当我尝试在codeigniter中进行此连接时
$this->db->join('ticketsale','ticketsale.id = 1');
我收到此错误:未知列' 1'在' on条款'
"SELECT * FROM (`users`) JOIN `ticketsale ` ON `ticketsale `.`id` = `1` JOIN `ticketsale_.."
如何让codeigniter不要把``大约1?
答案 0 :(得分:0)
因为在join语句中,您必须使用列名来匹配它们并加入其他查询。如果您想这样做,您应该添加Where语句。 例如:
"SELECT * FROM (`users`) JOIN `ticketsale ` ON `ticketsale `.`id` = `hereyourtable`.`hereis the column` JOIN `ticketsale_.. WHERE `ticketsale `.`id`=1"
<强>更新:强> 此外,在活动记录中;
$this->db->join("ticketsale","ticketsale.id=users.column")->where("ticketsale.id"=>"1");
this query will return same as native sql query