方法链接基于条件

Question

如何根据laravel 4中的条件进行方法链接？假设一个值不为false,那么内部方法将链接到if语句之前调用的方法.

是否有可能在laravel？

$data = User::where('username', $somevariable );

if(isset( $somevar_again ))
{
  $data->where('age', 21);
}
$data->orderBy('reg_date', 'DESC')->get();
return $data->first();

//尝试上面的代码,它给我错误的结果codeigniter我可以做到这一点

$this->db->select('e.*, v.name_en as v_name_en')
    ->from($this->table_name . ' e, ' . $this->ptc_venues . ' v');
  $this->db->where('e.venue_id_en = v.id'); 

  if(isset($search)){
   $this->db->where('(v.name_en LIKE "%'.$search.'%")');
  }

  $this->db->limit($limit, $start);
  $this->db->order_by('e.added_date_en', 'DESC');

Answer 1

我相信您的问题发生是因为您没有在每次查询构建器方法调用后存储返回的结果查询.

$query = User::query();

// Checking for username if exists
if (!empty($username)) {
    $query = $query->where('username', $username);
}

// Check for age if exists
if (isset($age)) {
    $query = $query->where('age', $age);
}

// Ordering
$query = $query->orderBy('reg_date', 'DESC');

// Get the first result
// After this call, it is now an Eloquent model
$user = $query->first();

var_dump($user);