Php CodeIgniter DB获取_位置并加入
这是编写codeigniter查询的最佳方法吗Php CodeIgniter DB获取_位置并加入,php,database,codeigniter,Php,Database,Codeigniter,这是编写codeigniter查询的最佳方法吗 $where = array('registration_status_fk' => 2, 'membership.membership_type_id' => 4); $join = array('membership', 'membership.id=members.id'); $query = $this->join($join[0], $join[1])->get_where($this->tbl_name,
$where = array('registration_status_fk' => 2, 'membership.membership_type_id' => 4);
$join = array('membership', 'membership.id=members.id');
$query = $this->join($join[0], $join[1])->get_where($this->tbl_name, $where);
还是有更好的方法来完成我正在做的事情
这就是我得到的错误
Error Number: 1054
Unknown column 'membership.membership_type_id' in 'where clause'
SELECT * FROM (`member`) WHERE `registration_status_fk` = 2 AND `membership`.`membership_type_id` = 4
Filename: C:\xampp\htdocs\OAWA\system\database\DB_driver.php
Line Number: 330
您可以这样编写联接查询
$this->db->select('*')->from('members')->join('membership', 'membership.id=members.id')->where($where)->get();
在$where
数组中,您需要表名和列名,否则如果两个表都有相同的列,则可能无法工作。您太接近了
我使用您的示例构造了自己的join/get_where语句,该语句生成正确的结果:
$where = array('as_id' => $id);
$join = array('as_types', 'as_types.as_type_id=assets.as_type');
$query = $this->db->join($join[0], $join[1])->get_where('assets', $where);
按以下方式重新编写查询:
$where = array('registration_status_fk' => 2, 'membership.membership_type_id' => 4);
$join = array('membership', 'membership.id=members.id');
$query = $this->join($join[0], $join[1])->get_where('members', $where);
请注意,唯一的更改是在第3行:
$this->tbl\u name
应该更改为一个简单的字符串,'members'
,您仍然可以链接->result()