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).
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
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
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
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.
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
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
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....
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)
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
)
来源：https://stackoverflow.com/questions/15380718/select-the-top-1-row-from-each-group