Question

从这个数据中我如何获取tbl_attendance中user_id与application_status 3和4没有当前日期的所有用户

attendance_id   user_id      date_in        attendance_status status 0=absent 1=present 3 = onleave 4 = onoff
    1            1          2017-02-05      1
    2            36         2017-02-11      4
    3            36         2017-02-11      4
    4            36         2017-02-11      3
    5            1          2017-02-02      1
    6            36         2017-02-01      1

我的代码就像这样

$date=date('Y-m-d');
$this->db->where('tbl_users.user_id NOT IN(SELECT user_id FROM tbl_attendance WHERE tbl_attendance.date_in= "'.$date.'" )');

$this->db->where_in( 'tbl_attendance.attendance_status', array( '3', '4' ) );

想要将这两行合并并产生输出，如果当前日期存在于3或4中请帮助我

Answer 1

你可以尝试这样。结合两个条件

$this->db->where("tbl_users.user_id NOT IN(SELECT user_id FROM tbl_attendance WHERE tbl_attendance.date_in='$date' AND tbl_attendance.attendance_status IN(3,4))");

Answer 2

看到您正在使用查询构建器...

$this->db->group_start()
    $this->db->where_in( 'tbl_attendance.attendance_status', array( '3', '4' ) );
    $this->db->where_not_in('tbl_users.user_id', 'SELECT user_id FROM tbl_attendance WHERE tbl_attendance.date_in = '. $date, false)
$this->db->group_end()

应该有效

嵌套在codeigniter中的条件

2 个答案: