1 each range interval data retrieval
Hi guys
I have this game of number, for example. 12 13 201 243 303 584 674 698 731 745 768 800. So what I want to do is to extract the highest number in each interval of 100. For example, in the range of 0 to 100, 13 will only be extracted from.
Finally, all the data in the result loop must be an array of 13 243 303 584 698 768 800.
I tried to use a nested loop and a false true case with table generation, but I couldn't get the setting I want. Please help me on this thank you.
In addition, each number on the top is accompanied by another number, meanwhile extract the data I have, I'd be the number that goes with it on the next column.
See you soon
PX
This assumes that your table is sorted.
If you use the rest quotient function between the current value and the following in your table and compare the quotients, you will know if the next value is a new hundred, making it the largest present value number in the current one hundred.
Tags: NI Software
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Oracle Database 11g Express Edition Release 11.2.0.2.0 - Production
Hi gurus
I'm stuck with a scenario and your employees need help to solve this problem.
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Insertion and table creation
drop table ident;
create table ident
(
agreement_id number (5),
ident_pk number (5),
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Date of Cov_termination_date
);
-------------------
Insert ident
(
Select 100,1,to_date('2013-01-01','YYYY-MM-DD'), double to_date('2013-01-31','YYYY-MM-DD')
Union of all the
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Union of all the
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create the table ident_dtl
(
ident_pk number (5),
ident_dtl_pk number (5),
date of effective_date,
date of termination_date
);
---------------------------------
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(
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);
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------------------------------------
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for example:
See the data in Ident_pk = 1, a = cov_effective_date 13/01/01 and its cov_termination_date = 13/01/31 and in the details table against column Ident_pk = 1, you can find that at lease 1 combinaisondes date match against Ident_pk = 1
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Same as 3
-----------
Summary is that there are date range in the ident against agreement_id and Ident_pk table and at least a range of dates even should be exist in detail table is ident_dtl against Ident_pk and if not then show that agreement_id and Ident_pk .
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Shu
Hello
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i.ident_pk
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AND d.effective_date = i.cov_effective_date
AND d.termination_date = i.cov_termination_date
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Another pure SQL approach to such problems is to use an EXISTS subquery.
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If you do any filtering, it may be more effective to do this first. Start the WITH clause with a subquery that performs filtering, then use this result placed instead of your full table in the view online within the cntr as well as in the main query.
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Hi friend
I have a doubt, how to get LASTDAY for each month between data dates...
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Something like this (assuming that the dates are originally VARCHAR2s):
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Hello
Database version:
DB: Oracle Database 11 g Enterprise Edition Release 11.2.0.3.0 - 64 bit Production
Operating system: HP - UX nduhi18 B.11.31 U ia64 1022072414 unlimited-license user
APP: SAP - ERP
I have to the partition of the RANGE on UPDATED_ON or PROFILE is a table that has a structure below:
Name Null? Type
-------------------- -------- --------------
------------------
MANDT NOT NULL VARCHAR2 (9)
MR_ID NOT NULL VARCHAR2 (60)
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(
"CREATED_ON".
)
SUBPARTITION BY HASH
(
'PROFILE '.
)
SUBPARTITION TEMPLATE
(
TABLESPACE SUBPARTITION 'PROF_SUB01"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB02"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB03"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB04"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB05"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB06"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB07"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB08"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB09"'PSAPISU."
TABLESPACE SUBPARTITION 'PROF_SUB10' 'PSAPISU '.
)
(
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"JAN12_CP01" VALUES LOWER PARTITION TO (20120201000000),
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)
works very well.
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Catch the range of Date values
Hi, guys.
I need to check the date range that my product was available. You can use as a basis the values in the table below, but this structure can be modified to add all the things that could support the solution:
ID_PRDCT DT_STRT DT_END
1 20/01/25 12/01/12
1 01/23/12-27/01/12
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1 28/01/12-21/04/12
1 20/04/12-25/05/12
1 24/05/12-28/05/12
1 26/05/12 05/29/12
1 01/07/12-03/08/12
1 01/08/12-16/08/12
1 18/08/12-30/08/12
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In this case, the result of my date range should be:
1 20/01/12-29/05/12
1 01/07/12-16/08/12
1 18/08/12-30/08/12
1 01/12/12-31/12/12Hello
An the Heres how:
WITH got_new_grp AS ( SELECT id_prdct, dt_strt, dt_end , CASE WHEN dt_strt <= MAX (dt_end) OVER ( PARTITION BY id_prdct ORDER BY dt_strt ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING ) THEN 0 ELSE 1 END AS new_grp FROM t ) , got_grp AS ( SELECT id_prdct, dt_strt, dt_end , SUM (new_grp) OVER ( PARTITION BY id_prdct ORDER BY dt_strt ) AS grp FROM got_new_grp ) SELECT id_prdct , MIN (dt_strt) AS grp_dt_strt , MAX (dt_end) AS grp_dt_end FROM got_grp GROUP BY id_prdct , grp ORDER BY id_prdct , grp ;
This assumes dt_strt<= dt_end="" on="" every="">=>
If you would care to post CREATE TABLE and INSERT statements for your sample data, then I could test it.You cannot use LAG or LEAD to see where each new group starts, because the order of the dt_strt is not the same as the order of the dt_end. In other words, if you ORDER BY dt_strt, two adjacent lines could be in the same group regardless of their values, because a previous line can have a dt_end, and as a result this line would be overlapp both of them. Similarly, if you ORDER BY dt_end, no matter what two adjacent lines contain, whether they are in the same group if som rank later has a sufficient dt_strt in advance.
-
How to get the number of days in a month of belonging to a range of dates
Hi, I'm going crazy around a problem, I have 2 dates and one month I wanto to retrieve the number of days belonging to the months that fall within the range.
for example:
month of January 2011. begin_date = January 11, 2011, May 30, 2011 end_date result is 21
month of January 2011. begin_date = December 11, 2010, result of end_date, January 10, 2011 10
month January 2011 .begin_date = 2 February 2011, end_date may 25, 2011 result 0
month of January 2011. begin_date = January 3, 2011, January 5, 2011 result DATE END 3
and so on...
I appreciate any suggestion
Thank you
AndreaSomething like that... ?
SQL> with t as 2 (select to_date('11/01/11','dd/mm/yy') from_dt, 3 to_date('30/05/11','dd/mm/yy') to_dt, 4 'Jan-11' mnth from dual) 5 select least(last_day(to_date(mnth,'Mon-yy')),to_dt) - 6 greatest(to_date(mnth,'Mon-yy'),from_dt) cnt 7 from t ; CNT ---------- 20
If 0 is also expected...
SQL> with t as 2 (select to_date('11/01/11','dd/mm/yy') from_dt, 3 to_date('30/05/11','dd/mm/yy') to_dt, 4 'Jan-11' mnth from dual union all 5 select to_date('11/01/11','dd/mm/yy') from_dt, 6 to_date('30/05/11','dd/mm/yy') to_dt, 7 'Jan-12' mnth from dual 8 ) 9 select from_dt,to_dt,mnth, 10 greatest( 11 least(last_day(to_date(mnth,'Mon-yy')),to_dt) 12 - 13 greatest(to_date(mnth,'Mon-yy'),from_dt) 14 ,0) cnt 15 from t; FROM_DT TO_DT MNTH CNT --------- --------- ------ ---------- 11-JAN-11 30-MAY-11 Jan-11 20 11-JAN-11 30-MAY-11 Jan-12 0
Published by: JAC on February 9, 2012 19:22
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