在 CodeIgniter 中将 SQL 转换为活动记录

Question

SELECT *, SUM(`Relaxation_rating` + `Food_rating`) AS Ratings
GROUP BY CODE
ORDER BY Ratings DESC
FROM destinations

如何在 CodeIgniter 中将此 sql 代码转换为 CctiveRecord？

我尝试了以下方法，但确实没有结果。

$this->db->select("'(SELECT *, SUM(Relaxation_rating + Food_rating) FROM database GROUP BY CODE') AS Ratings", FALSE);
$this->db->get("database"); 

Answer 1

尝试：

$data = $this->db->select('*, SUM(`Relaxation_rating` + `Food_rating`) AS Ratings', false)->group_by('CODE')->order_by('Ratings', 'desc')->get('destinations')->result_array();

或者，

$this->db->select('*, SUM(`Relaxation_rating` + `Food_rating`) AS Ratings', false)->group_by('CODE')->order_by('Ratings', 'desc')->get('destinations')->result();

Answer 2

尝试一下

$this->db->select('*, SUM(`Relaxation_rating` + `Food_rating`) AS Ratings');
$this->db->from('destinations');
$this->db->group_by('CODE');
$this->db->order_by('Ratings','desc');
$result = $this->db->get();

Answer 3

工作查询构建器方法脚本不显式注入特定于方言的标识符引号（例如反引号或双引号）。

• SELECT 子句中没有引号：

  SELECT *, SUM(relaxation_rating + food_rating) ratings
  FROM `database`
  GROUP BY `code`
  ORDER BY `ratings` DESC
  

  生成者：

  $this->db
      ->select('*, SUM(relaxation_rating + food_rating) ratings')
      ->group_by('code')
      ->order_by('ratings', 'DESC')
      ->get('database')
      ->result();
  

  或

  $this->db
      ->select()
      ->select('SUM(relaxation_rating + food_rating) ratings')
      ->group_by('code')
      ->order_by('ratings', 'DESC')
      ->get('database')
      ->result();
  

  
  

• 通过破解受保护的

  _protect_identifiers

  属性，在 SELECT 子句中的列上没有引号：

  SELECT *, SUM(relaxation_rating + food_rating) ratings
  FROM `database`
  GROUP BY `code`
  ORDER BY `ratings` DESC
  

  生成者：

  $this->db->select();
  (fn() => $this->_protect_identifiers = false)->call($this->db);
  $this->db->select_sum('relaxation_rating + food_rating', 'ratings');
  (fn() => $this->_protect_identifiers = true)->call($this->db);
  return $this->db
      ->group_by('code')
      ->order_by('ratings', 'DESC')
      ->get('database')
      ->result();
  

  黑客保护类属性的另一个演示可以在如何列出 CodeIgniter 应用程序中所有加载的库和帮助程序？

  
  

• 对 SELECT 子句中所有列/别名的引用：

  SELECT *, SUM(`relaxation_rating` + `food_rating`) `ratings`
  FROM `database`
  GROUP BY `code`
  ORDER BY `ratings` DESC
  

  生成者：

  $this->db
      ->select(
          sprintf(
              '*, SUM(%s + %s) %s',
              $this->db->escape_identifiers('relaxation_rating'),
              $this->db->escape_identifiers('food_rating'),
              $this->db->escape_identifiers('ratings')
          ),
          false
      )
      ->group_by('code')
      ->order_by('ratings', 'DESC')
      ->get('database')
      ->result();