条件codeigniter的多个

Question

如何将此查询转换为活动记录？

"UPDATE table_user 
 SET email = '$email', last_ip = '$last_ip' 
 where username = '$username' and status = '$status'";

我试图将上面的查询转换为:

$data = array('email' => $email, 'last_ip' => $ip);
$this->db->where('username',$username);
$this->db->update('table_user',$data);

如何使用where clausa状态？

# must i write db->where two times like this?
$this->db->where('username',$username);
$this->db->where('status',$status);

我也试过这个:

$this->db->where('username',$username,'status',$status);

Answer 1

你可以使用一个数组并传递数组.

Associative array method:
$array = array('name' => $name, 'title' => $title, 'status' => $status);

$this->db->where($array); 

// Produces: WHERE name = 'Joe' AND title = 'boss' AND status = 'active'

或者如果你想做除了比较之外的事情

$array = array('name !=' => $name, 'id <' => $id, 'date >' => $date);

$this->db->where($array);

Answer 2

是的,多次调用where()是实现此目的的完美有效方法.

$this->db->where('username',$username);
$this->db->where('status',$status);

http://www.codeigniter.com/user_guide/database/query_builder.html