将Sql查询转换为CodeIgniter Active Records

Question

我想只查询那些喜欢最高的项目。

这是我要转换为CodeIgnitor的Active Records的SQL查询：

SELECT *, SUM(like) as totalLikes
FROM tbl_like
GROUP BY uploadID
ORDER BY totalLikes DESC
LIMIT 2

笨：

public function get_cheezyPic(){
      $this->db->select('uploadID, SUM(like) as totalLikes');
      $this->db->from('tbl_like');
      $this->db->group_by('uploadID');
      $this->db->order_by('totalLikes DESC');
      $this->db->limit(2);

      $query= $this->db->get();

      return $query->result_array();}

但是当我尝试运行此代码时，我遇到了这个错误

您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册，以便在'like）附近使用正确的语法作为totalLikes FROM（ tbl_like ）GROUP BY uploadID ORDER BY totalLikes '在第1行

SELECT `uploadID`, SUM(like) as totalLikes FROM (`tbl_like`) GROUP BY `uploadID` ORDER BY `totalLikes` desc LIMIT 2

这段代码出了什么问题？

谢谢您的帮助。

Answer 1

这对你有用。 请注意，您使用保留关键字，所以应该用backtic``包围。

public function get_cheezyPic(){
      $this->db->select('uploadID, SUM(`like`) as totalLikes',false);
      $this->db->from('tbl_like');
      $this->db->group_by('uploadID');
      $this->db->order_by('totalLikes DESC');
      $this->db->limit(2);

      $query= $this->db->get();

      return $query->result_array();
}