问题描述
如果可能，请提供帮助。
Please need help if possible.
select a.* from (
select u.user_guid,u.user_name,u.user_email,u.user_type,
uinf.user_info_fname,uinf.user_info_lname,uinf.user_info_address,img.image_type,img.image_location,
v.visitor_datetime,v.visitor_duration,v.visitor_ip,v.media_guid
from
tb_visitors_log as v
left outer join tb_media as m on v.media_guid=m.media_guid left outer join
tb_users as u on v.user_guid=u.user_guid left outer join
tb_user_info as uinf on uinf.user_guid=u.user_guid
left outer join tb_images as img on img.image_guid=uinf.user_info_thumbnail
where v.media_guid=@media_guid and uinf.user_info_lname like concat('%',@searchvalue,'%')) as a
order by a.visitor_datetime desc
我希望输出按a.user_guid分组。在MySQL中我这样做：
I want the output to be grouped by a.user_guid. In MySQL I do this:
select a.* from (select u.user_guid,u.user_name,u.user_email,u.user_type,
uinf.user_info_fname,uinf.user_info_lname,uinf.user_info_address,img.image_type,img.image_location,
v.visitor_datetime,v.visitor_duration,v.visitor_ip,v.media_guid
from
tb_visitors_log as v
left outer join tb_media as m on v.media_guid=m.media_guid left outer join
tb_users as u on v.user_guid=u.user_guid left outer join
tb_user_info as uinf on uinf.user_guid=u.user_guid
left outer join tb_images as img on img.image_guid=uinf.user_info_thumbnail
where v.media_guid=P_media_guid and uinf.user_info_lname like concat('%',P_searchvalue,'%')
order by v.visitor_datetime desc
) as a group by a.user_guid;
但在SQL Server 2012中不能那样做
这是我的结果
https://plus.google.com/u/1/photos/104984108592274133808/albums/6044709663274953441/6044709665297999426?pid=6044709665297999426& oid = 104984108592274133808
这是我想要的那个
https：// plus。 google.com/u/1/photos/104984108592274133808/albums/6044709663274953441/6044709666972693378?pid=6044709666972693378&oid=104984108592274133808
But in SQL Server 2012 I can't do that
this my resultt
https://plus.google.com/u/1/photos/104984108592274133808/albums/6044709663274953441/6044709665297999426?pid=6044709665297999426&oid=104984108592274133808
and this is the one i want to have
https://plus.google.com/u/1/photos/104984108592274133808/albums/6044709663274953441/6044709666972693378?pid=6044709666972693378&oid=104984108592274133808
推荐答案
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