如何在Laravel 5.2中编写此查询？

Question

I am using Laravel 5.2 ,how to write this query?

There are two tables,user and articles,they have a one-to-many relationship.

I want to query users according to these conditions:
1、Show users who have articles,not show users who have not articles.
2、Articles contain two types,published and not published, "1" indicates published,show published articles ,not show articles which not published.
3、30 users are shown per page.

Like this, it's not right ,how to modify it?

HomeController:

public function index()
{
    $users = User::with('articles')->where('is_published','=',1)->paginate(30);
    return view('index', compact('users'));
}

User:

class User extends Model implements AuthenticatableContract, CanResetPasswordContract, HasRoleAndPermissionContract

{
    use Authenticatable, CanResetPassword, HasRoleAndPermission;

    protected $fillable = [
        'name', 'email', 'password',
    ];

    protected $hidden = [
        'password', 'remember_token',
    ];

    public function articles()
    {
        return $this->hasMany(Article::class);
    }

}

Article:

class Article extends Model
{
    public function user()
    {
        return $this->belongsTo(User::class);
    }

    //edit-1:         
    //Scope a query to only include status=1.
    public function scopeStatus($query)
    {
        return $query->where('status',1);
    }

}

edit:
@SSuhat @RamilAmr Thanks! If there is a Local Scope in Model "Article",how to modify the answer:

$query = User::with(['articles' => function ($q) {
                $q->where('is_published', 1);
            }])
         ->has('articles') //<<<<<<<<<
         ->paginate(30);
return $query;

Answer 1

Try this:

$query = User::with(['articles' => function ($q) {
                $q->where('is_published', 1);
            }])->paginate(30);
return $query;