如何选择具有相同column_a但column_b不同的行？

Question

我想在sql server中选择行，下面是我的问题：

Table1
-------------------------- 
|     Name   |    Type   |
--------------------------
|      A     |     1     |
|      A     |     2     |
|      B     |     1     |
|      B     |     3     |
|      A     |     3     |
|      C     |     1     |
|      C     |     3     |
|      D     |     1     |
|      D     |     2     |
|      D     |     3     |
|      .     |     .     |
|      .     |     .     |

选择如下行：

Table2
-------------------------- 
|     Name   |    Type   |
--------------------------
|      A     |     1     |
|      A     |     2     |
|      A     |     3     |
|      D     |     1     |
|      D     |     2     |
|      D     |     3     |
|      .     |     .     |
|      .     |     .     |

选择规则是……

显示名称和类型，其中类型必须有 1,2 和 3。

示例：A 有 1,2,3 种类型，所以我会选择它。

例子：B只有1,2个类型，所以我不会选择它。

Answer 1

您可以为此使用窗口函数：

select name, type
from (
    select 
        t.*,
        sum(case when type in (1, 2, 3) then 1 else 0 end)
                over(partition by name) cnt
    from mytable t
) t
where cnt = 3

这假设每个（名称，类型）元组在原始表中只出现一次，这与您的示例数据一致。

Demo on DB Fiddle：

姓名 |类型 :--- | ---: 一个 | 1 一个 | 2 一个 | 3 D | 1 D | 2 D | 3

Answer 2

您可以在三个 Type 列上使用 INNER JOINs 来实现此目的：

SELECT Table1.[Name],
       Table1.[Type]
  FROM Table1
    INNER JOIN (
               SELECT [Name]
                 FROM Table1
                 WHERE ([Type] = 1)
               ) A ON A.[Name] = Table1.[Name]
    INNER JOIN (
               SELECT [Name]
                 FROM Table1
                 WHERE ([Type] = 2)
               ) B ON B.[Name] = A.[Name]
    INNER JOIN (
               SELECT [Name]
                 FROM Table1
                 WHERE ([Type] = 3)
               ) C ON C.[Name] = A.[Name]

这个输出：

Name    Type
A       1
A       2
A       3
D       1
D       2
D       3

匹配的sqlfiddle。

这通过返回包含[Type] = 1 的行，然后只匹配[Type] = 2 和[Type] = 3 的行来工作。然后将其连接回您的主表并返回结果。

Answer 3

获取带有group by name 的名称并在having 子句中设置条件：

select * from Table1
where name in (
  select name 
  from Table1
  group by name
  having count(distinct type) = 3
)

如果Type 列有除1, 2, 3 以外的其他值，则：

select * from Table1
where type in (1, 2, 3) and name in (
  select name 
  from Table1
  where type in (1, 2, 3)
  group by name
  having count(distinct type) = 3
)

请参阅demo。

结果：

> Name | Type
> :--- | ---:
> A    |    1
> A    |    2
> A    |    3
> D    |    1
> D    |    2
> D    |    3

Answer 4

如果是 sql server 2017 及以上版本或 Azure SQL，则可以使用 string_agg，如下所示：

Select * from #yourTable yt join (
    select [name], string_agg([Type], ',') as st_types
    from #YourTable 
    group by [name] ) a 
on yt.name = a.[name] and a.st_types like '%1,2,3%'

Answer 5

我给你这个，如果你有这将工作：

A 1
A 2
A 3
A 2

它只会给你B。

SELECT * 
FROM Table1 
WHERE Name in (
SELECT Name from 
(
   SELECT Name, Type, count(Name) c from Table1 where Type = 1
   GROUP BY Name, Type 
   HAVING count(Name) = 1
   UNION
   SELECT Name, Type, count(Name) c from Table1 where Type = 2
   GROUP by Name, Type 
   HAVING count(Name) = 1
   UNION
   SELECT Name, Type, count(Name) c from Table1 where Type = 3
   GROUP by Name, Type 
   HAVING  count(Name) = 1) t 
GROUP by name
HAVING count(c) = 3)

这里是DEMO