SQL的毗连，PHP

Question

这是我的代码

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

错误： 一个错误时遇到

Unknown column '1' in 'field list' 

请帮助我！

Answer 1

似乎你忘了指定你的表名。

请您表的名称替换[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

尝试这种方式

$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;