Question

您好我有这个查询，而我正在尝试实现列NextColumn中的行数将是列z中具有相同值的行数。我认为可以这样做：

(SELECT COUNT(U1.z) GROUP BY U1.z)

但我得到了这个：

Each GROUP BY expression must contain at least one column that is not an outer reference.

 SELECT K.name AS Name, K.surname AS Surname, U1.akce AS Event, 
    U2.[text] AS [Scheme], U1.[text]  AS [Registered under], u1.z, NextColumn
          FROM klient K
          INNER JOIN ubytov U1 ON U1.[text] = K.ubytov
          LEFT OUTER JOIN ubytov U2 ON U1.z = U2.id WHERE U1.akce = '140012-02'
        ORDER BY U1.[text]

有人会帮我解决这个问题吗？

感谢您的时间。

My data now:

Name z NextColumn

1 8 1

2 8 1 

3 7 1

4 7 1

我想根据z值

获取

Name z NextColumn

1 8 2

2 8 2 

3 7 2

4 7 2

使用此查询

    with CTE as
(
select Count(u.z) as NextColumn, u.z as Z
from ubytov u
group by u.z
)
select * from CTE

我得到了这些结果：

NextColumn Z

0 NULL

1 34

1 36 

1 37

Answer 1

尝试将NextColumn改为相关子查询：

SELECT K.name AS Name, K.surname AS Surname, U1.akce AS Event, 
    U2.[text] AS [Scheme], U1.[text]  AS [Registered under], u1.z, 
    (SELECT COUNT(*) FROM ubytov u3 WHERE u3.z = u1.z) NextColumn
          FROM klient K
          INNER JOIN ubytov U1 ON U1.[text] = K.ubytov
          LEFT OUTER JOIN ubytov U2 ON U1.z = U2.id WHERE U1.akce = '140012-02'
        ORDER BY U1.[text]

Answer 2

另一种可能的解决方案：

with CTE as
(
  select Count(u.z) as NextColumn, u.z as Z
  from ubytov u
  inner join Klients K on  u.[text] = K.ubytov
  LEFT OUTER JOIN ubytov U2 ON u.z = U2.id WHERE u.akce = '140012-02'
  group by u.z
)

SELECT K.name AS Name, K.surname AS Surname, U1.akce AS Event, 
    U2.[text] AS [Scheme], U1.[text]  AS [Registered under], u1.z, 
   CTE.NextColumn
          FROM klient K
          INNER JOIN ubytov U1 ON U1.[text] = K.ubytov
          INNER JOIN CTE on U1.z = CTE.Z
          LEFT OUTER JOIN ubytov U2 ON U1.z = U2.id WHERE U1.akce = '140012-02'
        ORDER BY U1.[text]

根据table2中的值计算表中的值

2 个答案: