Question

我有这张桌子：

Docenza(id, id_facolta, ..., orelez)
Facolta(id, ...)

我想为每个facolta都获取完成最大orelez数量和orelez数量的Docenza的ID：

id_docenzaP  facolta1   max(orelez)
id_docenzaQ  facolta2   max(orelez)
...
id_docenzaZ  facoltaN   max(orelez)

我该怎么做？这就是我要做的：

SELECT DISTINCT ... F.nome, SUM(orelez) AS oreTotali
FROM Docenza D 
  JOIN Facolta F ON F.id = D.id_facolta
GROUP BY F.nome

我得到类似的东西

 docenzaP  facolta1   maxValueForidP
 docenzaQ  facolta1   maxValueForidQ
 ...
 docenzaR  facolta2   maxValueForidR
 docenzaS  facolta2   maxValueForidS
 ...
 docenzaZ  facoltaN   maxValueForFacoltaN

我怎样才能仅取每个facolta的最大值？

Answer 1

大概是您想要的：

SELECT F.nome, sum(orelez) AS oreTotali
FROM Docenza D JOIN
     Facolta F
     ON F.id = D.id_facolta
GROUP BY F.nome;

我不确定SELECT DISTINCT应该做什么。它几乎从未与GROUP BY一起使用。 . . .建议您选择其他列，这些列对于您想要的结果是不需要的。

Answer 2

这是未经测试的，并且由于您没有提供具有预期结果的样本数据，因此我无法确定它确实是您所需要的。

这有点难看，我敢肯定有一些聪明的相关子查询方法，但是我从来都不擅长这些。

SELECT st.focolta,
       s_orelez,
       TMP3.id_docenza
  FROM some_table AS st
 INNER
  JOIN (SELECT * 
          FROM (SELECT focolta,
                       s_orelez,
                       id_docenza,
                       ROW_NUMBER() OVER -- Get the ranking of the orelez sum by focolta.
                         ( PARTITION BY focolta
                               ORDER BY s_orelez DESC
                         ) rn_orelez
                  FROM (SELECT focolta,
                               id_docenza,
                               SUM(orelez) OVER -- Sum the orelez by focolta
                                ( PARTITION BY focolta
                                ) AS s_orelez 
                          FROM some_table
                       ) TMP
               ) TMP2 
         WHERE = TMP2.rn_orelez = 1 -- Limit to the highest rank value
       ) TMP3
    ON some_table.focolta = TMP3.focolta; -- Join to focolta to the id associated with the hightest value.

计算SQL中每一行的MAX

2 个答案: