使用 Propel ORM 左连接 select
Left join select using Propel ORM
我有3个table
主要 table:
+----+------------+
| id | major |
+----+------------+
| 1 | Computer |
| 2 | Architect |
| 3 | Designer |
+----+------------+
教室table:
+----+----------+-------+
| id | major_id | name |
+----+----------+-------+
| 1 | 1 | A |
| 2 | 1 | B |
| 3 | 1 | C |
| 4 | 2 | A |
| 5 | 2 | B |
| 6 | 3 | A |
+----+----------+-------+
最后,student_classroom table
+----+------------+--------------+----------+
| id | student | classroom_id | status |
+----+------------+--------------+----------+
| 1 | John | 1 | Inactive |
| 2 | Defou | 2 | Active |
| 3 | John | 2 | Active |
| 4 | Alexa | 1 | Active |
| 5 | Nina | 1 | Active |
+----+------------+--------------+----------+
我如何使用 propel 来构建下面的查询
select
a.id,
a.major,
b.number_of_student,
c.number_of_classroom
from major a
left join (
select
major.major_id,
count(student_classroom.id) as number_of_student
from major
left join classroom on classroom.major_id = major.id
left join student_classroom on student_classroom.classroom_id = classroom.id
where student_classroom.`status` = 'Active'
group by major_id
) b on b.major_id = a.major_id
left join (
select
major.major_id,
count(classroom.id) as number_of_classroom
from major
left join classroom on classroom.major_id = major.id
group by major_id
) c on c.major_id = a.major_id
因为我希望最终结果是这样的,所以我花了几个小时试图弄清楚但没有成功。
+----+------------+-------------------+---------------------+
| id | major | number_of_student | number_of_classroom |
+----+------------+-------------------+---------------------+
| 1 | Computer | 4 | 3 |
| 2 | Architect | 0 | 2 |
| 3 | Designer | 0 | 1 |
+----+------------+-------------------+---------------------+
试试这个
select
m.id,
m.major,
count(distinct s.id) as number_of_student ,
count(distinct c.id) as number_of_classroom
from major m
left join classroom c on
(m.id = c.major_id)
left join student_classroom s
on (s.classroom_id = c.id and c.major_id = m.id and s.status = 'active')
group by m.id
order by m.id
我有3个table
主要 table:
+----+------------+
| id | major |
+----+------------+
| 1 | Computer |
| 2 | Architect |
| 3 | Designer |
+----+------------+
教室table:
+----+----------+-------+
| id | major_id | name |
+----+----------+-------+
| 1 | 1 | A |
| 2 | 1 | B |
| 3 | 1 | C |
| 4 | 2 | A |
| 5 | 2 | B |
| 6 | 3 | A |
+----+----------+-------+
最后,student_classroom table
+----+------------+--------------+----------+
| id | student | classroom_id | status |
+----+------------+--------------+----------+
| 1 | John | 1 | Inactive |
| 2 | Defou | 2 | Active |
| 3 | John | 2 | Active |
| 4 | Alexa | 1 | Active |
| 5 | Nina | 1 | Active |
+----+------------+--------------+----------+
我如何使用 propel 来构建下面的查询
select
a.id,
a.major,
b.number_of_student,
c.number_of_classroom
from major a
left join (
select
major.major_id,
count(student_classroom.id) as number_of_student
from major
left join classroom on classroom.major_id = major.id
left join student_classroom on student_classroom.classroom_id = classroom.id
where student_classroom.`status` = 'Active'
group by major_id
) b on b.major_id = a.major_id
left join (
select
major.major_id,
count(classroom.id) as number_of_classroom
from major
left join classroom on classroom.major_id = major.id
group by major_id
) c on c.major_id = a.major_id
因为我希望最终结果是这样的,所以我花了几个小时试图弄清楚但没有成功。
+----+------------+-------------------+---------------------+
| id | major | number_of_student | number_of_classroom |
+----+------------+-------------------+---------------------+
| 1 | Computer | 4 | 3 |
| 2 | Architect | 0 | 2 |
| 3 | Designer | 0 | 1 |
+----+------------+-------------------+---------------------+
试试这个
select
m.id,
m.major,
count(distinct s.id) as number_of_student ,
count(distinct c.id) as number_of_classroom
from major m
left join classroom c on
(m.id = c.major_id)
left join student_classroom s
on (s.classroom_id = c.id and c.major_id = m.id and s.status = 'active')
group by m.id
order by m.id