Question

ID  GROUP       NAME
1   1,2,      Supreeth
2   1,2,5,      Aishu
3   3,         Arvi
4   4,5,       Gani
5   4,3,2,      Jyo
6   3,2,1,      Upi
7   2,3,4,1,5,  Savi

我有这样的表，我正试图得到小组的数量

    DECODE    Count
1   A         4
2   B         5
3   C         4
4   D         3
5   E         3

我想将group的值解码为DECODE值，例如1到A，2到B，3到C，4到D，5到E ..

我试着这样但是得到错误

select  count(*) from s_c where age like '%1%' and '%2%' and '%3%';

Answer 1

一种简单的（不一定有效的）方法是获得每个组值的单独计数。请注意，列名Group正在使用保留字，因此您必须使用适当的分隔标识符表示法。当您使用便携式（SQL标准）双引号时，您必须得到标识符正确的情况;我假设小写 - YMMV。

SELECT 1 AS GroupID, 'A' AS Decode, COUNT(*) AS GroupCount
  FROM AnonymousTable
 WHERE "group" LIKE '%1%'
 GROUP BY GroupID, Decode
UNION
SELECT 2 AS GroupID, 'B' AS Decode, COUNT(*) AS GroupCount
  FROM AnonymousTable
 WHERE "group" LIKE '%1%'
 GROUP BY GroupID, Decode
UNION
SELECT 3 AS GroupID, 'C' AS Decode, COUNT(*) AS GroupCount
  FROM AnonymousTable
 WHERE "group" LIKE '%1%'
 GROUP BY GroupID, Decode
UNION
SELECT 4 AS GroupID, 'D' AS Decode, COUNT(*) AS GroupCount
  FROM AnonymousTable
 WHERE "group" LIKE '%1%'
 GROUP BY GroupID, Decode
UNION
SELECT 5 AS GroupID, 'E' AS Decode, COUNT(*) AS GroupCount
  FROM AnonymousTable
 WHERE "group" LIKE '%1%'
 GROUP BY GroupID, Decode

但这不能很好地扩展;添加另外5组，这是非常不愉快的;加500，这是无法管理的。您最好将数据存储在正确规范化的表中，然后可以使用更简单的SQL进行分析。

替代架构设计和查询

Users                     UserGroups          Groups
ID        Name            UserID GroupID      ID         Decode
1         Supreeth        1      1            1          A
2         Aishu           1      2            2          B
3         Arvi            2      1            3          C
4         Gani            2      2            4          D
5         Jyo             2      5            5          E
6         Savi            3      3
                          4      4
                          4      5
                          ...

以下是简化查询，它可能比原始查询执行得更好，并且可以扩展到任意数量的组（如果需要，可以扩展到数百万）：

SELECT u.GroupID, g.Decode, COUNT(*) AS Count
  FROM UserGroups AS u
  JOIN Groups     AS g ON u.GroupID = g.ID
 GROUP BY u.GroupID, g.Decode

标准化使生活更轻松 - 这是做这件事的一个原因！

Answer 2

此解决方案可能不太好或不灵活，因为它取决于硬编码值，但它可以实现所需的输出，如问题中所述：

SELECT 1 AS [Num], 'A' AS Decode, COUNT(*) AS [Count] FROM my_table WHERE [Group] LIKE '%1%'
UNION
SELECT 2 AS [Num], 'B' AS Decode, COUNT(*) AS [Count] FROM my_table WHERE [Group] LIKE '%2%'
UNION
SELECT 3 AS [Num], 'C' AS Decode, COUNT(*) AS [Count] FROM my_table WHERE [Group] LIKE '%3%'
UNION
SELECT 4 AS [Num], 'D' AS Decode, COUNT(*) AS [Count] FROM my_table WHERE [Group] LIKE '%4%'
UNION
SELECT 5 AS [Num], 'E' AS Decode, COUNT(*) AS [Count] FROM my_table WHERE [Group] LIKE '%5%'

Answer 3

我知道这已经得到了解答，但我想分享我的观点。

我所做的实际上是将逗号分隔列拆分为行然后我只应用计数。至于解码，我只需将数字添加到64并应用将返回ASCII表的大写字母的CHAR函数，而不是手动执行作业...

SELECT Decode,
       COUNT(*)
FROM
(
    SELECT ID,
           CHAR(64+CONVERT(INT,String)) AS Decode,
           string
    FROM
    (
           SELECT A.[id],  
                Split.a.value('.', 'VARCHAR(100)') AS String  
           FROM  
               (
               SELECT id,
                      CAST ('<M>' + REPLACE(SUBSTRING([GROUP],0,LEN([GROUP])), ',', '</M><M>') + '</M>' AS XML) AS String  
               FROM  Table1
               ) AS A CROSS APPLY String.nodes ('/M') AS Split(a)
     ) T1
  ) T2
 GROUP BY Decode
 ORDER BY Decode

只是为了让你知道我到了Column To Rows。

我还创建了一个小提琴here

我昨天会给出答案，但我全神贯注......对不起

Giannis

如何获得GROUP的数量

3 个答案:

替代架构设计和查询