grouping sequenced data

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

WAeberhardtAsked:
Who is Participating?
 
sdstuberConnect With a Mentor Commented:
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
 
cjonlineCommented:
Can you explain the ordering?
0
 
ChristoferDutzCommented:
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
Improve Your Query Performance Tuning

In this FREE six-day email course, you'll learn from Janis Griffin, Database Performance Evangelist. She'll teach 12 steps that you can use to optimize your queries as much as possible and see measurable results in your work. Get started today!

 
sdstuberCommented:
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

Open in new window

0
 
WAeberhardtAuthor Commented:
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
 
WAeberhardtAuthor Commented:
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
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.