如何连接sql [duplicate]中的所有行

Question

这个问题在这里已有答案：

• Simulating group_concat MySQL function in Microsoft SQL Server 2005? 10个答案

我得到了以下练习：确定所有地址的平均距离，并在csv列表“CSV_Avg”中按地址连接平均距离，如表Address _Distances中所示。

这就是表格的外观

这是查询应该做的

到目前为止，我已经完成了这个，但我找不到如何连接单个列中的所有行的方法

这就是我的查询的样子

SELECT DISTINCT 

      [id_address]    
      ,sum([distance])/count([distance]) as "AVGDistance"
      ,CONCAT([id_address],'=',[distance]) as "CSV_AVG"
  FROM [Test].[dbo].[View_2]
  WHERE [id_address] IS NOT NULL

  Group by [id_address],[distance]

Answer 1

你可以这样做：

WITH CTE
AS
(
  SELECT 
      [id_address]    
      ,sum([distance])/count([distance]) as "AVGDistance"
  FROM view_2
  WHERE [id_address] IS NOT NULL
  Group by [id_address]
)
SELECT *, (SELECT  STUFF((SELECT ',' + CONCAT([id_address],'=',AVGDistance) 
                         FROM CTE
                        FOR XML PATH(''), TYPE
                       ).value('.', 'NVARCHAR(MAX)') 
                          , 1, 1, '')) AS CSV_AVG
FROM CTE;

• demo

这将为您提供与您正在寻找的完全相同的结果：

| id_address | AVGDistance |                               CSV_AVG |
|------------|-------------|---------------------------------------|
|         21 |       72.84 | 21=72.840000,35=49.195000,95=1.690000 |
|         35 |      49.195 | 21=72.840000,35=49.195000,95=1.690000 |
|         95 |        1.69 | 21=72.840000,35=49.195000,95=1.690000 |