标签: sql-server-group-concat

我有一个包含〜百万个条目的表,如下所示:

customer_id | purchased_at     | product
1           | 2012-06-01 00:00 | apples
1           | 2012-09-02 00:00 | apples
1           | 2012-10-01 00:00 | pears
2           | 2012-06-01 00:00 | apples
2           | 2012-07-01 00:00 | apples
3           | 2012-09-02 00:00 | pears
3           | 2012-10-01 00:00 | apples
3           | 2012-10-01 01:00 | bananas

我想将产品连接到一行,DISTINCT和purchase_at的顺序

在MySQL中我只是使用

select customer_id, min(purchased_at) as first_purchased_at, 
group_concat(DISTINCT product order by purchased_at) as all_purchased_products
from purchases group by customer_id;

要得到

customer_id | first_purchased_at | all_purchased_products
1 …

我有一张看起来像这样的表:

这个表包含343行.

我正在尝试对它运行此查询:

create table newTest2 
select function_name, service_name, min(concurrency), substring_index(group_concat(date order by concurrency ), ',',1) as minDate, 
 max(concurrency), substring_index(group_concat(date order by concurrency desc), ',',1) as maxDate , avg(concurrency)
        from conc_intermidate
        group by function_name,service_name;

当我运行查询时,它给了我:"行203被GROUP_CONCAT()剪切",我不知道为什么它给了我这个错误.请帮忙!谢谢...

我有下表：

Code    Ref     Value
A1      Car     A
A1      Car     -
A1      Car     B
B2      Truck   CC
B2      Truck   D
B2      Truck   -
C3      Van     E
C3      Van     F
C3      Van     -
C3      Van     G

我试图实现的目标是一个将所有值组合在一起的连接字符串，如下所示：

Code   Ref    Value
A1     Car    A-B
B2     Truck  CCD-
C3     Van    EF-G

我离开了这里的例子，但一无所获。这是我想出的：

SELECT [Table].[Code]
, [Table].[Ref]
, STUFF((SELECT DISTINCT [Value]
  FROM [Table2]
  FOR XML PATH ('')),1, 1,'') AS Values
FROM [Table]
LEFT JOIN [Table2] ON
[Table2].[Code] = [Table].[Code]

我哪里错了？有没有更有效的方法来做到这一点？

我去桌子了。

TABLE 1: [..fields..] [CATEGORIE] [..fields..]

TABLE 2: [..fields..] [ID] [CATEGORIE] [..fields..]

我想连接有点特殊，并尝试这样：

SELECT [..other fields..], CATEGORIE, (SELECT ID FROM TABLE2 WHERE TABLE2.CATEGORIE = TABLE1.CATEGORIE) FROM TABLE1;

我想在主查询的列中有子查询的ID

这样（[]代表列）

[resultfield1] [resultfield2] [resultfield3] [ID1，ID2，ID3，ID4，...]

有没有办法负担得起？

非常感谢您的帮助，

提前致谢