如何写这样的选择:
SELECT filename, username, date_time, field1, field2... FROM... JOIN... WHERE...
UNLESS user_downloaded_this_file_today_already
我想忽略同一用户在同一天下载的文件。数据示例:
12 | file1.jpg | barney | 2012-03-15 12:50:10 | ...
13 | file1.jpg | roger | 2012-03-15 13:50:10 | ...
14 | file2.jpg | barney | 2012-03-15 14:50:10 | ...
15 | file1.jpg | barney | 2012-03-15 15:50:10 | ...
如何编写一个忽略第4行的SELECT?相同的文件名,相同的用户,日差异<这有可能吗?
答案 0 :(得分:4)
为了在当天允许重复,我认为以下就足够了:
select filename, username, date_time, field1, field2
from yourtable t
where not exists (
select 1
from yourtable
where t.filename = filename
and t.username = username
and trunc(t.date_time) = trunc(date_time)
and t.date_time > date_time)
order by t.date_time
这将每天返回每个文件/用户组合的第一个请求。
答案 1 :(得分:2)
只需使用SELECT DISTINCT
(假设您不需要数字列,因为它不在您的选择中):
SELECT DISTINCT FileName,
UserName,
Date_Time
FROM [YourTable]
如果您的Date_Time
列包含时间而不只是00:00:00
,则您需要删除DISTINCT
正常工作的时间:
SELECT DISTINCT FileName,
UserName,
CONVERT(DATETIME, FLOOR(CONVERT(FLOAT, Date_Time))) AS Date_Time
FROM [YourTable]
答案 2 :(得分:1)
SELECT filename, username, date_time, field1, field2
FROM (
SELECT filename, username, date_time, field1, field2,
ROW_NUMBER() OVER (ORDER BY username,filename, CONVERT(date,date_time)) AS RN,
RANK() OVER (ORDER BY username,filename, CONVERT(date,date_time)) AS R
FROM [table]
) A
WHERE A.RN = A.R