Reputation: 5736
Hi I have a table of data I want to output the dense_rank of the names starting from the first group of names according to sorted dates order. e.g.
DROP TABLE MyTable
SELECT * INTO MyTable FROM (
VALUES ('2015-12-23', 'ccc'),('2015-12-21', 'aaa'),
('2015-12-20', 'bbb'),('2015-12-22', 'aaa')
) t (date, name)
SELECT DENSE_RANK() OVER (ORDER BY name) rank, * FROM MyTable ORDER BY date
For the query above I got
rank date name
2 2015-12-20 bbb
1 2015-12-21 aaa
1 2015-12-22 aaa
3 2015-12-23 ccc
You can see the dates are sorted (good), ranks are assigned to names in group (good), but the ranks do not start from the first group of name, e.g. I want
rank date name
1 2015-12-20 bbb
2 2015-12-21 aaa
2 2015-12-22 aaa
3 2015-12-23 ccc
How would you correct the query? If there are multiple working answers, the simplest/shortest one will be picked as the answer. Thanks.
Thanks @lad2025 and @GiorgosBetsos for clarifying my question. Sorting is straightly according to dates and dates are unique in my case. Names can repeat and appear non-consecutively. So with ('2015-12-24', 'aaa')
, the output is
rank date name
1 2015-12-20 bbb
2 2015-12-21 aaa
2 2015-12-22 aaa
3 2015-12-23 ccc
4 2015-12-24 aaa
Upvotes: 8
Views: 7332
Reputation: 72175
You can use:
SELECT DENSE_RANK() OVER (ORDER BY minGrpDate),
[date], name
FROM (
SELECT MIN([date]) OVER (PARTITION BY grp) AS minGrpDate,
[date], name
FROM (
SELECT [date], name,
ROW_NUMBER() OVER (ORDER BY [date])
-
ROW_NUMBER() OVER (PARTITION BY name ORDER BY [date]) AS grp
FROM mytable) AS t ) AS s
ORDER BY Date
Explanation:
grp
field identifies islands of consecutive records having the same name
. minGrpDate
, which is calculated using grp
, is the minimum date of each island.minGrpDate
we can now apply DENSE_RANK()
to get required rank.Note1: The above query handles discontinuities in name
field, i.e. the case of non-consecutive fields having the same name.
Note2: The query does not handle the case of different name
values sharing the same date
value.
Upvotes: 4
Reputation: 35780
First rank distinct names ordered by date and then join on the table:
;WITH cte AS(SELECT name, ROW_NUMBER() OVER(ORDER BY MIN(date)) rn
FROM dbo.MyTable
GROUP BY name)
SELECT c.rn, m.date, m.name
FROM cte c
JOIN dbo.MyTable m ON m.name = c.name
ORDER BY m.date
Upvotes: 3