($user_id, $name, $value)
    {

        if (!User::find($user_id)) {
            return error('NoForeign_User');
        }

        self::updateOrCreate(
            ['user_id' => $user_id, 'name' => $name],
            ['setting_value' => $value]
        );
    }
}

And I want to use it this way inside UserSettingController.php:

public function user(Request $request)
{
    Validator::make($request->all(), [
        'user_id'       => 'required|int',
        'name'          => 'required|string',
        'setting_value' => 'required|string',
    ], $this->messages)->validate();

    // HERE IS THE CALL
    UserSetting::set($request->user_id, $request->name, $request->setting_value);

    return saved();
}

I need to call UserSetting statically but NOT with a return:

return UserSetting::set(...)

But when the static function gets to the if(!User::find($user_id)) it carries on and shows the saved() helper instead of the return error('NoForeign_User')

While if I do it with a return return UserSetting::set(...) it correctly shows the error.

Is this correct? Do I have any other option other then returning the static class?

EDIT:

My error() function is this:

function error($message, $code = 422)
{
    $response = ['message' => $message];
    if ($errors) {
        $response += [
            'errors' => $errors,
        ];
    }

    return response()->json([
        'message' => $message,
    ], $code);
}

nakov · Accepted Answer · 2019-04-28 21:20:56Z

1

If you don't check the result that is being returned when calling the static method, then for sure the program will continue to execute the next line which is the saved() method in your case.

In order to stop execution a better approach is to throw an exception which will propagate to the Exception handler and stop the function execution.

So instead of this:

if (!User::find($user_id)) {
    return error('NoForeign_User');
}

try this:

if (!User::find($user_id)) {
    throw new \Exception('NoForeign_User');
}

or yet another and the best approach in my opinion that I just remembered is to simply use the findOrFail function:

User::findOrFail($user_id);

This will also throw an exception if the user has not been found.

edited Apr 28, 2019 at 21:20

answered Apr 28, 2019 at 20:06

nakov

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