Question

我试图生成一份报告，显示所有建立和生活在特定机构的学生人数。（三个表 - 建立表，房间表和床位表）

这是我的模特：

        public function report_Management(){

          $query="SELECT e.owner, e.id as 'Establishment_ID', e.name, (SELECT COUNT(b.bed_no) FROM beds b, users u, establishment e, room r WHERE r.establishment_id = Establishment_ID) as 'Number Of Ateneans' FROM establishment e GROUP BY e.id";


  return $this->db->query($query);

        }

我试图将e.id用作＆＃39; Establishment_ID＆＃39;在内部的Select语句中

Answer 1

$ query =＆＃34; SELECT e.owner，e.id as＆＃39; Establishment_ID2＆＃39;，e.name，（SELECT COUNT（b.bed_no）FROM beds b，registrar u，room r WHERE Establishment_ID2 = r.establishment_id AND r.room_id = b.room_id AND u.id = b.student_id）as＆＃39; Number of Ateneans＆＃39;从建立e GROUP BY e.id＆＃34 ;;

是否可以在codeigniter中将外部选择变量用于内部选择？

1 个答案: