How to get rank?

[jim_geissman]

I would like to write a query that gives me the values of a set of
observations, and their rank.

CREATE TABLE #Values(val int)
INSERT #Values SELECT 1
INSERT #Values SELECT 5
INSERT #Values SELECT 10

I would like to select this:

1 10 -- rank 1, value 10
2 5
3 1

I can put them into a temp table with an identity column, ordered by
the column I'm interested in, and then retrieve in order by the
identity column. I'm wondering if there's a way to do that with a
subquery.

Thanks,
Jim

[SQL Menace]

Running count is one way

CREATE TABLE #Values(val int)
INSERT #Values SELECT 1
INSERT #Values SELECT 5
INSERT #Values SELECT 10

select (select count(*) from #Values v where val <= v2.Val) as Rank,*
from #Values v2
order by 2

Denis the SQL Menace
http://sqlservercode.blogspot.com/
[e-mail] wrote:

I would like to write a query that gives me the values of a set of
observations, and their rank.

CREATE TABLE #Values(val int)
INSERT #Values SELECT 1
INSERT #Values SELECT 5
INSERT #Values SELECT 10

I would like to select this:

1 10 -- rank 1, value 10
2 5
3 1

I can put them into a temp table with an identity column, ordered by
the column I'm interested in, and then retrieve in order by the
identity column. I'm wondering if there's a way to do that with a
subquery.

Thanks,
Jim

[Alexander Kuznetsov]

Jim,

if you are using SS 2005 , use row_number() or rank() OLAP function

[Madhivanan]

Where do you want to show data?
If you use front end application, do Ranking there

Madhivanan


ji**********@countrywide.com wrote:

> I would like to write a query that gives me the values of a set of
observations, and their rank.

CREATE TABLE #Values(val int)
INSERT #Values SELECT 1
INSERT #Values SELECT 5
INSERT #Values SELECT 10

I would like to select this:

1 10 -- rank 1, value 10
2 5
3 1

I can put them into a temp table with an identity column, ordered by
the column I'm interested in, and then retrieve in order by the
identity column. I'm wondering if there's a way to do that with a
subquery.

Thanks,
Jim