我有一张桌子,我的数据如下:
ID name contact
1234 company XYZ jane@xyz.com
1234 company XYZ john@xyz.com
1234 company XYZ sue@xyz.com
1234 company XYZ mark@xyz.com
1234 company XYZ jack@xyz.com
1234 company XYZ alex@xyz.com
但我需要这个结果集:
ID name contact
12345 company XYZ jane@xyz.com,john@xyz.com,sue@xyz.com,mark@xyz.com,jack@xyz.com,alex@xyz.com
答案 0 :(得分:0)
以下是示例:
SELECT ID, Name,
STUFF((SELECT ',' + Contact
FROM SomeTable
WHERE ID = t.ID
FOR XML PATH('')),1,1,'') Contacts
FROM SomeTable t
GROUP BY ID, Name