SQL的毗連，PHP

Question

這是我的代碼

$this->db->select('*, CONCAT(left(card_id, 1), right(card_id, 6)) AS character_id', FALSE); 

錯誤： 一個錯誤時遇到

Unknown column '1' in 'field list' 

請幫助我！

Answer 1

似乎你忘了指定你的表名。

請您表的名稱替換[表名]：

$this->db->query('SELECT *, CONCAT(left(card_id, 1), right(card_id, 6)) AS character_id FROM [tablename]'); 

或：

$this->db->select("*, CONCAT(left(card_id, 1), right(card_id, 6)) AS character_id"); 
$this->db->from("[tablename]"); 
$res = $this->db->get(); 

Answer 2

嘗試這種方式

$this->db->select("*, CONCAT(left(card_id, 1), right(card_id, 6)) AS character_id",FALSE); 
$this->db->from("table"); 
$query = $this->db->get(); 
return $query;