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Solved

grouping sequenced data

Posted on 2009-07-08
6
212 Views
Last Modified: 2013-11-05
From the table

Start_date    end_date  value
2009/01/01  2009/30/04   A
2009/01/05  2009/31/06   A
2009/01/06  2009/31/08   B
2009/01/09  2009/31/12   A

I need the table
2009/01/01  2009/31/05   A
2009/01/06  2009/31/08   B
2009/01/09  2009/31/12   A

Werner

0
Comment
Question by:WAeberhardt
6 Comments
 

Expert Comment

by:cjonline
ID: 24801869
Can you explain the ordering?
0
 
LVL 20

Expert Comment

by:ChristoferDutz
ID: 24801876
I doubt this will be possible using standard SQL since in SQL statements there is no such concept of "previous record" or "next record" ... I think you will have to do the processing in the programming language that you are using to access the database.
0
 
LVL 74

Expert Comment

by:sdstuber
ID: 24803114
if your database supports LEAD and LAG you can do this with sql.

I think there is a problem in your sample data...

2009/31/06   should be
2009/31/05  I think,  based on your expected results.

I've tested the query below in Oracle and it produces the expected results.
SELECT   DISTINCT
         CASE WHEN start_date - 1 <= prevend THEN prevstart ELSE start_date END start_date,
         CASE WHEN end_date + 1 >= nextstart THEN nextend ELSE end_date END end_date,
         VALUE
FROM     (SELECT start_date,
                 end_date,
                 VALUE,
                 LAG(start_date) OVER (PARTITION BY VALUE ORDER BY start_date) prevstart,
                 LAG(end_date) OVER (PARTITION BY VALUE ORDER BY start_date) prevend,
                 LEAD(start_date) OVER (PARTITION BY VALUE ORDER BY start_date) nextstart,
                 LEAD(end_date) OVER (PARTITION BY VALUE ORDER BY start_date) nextend
          FROM   your_table) x
ORDER BY start_date, end_date, VALUE

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Author Comment

by:WAeberhardt
ID: 24804249
sdstuber,

Regarding data, You are right. (mistake in my sample)

Can You give me the syntax of the LAG and LEAD scalar functions? I'll try to write them for my dialect.
Werner

0
 
LVL 74

Accepted Solution

by:
sdstuber earned 500 total points
ID: 24804337
the syntax is in the code snippet

if your database doesn't support analytics you probably won't be able to write them

Instead you will probably need to do self-joins to find previous/next values
or iterate through the values with a procedural language.
0
 

Author Comment

by:WAeberhardt
ID: 24808124
Two benefits of answer

firstly, it was important to know, that there is no solution in standard SQL
secondly, the path to solve the problem in now clear (working with cursor. )
--> I can rebuild the solution in my dialect.

Werner
0

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