可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试):
问题:
I have a table that lists the versions of software that are installed:
id | userid | version | datetime
----+--------+---------+------------------------
111 | 75 | 10075 | 2013-03-12 13:40:58.770
112 | 75 | 10079 | 2013-03-12 13:41:01.583
113 | 78 | 10065 | 2013-03-12 14:18:24.463
114 | 78 | 10079 | 2013-03-12 14:22:20.437
115 | 78 | 10079 | 2013-03-12 14:24:01.830
116 | 78 | 10080 | 2013-03-12 14:24:06.893
117 | 74 | 10080 | 2013-03-12 15:31:42.797
118 | 75 | 10079 | 2013-03-13 07:03:56.157
119 | 75 | 10080 | 2013-03-13 07:05:23.137
120 | 65 | 10080 | 2013-03-13 07:24:33.323
121 | 68 | 10080 | 2013-03-13 08:03:24.247
122 | 71 | 10080 | 2013-03-13 08:20:16.173
123 | 78 | 10080 | 2013-03-13 08:28:25.487
124 | 56 | 10080 | 2013-03-13 08:49:44.503
I would like to display all fields of one record from each userid
but only the highest version (also version is a varchar
).
回答1:
You're not specifying how you want ties handled, but this will do it if you want the duplicates displayed;
SELECT a.* FROM MyTable a
LEFT JOIN MyTable b
ON a.userid=b.userid
AND CAST(a.version AS INT) < CAST(b.version AS INT)
WHERE b.version IS NULL
An SQLfiddle to test with.
If you want to eliminate duplicates and if they exist pick the newest of them, you'll have to extend the query somewhat;
WITH cte AS (SELECT *, CAST(version AS INT) num_version FROM MyTable)
SELECT a.id, a.userid, a.version, a.datetime
FROM cte a LEFT JOIN cte b
ON a.userid=b.userid
AND (a.num_version < b.num_version OR
(a.num_version = b.num_version AND a.[datetime]<b.[datetime]))
WHERE b.version IS NULL
Another SQLfiddle.
回答2:
If you use SQL-Server (minimum 2005) you can use a CTE
with the ROW_NUMBER
function. You can use CAST
for version to get the correct order:
WITH cte
AS (SELECT id,
userid,
version,
datetime,
Row_number()
OVER (
partition BY userid
ORDER BY Cast(version AS INT) DESC) rn
FROM [dbo].[table])
SELECT id,
userid,
version,
datetime
FROM cte
WHERE rn = 1
ORDER BY userid
Demo
ROW_NUMBER
returns always one record even if there are multiple users with the same (top) version. If you want to return all "top-version-user-records", you have to replace ROW_NUMBER
with DENSE_RANK
.
回答3:
WITH records
AS
(
SELECT id, userid, version, datetime,
ROW_NUMBER() OVER (PARTITION BY userID
ORDER BY version DESC) rn
FROM tableName
)
SELECT id, userid, version, datetime
FROM records
WHERE RN =1
回答4:
I think this may solve your problem :
SELECT id,
userid,
Version,
datetime FROM (
SELECT id,
userid,
Version,
datetime ,
DENSE_Rank() over (Partition BY id order by datetime asc) AS Rankk
FROM [dbo].[table]) RS
WHERE Rankk<2
I used RANK function for ur requirement....
回答5:
select l.* from the_table l
left outer join the_table r
on l.userid = r.userid and l.version < r.version
where r.version is null
回答6:
The following code will display what you want and is great for performance!
select * from the_table t where cast([version] as int) =
(select max(cast([version] as int)) from the_table where userid = t.userid)
回答7:
If my experience tuning has taught me anything, generalities are bad bad bad.
BUT, If the table your getting the Top X
from is large (i.e. hundreds of thousands or millions). CROSS APPLY
is almost universally the best. In fact, if you benchmark it, cross apply performs consistently & admirably at smaller scales as well (in the tens of thousands) And ever covers the with ties potential requirement.
Something like:
select
id
,userid
,version
,datetime
from
TheTable t
cross apply
(
select top 1 --with ties
id
from
TheTable
where
userid = t.userid
order by
datetime desc
)