试试这个：

public function hasAnyRoles(array $roles) {
    return $this->whereHas('roles', function ($query) use ($roles) {
         return $query->whereIn('name', $roles);
    })->isNotEmpty();
}

我的建议是，一旦使用关系，就为用户加载所有角色。当您通过像属性一样调用关系来加载关系时，laravel会将其缓存在引擎盖下，这样它就不会重复任何查询。然后，您可以简单地使用加载的角色来实现两种方法：

public function hasAnyRoles(array $roles)
{
    // When you call $this->roles, the relationship is cached. Run through
    // each of the user's roles.
    foreach ($this->roles as $role) {
        // If the user's role we're looking at is in the provided $roles
        // array, then the user has at least one of them. Return true.
        if (in_array($role->name, $roles) {
            return true;
        }
    }

    // If we get here, the user does not have any of the required roles. 
    // Return false.
    return false;
}

第二种方法是：

public function hasRoles(array $roles)
{
    // Run through each of the roles provided, that the user MUST have
    foreach ($roles as $role) {
        // Call the $this->roles relationship, which will return a collection
        // and be cached. Find the first role that has the name of the role
        // we're looking at.
        $found = $this->roles->first(function ($r) use ($role) {
            return $r->name == $role;
        });

        // If a role could not be found in the user's roles that matches the
        // one we're looking, then the user does not have the role, return 
        // false
        if (!$found) {
            return false;
        }
    }

    // If we get here, the user has all of the roles provided to the method
    return true;
}

当然，有许多不同的方法来实现这些方法，特别是在 Collection 类将帮助您，但关键是使用 $this->roles 只产生一个查询。