PL/SQL code help
Morning all,
I need assistance with regard to the PL/SQL code.
Question: How many cases was activated from Direct treatment information/advice within 8 weeks?
Sample data:
MEMBER_ID | CASE_ID | SP_CODE | SP_NAME | SP_SUBTYPE_CODE | SP_SUBTYPE_NAME | REFERRAL_DATE | SERVICE_DATE |
000000001 | 2013858 | 1001 | Info & advice | 1001 | Information | - | 25/09/2012 |
000000001 | 2013858 | 1005 | Direct treatment | 1022 | Seamless | 10/01/2012 | 10/01/2012 |
I need to count the number of the place where CASE_ID SP_CODE = '1001 ' and then a SP_CODE = '1005' as the program installation, then the difference between SERVICE_DATE for ' 1001' and for '1005' REFERRAL_DATE is within 8 weeks? I hope this makes sense?
Hello
One way is to use a Sun-query EXISTS, like this:
SELECT COUNT (case_id) AS cnt - or COUNT (DISTINCT case_id)
FROM table_x m
WHERE sp_code = 1005
AND THERE ARE)
SELECT 1
FROM table_x
WHERE sp_code = 1001
AND case_id = m.case_id
AND service_date > = m.service_date - (8 * 7)
AND service_date< > >
)
;
I hope that answers your question.
If not, post a small example of data (CREATE TABLE and INSERT statements) and the results desired from these data. Point where the query above will not and explain how to get good results in these places.
See the FAQ forum: https://forums.oracle.com/message/9362002#9362002
The combination (case_id, sp_code) is unique? What happens if a case_id has, say, 1001 multiple s all less than 8 weeks before the same 1005?
Member_id (or one of the other columns not in the above query) plays no role in this problem?
Tags: Database
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============================
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=============================
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END total_info_dept;Regards. Al
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hope responds to your question.mark points.
See you soon,.
KKPublished by: Jocelyn on January 24, 2011 22:25
Published by: Jocelyne 24 January 2011 22:27
Published by: Jocelyn on January 25, 2011 02:13
Published by: Jocelyne 25 January 2011 05:26
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