CODEWITHSUNDEEP
phplaravellaravel-5laravel-5.4
I'll-Be-Back
I'll-Be-Back

Reputation: 10828

Laravel - Type error: Too few arguments?

I get an error:

Type error: Too few arguments

I thought Laravel do some magic to if arguments is not fully passed?

For example:

In the Controller I have:

public function create(CreateRequest $request)
{
    return $this->todoService->createList($request);
}

In the todoService class:

use App\Plan

class todoService {   

    public function createList($request, Plan $plan)
    {
      //
    }
}

As you can see I did not pass Plan class object. Do I have to bind or something?

Upvotes: 11

Views: 77576

Answers (4)

edfgdfgdfg
edfgdfgdfg

Reputation: 1

1

/**
 * Create a new personal access token for the user.
 *
 * @param  string  $name
 * @param  array  $scopes
 * @return \Laravel\Passport\PersonalAccessTokenResult
 */
public function createToken($name, array $scopes = [])
{
    return Container::getInstance()->make(PersonalAccessTokenFactory::class)->make(
        $this->getKey(), $name, $scopes
    );
}

/**
 * Set the current access token for the user.
 *
 * @param  \Laravel\Passport\Token  $accessToken
 * @return $this

Upvotes: 0

Hafiz
Hafiz

Reputation: 4277

If you are calling createList() by yourself so you will need to pass both parameters by yourself. You can bind something with Plan but still if you will call something not Laravel, then you will be responsible to pass that function parameters.

This type hinting only works if Laravel is calling that function. So you need to call these functions through Laravel.

If you are trying to automatically injecting in a class's constructor, you can simply do this:

$service = $this->app->make('App\Plan\todoservice');

or 

$service = App::make('App\Plan\todoservice');

or 

$service = resolve('App\Plan\todoservice');

But this will only work for Constructors. Please note that you can also provide parameters as next arguments of make() or resolve() function.

In fact, different methods can also be called that way.

You can simply do:

    $service = App::make('App\Plan\todoservice');
    $container->call([$service, 'createList'], ['request' => $request] );

Here $container is object of Illuminate\Contracts\Container\Container.

Upvotes: 10

wujt
wujt

Reputation: 1328

You have to bind classes only if they depend on interfaces. If you specify particular class, reflection will do the job for you. documentation

The only way this will work, is to set the default value of second parameter. In any other situation, syntax exception will be thrown.

use App\Plan

class todoService    
{  
    public function createList($request, Plan $plan = null)
    {
      //
    }
}

public function create(CreateRequest $request)
{
    return $this->todoService->createList($request);
}

It will work, but will that make any sense?

Upvotes: 2

Niels Keurentjes
Niels Keurentjes

Reputation: 41968

Laravel cannot do any magic on this level as your coding error is simply a PHP syntax error. You're indicating that the second parameter is of type Plan, which makes it mandatory implicitly. Laravel cannot 'patch' simple function calls like this.

Your confusion is likely in that, depending on your routing, Laravel can inject the correct Plan parameter into the create controller method, thus allowing you to forward it into the service.

Upvotes: 1

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