SQL中表之间的连接答案

【问题标题】：Join between Tables in SQLSQL中表之间的连接
【发布时间】：2016-06-22 11:50:27
【问题描述】：

我有 5 张桌子

1. SCHOOL[id(bigInt, primary), name(varchar)]
2. SELECTED_INDICATOR[id(bigInt, primary), school_id(bigint)]
3. TEACHER[id(bigint, primary), indicator_id(bigInt), attendance_id(int)]
4. STUDENT[id(bigint, primary), indicator_id(bigInt), attendance_id(int)]
5. MIDDAY_MEAL[id(bigint,primary), indicator_id(bigint), served(boolean), consumed_number(int)]

在 TEACHER 表中，出席的 ID 可以有值：1 或 2 或 3。同样，在 STUDENT 表中，出席的 ID 可以有值：1 或 2。

我必须根据 SELECTED_INDICATOR id 生成报告，格式如下：

School_id |学校名称 | Total_教师 |老师_1 |老师_2 |老师_3 | Total_学生 |学生_1 |学生_2 |服务 |消费数

为此我尝试过：

select A.id, A.school_id, SC.name, 
 SUM(CASE WHEN T.attendance_id IN (1,2,3) THEN 1 ELSE 0 END) as TOTAL_TEACHER,
 SUM(CASE WHEN T.attendance_id IN (1) THEN 1 ELSE 0 END) as TEACHERS_1,
 SUM(CASE WHEN T.attendance_id IN (2) THEN 1 ELSE 0 END) as TEACHERS_2,
 SUM(CASE WHEN T.attendance_id IN (3) THEN 1 ELSE 0 END) as TEACHERS_3,
 SUM(CASE WHEN S.attendance_id IN (1,2) THEN 1 ELSE 0 END) as TOTAL_STUDENT,
 SUM(CASE WHEN S.attendance_id IN (1) THEN 1 ELSE 0 END) as STUDENTS_1,
 SUM(CASE WHEN S.attendance_id IN (2) THEN 1 ELSE 0 END) as STUDENTS_2,
 M.served, M.consumed_number
from SELECTED_INDICATOR A
join SCHOOL SC on A.school_id = SC.id
join TEACHER T on A.id = T.indicator_id
join STUDENT S on A.id = S.indicator_id
join MIDDAY_MEAL M on A.id = M.indicator_id
WHERE A.STATUS = 'COMPLETED' group by A.id;

当我使用 SELECTED_INDICATOR 一次加入 TEACHER 或 STUDENT 时，它会为我提供正确的数据。但是，当我在上面的查询中使用 SELECTED_INDICATOR 同时加入 TEACHER 和 STUDENT 时，我会得到大量与教师和学生相关的字段。

我的查询有什么问题？请帮助更正它，或提供任何替代查询。

【问题讨论】：

midday_meal 的相关性是什么？为什么要包括它？坦率地说，数据模型看起来很奇怪，我想从每个表中查看几行数据。
另外，不要依赖 MySQL 对 GROUP BY 的非标准方法，您应该在 GROUP BY 子句中指定每个非聚合列。
将您的组更改为此；按 A.id、A.school_id、SC.Name、M.served、M.consumed_number 分组
感谢您的支持，我在#sagi answer中得到了解决方案。

标签： mysql sql join inner-join

【解决方案1】：

尝试使用可以选择使用不同值的COUNT()。问题是表格使结果相乘。

    select A.id, A.school_id, SC.name, 
     COUNT(DISTINCT CASE WHEN T.attendance_id IN (1,2,3) THEN t.TeacherID  END) as TOTAL_TEACHER,
     COUNT(DISTINCT CASE WHEN T.attendance_id IN (1) THEN t.TeacherID END) as TEACHERS_1,
    ....
    FROM ....

【讨论】：

【解决方案2】：

    SELECT A.id, A.school_id, SC.name,
    (SELECT COUNT(T.attendance_id) FROM TEACHER T GROUP BY T.indicator_id HAVING A.id = T.indicator_id) AS TOTAL_TEACHER,
    (SELECT COUNT(T.attendance_id) FROM TEACHER T WHERE T.attendance_id = 1 GROUP BY T.indicator_id HAVING A.id = T.indicator_id) AS TEACHERS_1,
    (SELECT COUNT(T.attendance_id) FROM TEACHER T WHERE T.attendance_id = 2 GROUP BY T.indicator_id HAVING A.id = T.indicator_id) AS TEACHERS_2,
    (SELECT COUNT(T.attendance_id) FROM TEACHER T WHERE T.attendance_id = 3 GROUP BY T.indicator_id HAVING A.id = T.indicator_id) AS TEACHERS_3,
    (SELECT COUNT(S.attendance_id) FROM STUDENT S GROUP BY S.indicator_id HAVING A.id = S.indicator_id) AS TOTAL_STUDENT,
    (SELECT COUNT(S.attendance_id) FROM STUDENT S WHERE S.attendance_id = 1 GROUP BY S.indicator_id HAVING A.id = S.indicator_id) AS STUDENTS_1,
    (SELECT COUNT(S.attendance_id) FROM STUDENT S WHERE S.attendance_id = 2 GROUP BY S.indicator_id HAVING A.id = S.indicator_id) AS STUDENTS_2,
    M.served, M.consumed_number
    FROM SELECTED_INDICATOR A
    JOIN SCHOOL SC on A.school_id = SC.id
    JOIN MIDDAY_MEAL M on A.id = M.indicator_id

【讨论】：

感谢您的支持。但它为子查询提供了多行。
现在可以查看了。