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MySQL如何使其完全外露
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Stack Overflow用户
提问于 2016-01-05 09:01:34
回答 1查看 49关注 0票数 1

我有四张桌子,如下所示。

表:

代码语言:javascript
复制
ClassID     |   ClassSTD
--------------------------------
1           |   STD-1
2           |   STD-2
3           |   STD-3
4           |   STD-4

表:

代码语言:javascript
复制
SectionId   |   SectionName | ClassId
--------------------------------------------
1           |   sec-A       | 1
2           |   sec-B       | 1
3           |   sec-C       | 1
4           |   sec-A       | 2
5           |   sec-B       | 2
6           |   sec-C       | 2
7           |   sec-A       | 3

表:Subject

代码语言:javascript
复制
subjectId   |   subjectName
------------------------------------
1           |   Art
2           |   Music
3           |   Play

SubjectAllocationToClass

代码语言:javascript
复制
classId     |   sectionID           |   subjectId   | type
-----------------------------------------------------------------------
1(STD-1)        |   1(sec-A)            |   1(Art)      | main
1(STD-1)        |   2(sec-B)            |   1(Art)      | main
1(STD-1)        |   3(sec-C)            |   1(Art)      | optional
1(STD-1)        |   1(sec-A)            |   2(Music)    | main
1(STD-1)        |   2(sec-B)            |   2(Music)    | optional

上表"SubjectAllocationToClass“显示了类的两种主题类型(主主题和可选主题)的分布情况。

需要所有类而不分区段,subjectAllocation需要所有区段而不需要subjectAllocation

我试着做左连接,右连接,但没能得到欲望。

如何从SELECT语句中获得以下结果?

代码语言:javascript
复制
    classSTD |  sectionName | Main subjectName   | Optional subjectName
    ------------------------------------------------------------------------
    STD-1    |  sec-A       | Art, Music         |
    STD-1    |  sec-B       | Art                |  Music
    STD-1    |  sec-C       |                    |  Art
    STD-2    |              |                    |
    STD-3    |  sec-A       |                    |
    STD-4    |              |                    |



select 
  ClassSTD as ClassSTD, 
  sectionname AS SectionName, 
  COALESCE(GROUP_CONCAT(CASE WHEN sac.type = 'main' THEN subjectName END), '')  as 'Main subjectname',
  COALESCE(GROUP_CONCAT(CASE WHEN sac.type = 'optional' THEN subjectName END), '')  as 'Optional subjectname'
FROM SubjectAllocationToClass sac
JOIN  Class c  ON c.classid = sac.classid
Left JOIN Section sc ON sc.sectionid = sac.sectionid
Left JOIN Subject sj ON  sj.subjectid = sac.subjectid
GROUP BY ClassSTD, SectionName;
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2016-01-05 09:10:42

没有必要使用完全外部连接,您可以直接实现使用左连接SUBQUERY

试试这个:

代码语言:javascript
复制
SELECT C.ClassSTD, 
       COALESCE(A.SectionName, '') AS SectionName, 
       COALESCE(A.Mainsubjectname, '') AS 'Main subjectname',
       COALESCE(A.Optionalsubjectname, '') AS 'Optional subjectname'
FROM Class C
LEFT JOIN ( SELECT sc.classId AS classId, 
                   sc.sectionname AS SectionName, 
                   COALESCE(GROUP_CONCAT(CASE WHEN sac.type = 'main' THEN subjectName END), '')  AS Mainsubjectname,
                   COALESCE(GROUP_CONCAT(CASE WHEN sac.type = 'optional' THEN subjectName END), '')  AS Optionalsubjectname
            FROM Section sc
            LEFT JOIN SubjectAllocationToClass sac ON sc.sectionid = sac.sectionid
            LEFT JOIN SUBJECT sj ON  sj.subjectid = sac.subjectid
            GROUP BY sc.classId, sc.SectionName
          ) AS A ON C.classId = A.classId;

对于第二个查询::

代码语言:javascript
复制
SELECT C.ClassSTD, 
       COALESCE(A.SectionName, '') AS SectionName, 
       COALESCE(A.Mainsubjectname, '') AS 'Main subjectname',
       COALESCE(A.Optionalsubjectname, '') AS 'Optional subjectname'
FROM Class C
LEFT JOIN ( SELECT sc.classId AS classId, 
                   sc.sectionname AS SectionName, 
                   COALESCE(GROUP_CONCAT((CASE WHEN sac.type = 'main' THEN subjectName END) ORDER BY subjectName), '')  AS Mainsubjectname,
                   COALESCE(GROUP_CONCAT((CASE WHEN sac.type = 'optional' THEN subjectName END) ORDER BY subjectName), '')  AS Optionalsubjectname
            FROM Section sc
            LEFT JOIN SubjectAllocationToClass sac ON sc.sectionid = sac.sectionid
            LEFT JOIN SUBJECT sj ON  sj.subjectid = sac.subjectid
            GROUP BY sc.classId, sc.SectionName
          ) AS A ON C.classId = A.classId;
票数 1
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原文链接:

https://stackoverflow.com/questions/34607712

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