T-SQL选择具有相同列值的完整行

Question

我想显示具有在另一行中存在的序列的所有行。

如果我喜欢它，它会起作用

SELECT 
      [Serial]
FROM [x].[dbo].[Devices]
GROUP BY Serial
HAVING COUNT(*) > 1

但是当我添加更多选择列时

SELECT [ID]
      ,[UUID]
      ,[Serial]
FROM [x].[dbo].[Devices]
GROUP BY Serial
HAVING COUNT(*) > 1

我懂了

“ ID”在选择列表中无效，因为它不包含在聚合函数或GROUP BY子句中。

为什么我不能选择更多列？

我应该如何显示完整的行？

Answer 1

您可以使用窗口函数执行此操作：

select *
from (
   select *, 
          count(*) over (partition by [Serial]) as serial_count
   from [x].[dbo].[Devices]
)  t
where serial_count > 1;

通常，这要比通过聚合加入子选择更快。

Answer 2

希望您需要此查询，它将显示所有具有另一行中存在的序列的行。

SELECT  D1.[ID]
        ,D1.[UUID]
        ,D1.[Serial]
FROM [x].[dbo].[Devices] D1
JOIN (  SELECT [Serial]
        FROM [x].[dbo].[Devices]
        GROUP BY Serial
        HAVING COUNT(*) > 1 ) D2 ON D1.[Serial] = D2.[Serial]

Answer 3

只要ID和UUID对于序列而言是唯一的，请尝试按所有列进行分组。

SELECT [ID]
  ,[UUID]
  ,[Serial]
FROM [x].[dbo].[Devices]
GROUP BY Serial
  ,[ID]
  ,[UUID]
HAVING COUNT(*) > 1