Query relationships about laravel-graphql

[chrissm79]

In the resolve method of your query, wouldn't it just be:

public function resolve($root, $args, ResolveInfo $info)
{
    $fields = $info->getFieldSelection($depth = 3);

    $me = Auth::user(); // Or however you're handling finding "me"

    foreach ($fields as $field => $keys) {
        if ($field === 'friends') {
            $me->with('friends');
        }
    }

    return $me;
}

[joshhornby]

@chrissm79 How would I then call name on the friends?

[chrissm79]

You don't need to call "name" on friends, it just needs to be included in the array that is sent back from the resolve function (which you can do by eager loading the relationship).

If "friends" is a collection of the "User" model and it has a "name" attribute, it will automagically be sent back with the request. If "name" is actually the first_name and last_name attributes of the User model, then you could either use a mutator on the model or create a custom field.

If neither is the case I would need to see your Query and Type classes.

[joshhornby]

In this case in the me GraphQLType class how should I define friends in the fields() function?

[chrissm79]

You could do something like this (if it makes sense for your app):

MeType (which is the type your MeQuery would return)

namespace App\GraphQL\Types;

use GraphQL;
use GraphQL\Type\Definition\Type;
use Folklore\GraphQL\Support\Type as GraphQLType;

class MeType extends GraphQLType
{
    /**
     * Attributes of Type.
     *
     * @var array
     */
    protected $attributes = [
        'name' => 'Me',
        'description' => 'Authenticated user details.'
    ];

    /**
     * Available Field Types.
     *
     * @return array
     */
    public function fields()
    {
        return [
            'id' => [
                'type' => Type::id(),
                'description' => 'ID of user.'
            ],
            'name' => [
                'type' => Type::string(),
                'description' => 'Name of authenticated user.'
            ],
            'email' => [
                'type' => Type::string(),
                'description' => 'Email address of user.'
            ],
            'friends' => [
                // This is saying that friends is a "list" of the type "user"
                'type' => Type::listOf(GraphQL::type('user')),
                'description' => 'List of friends.'
            ]
        ];
    }
}

UserType

namespace App\Http\GraphQL\Types;

use GraphQL;
use GraphQL\Type\Definition\Type;
use Folklore\GraphQL\Support\Type as GraphQLType;

class UserType extends GraphQLType
{
    /**
     * Attributes of Type.
     *
     * @var array
     */
    protected $attributes = [
        'name' => 'User',
        'description' => 'User details.'
    ];

    /**
     * Available Field Types.
     *
     * @return array
     */
    public function fields()
    {
        return [
            'id' => [
                'type' => Type::id(),
                'description' => 'ID of user.'
            ],
            'name' => [
                'type' => Type::string(),
                'description' => 'Name of user.'
            ],
            'email' => [
                'type' => Type::string(),
                'description' => 'Email address of user.'
            ]
        ];
    }
}

In you graphql.php config file

'query' => [
    'me' => App\GraphQL\Queries\MeQuery::class,
    // ...
]
'types' => [
    'me' => App\GraphQL\Types\MeType::class,
    'user' => App\GraphQL\Types\UserType::class,
    // ...
]

[joshhornby]

@chrissm79 Thanks for the example. I seem to be getting a User Error: expected iterable, but did not find one. error. I will paste the code I have, my example is a user and a team:

The relationship is set in the User model:

public function team()
{
    return $this->belongsTo(Team::class);
}

UserType

class UserType extends GraphQLType
{
    protected $attributes = [
        'name' => 'User',
        'description' => 'User details.'
    ];

    public function fields()
    {
        return [
            'id' => [
                'type' => Type::nonNull(Type::int()),
                'description' => 'The id of the user'
            ],
            'email' => [
                'type' => Type::nonNull(Type::string()),
                'description' => 'The email of user'
            ],
            'team' => [
                'type' => Type::listOf(GraphQL::type('team')),
                'description' => 'Customer orders.',
            ]
        ];
    }
}

TeamType

class TeamType extends GraphQLType
{
    protected $attributes = [
        'name' => 'Team',
        'description' => 'Team the user belongs to.'
    ];

    public function fields()
    {
        return [
            'id' => [
                'type' => Type::nonNull(Type::int()),
                'description' => 'The id of the team'
            ],
            'name' => [
                'type' => Type::nonNull(Type::string()),
                'description' => 'The name of team'
            ]
        ];
    }
}

UsersQuery

class UsersQuery extends Query
{
    protected $attributes = [
        'name' => 'Users query'
    ];

    public function type()
    {
        return Type::listOf(GraphQL::type('users'));
    }

    public function args()
    {
        return [
            'id' => ['name' => 'id', 'type' => Type::string()],
            'email' => ['name' => 'email', 'type' => Type::string()]
        ];
    }

    public function resolve($root, $args, ResolveInfo $info)
    {
        $fields = $info->getFieldSelection($depth = 5);

        $users = User::query();

        foreach ($fields as $field => $keys) {
            if ($field === 'team') {
                $users->with('team');
            }
        }

        if(isset($args['id']))
        {
            return User::where('id' , $args['id'])->get();
        }
        else if(isset($args['email']))
        {
            return \User::where('email', $args['email'])->get();
        }
        else
        {
            //return User::all();
        }
        // If I dd($users->get())); here I see the data loaded in the relationship
        return $users->get();
    }
}

This returns an error saying User Error: expected iterable, but did not find one.

[joshhornby]

I fixed this by looking at the code example by @chrissm79 in this issue #16.

My fields() now looks like:

'team' => [
    'type' => Type::listOf(GraphQL::type('team')),
    'description' => 'Customer orders.',
    'resolve' => function($data, $args) {
        return $data->team()->get();
     }
],

[kendrawalker]

Hello @joshhornby I am new to graphql. I am currently working on a project with laravel and graphql as well. I am having issues querying a table based on a foreign key value. I see you had a similar problem.

I see in your project, you wrote this:

query FetchUser{
    user(id: 123456789) {
        id
        posts(id: 987654321) {
            id
        }
    }
}

In my project, I wrote a similar query:

plenty_systems{
    id,
    url,
    username,
    password,
    customer (id:${custID}) {
      id
    }
  }

However, my system is returning errors. Do you have any advice for me?