Question

我有三张表，如出勤率，时间和地位。为了得到我的输出，我写了这个查询

SELECT * FROM status s 
LEFT JOIN (
    SELECT period_id,p_name,student_id,date,status 
    FROM period p 
    LEFT JOIN attendance a 
    ON p.period_id=a.period_id)t1
ON s.status=t1.status

但它没有提供所需的输出......任何人都可以提供帮助

出勤：

--------------------------------------------------------- 
id | student_id | period_id    |date       | Status     |
--------------------------------------------------------
1  | 1          | 1            |2016-12-24 | Present
--------------------------------------------------------|
2  | 1          | 3            |2016-12-24 | Absent     
--------------------------------------------------------|
3  | 2          | 1            |2016-12-24 | Late    
--------------------------------------------------------|
4  | 2          | 2            |2016-12-24 | Present

期间：

-----------------------
period_id  |  p_name   | 
-----------------------
1          |  period1  |
-----------------------
2          |  period2  |
-----------------------
3          |  period3  |

状态：

-------------------
id    |  status   |
-------------------
1     |  Present  |
-------------------
2     |  Absent   |
-------------------
3     |  Late

输出

-----------------------------------------------------
student_id   |  period_id |  date        |  Status  |
-----------------------------------------------------
1            |  1         |  2016-12-24  |  Present |
-----------------------------------------------------
1            |  2         |  2016-12-24  |  null    |
-----------------------------------------------------
1            |  3         |  2016-12-24  |  Absent  |
-----------------------------------------------------
2            |  1         |  2016-12-24  |  Late    |
-----------------------------------------------------
2            |  2         |  2016-12-24  |  Present |
-----------------------------------------------------
2            |  3         |  2016-12-24  |   null

Answer 1

LEFT JOIN中的第一个表必须是包含结果中所需的所有行的表。在您的情况下，这应该是所有不同学生，日期和时期的完整交叉产品。所以查询应该是：

SELECT s.student_id, p.period_id, d.date, a.status
FROM (SELECT DISTINCT student_id FROM attendance) AS s
CROSS JOIN period AS p
CROSS JOIN (SELECT DISTINCT date FROM attendance) AS d
LEFT JOIN attendance AS a
ON a.student_id = s.student_id AND a.period_id = p.period_id AND a.date = d.date
ORDER BY d.date, s.student_id, p.period_id

DEMO

似乎不需要status表，因为您在attendance表中冗余地具有状态名称，而不是status_id。

获得每个学生的所有期间出勤率

1 个答案: