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Archive for August, 2011

TO_CHAR(…, ‘D’)

Tuesday, 23 August, 2011 4 comments
to_char(nchar-clob-or-nclob)

to_char(datetime-or-interval)
to_char(datetime-or-interval, 'format-string')
to_char(datetime-or-interval, 'format-string', 'nlsparam')

to_char(number)
to_char(number,'format-string')
to_char(number,'format-string', 'nlsparam')

How do I get than MONDAY=1, TUESDAY=2, WEDNESDAY=3 … ?
solution1
With to_char()

alter session set nls_territory=germany;
select to_char(sysdate,’DAY D’) from dual;
TUESDAY 2

solution2
With decode()

The abbrivation of a day’s name (such as SUN, MON … can be retrieved with the format specifier DY. As to_char is sensitive to the actual nls setting, I use nls_date_language=english to make sure that the english abbreviation is returned, regardless of the nls setting.

select decode(to_char(sysdate, ‘FMDAY’, ‘NLS_DATE_LANGUAGE=american’),’MONDAY’, ’1′, ‘TUESDAY’, ’2′, ‘…’)) from dual;

solution3
With mod()
As a reference, I take monday Jan 1st, 1000.

select mod(trunc(sysdate)-date ’1000-01-01′,7)+1 from dual;
  • bonus )))
  • How do I trunc date to current monday?

    trunc(date, ‘D’) or here is with my solution with 1000-01-01:

    select trunc((sysdate-date ’1000-01-01′)/7)*7+date ’1000-01-01′ from dual;
    

    thanks to jan-marcel idea, I found that one
    trunc(date,’IW’) for current monday and date-trunc(date,’IW’)+1 for day number

    Spelling a number

    select to_char(to_date('429','J'),'Jsp') from dual;
    TO_CHAR(TO_DATE('429','J
    ------------------------
    Four Hundred Twenty-Nine
    

    from1
    from2

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    Categories: PL/SQL Tags: , ,

    Extraction of consecutive periods (without regard to the OverLap)

    Thursday, 18 August, 2011 Leave a comment

    kikanT
    ID StaD EndD
    -- ---------- ----------
    10 2010-06-01 2010-06-12
    10 2010-06-13 2010-06-14
    10 2010-06-15 null
    20 2010-06-01 2010-06-11
    20 2010-06-13 2010-06-15
    50 2010-11-11 2010-11-13
    50 2010-11-14 2010-11-20
    50 2010-12-22 2010-12-30

    Summarized in consecutive periods every ID.

    ID StaD EndD
    -- ---------- ----------
    10 2010-06-01 null
    20 2010-06-01 2010-06-11
    20 2010-06-13 2010-06-15
    50 2010-11-11 2010-11-20
    50 2010-12-22 2010-12-30

    Data creation script

    create table kikanT(ID,StaD,EndD) as
    select 10,date '2010-06-01',date '2010-06-12' from dual union
    select 10,date '2010-06-13',date '2010-06-14' from dual union
    select 10,date '2010-06-15',null              from dual union
    select 20,date '2010-06-01',date '2010-06-11' from dual union
    select 20,date '2010-06-13',date '2010-06-15' from dual union
    select 50,date '2010-11-11',date '2010-11-13' from dual union
    select 50,date '2010-11-14',date '2010-11-20' from dual union
    select 50,date '2010-12-22',date '2010-12-30' from dual;
    

    Solution1
    How to use hierarchical queries (> 10g)

    select ID,min(connect_by_root StaD) as StaD,EndD
      from kikanT
     where connect_by_IsLeaf = 1
    connect by prior ID=ID
           and prior EndD+1=StaD
    group by ID,EndD
    order by ID,EndD;
    

    Solution2
    Analysis of the relevant number

    select ID,min(StaD) as StaD,
    max(EndD) Keep(Dense_Rank Last order by EndD) as EndD
      from (select ID,StaD,EndD,
            sum(willSum) over(partition by ID order by StaD) as GID
            from (select ID,StaD,EndD,
                  case when StaD = 1+Lag(EndD) over(partition by ID order by StaD)
                       then 0 else 1 end as willSum
                    from kikanT))
    group by ID,GID
    order by ID,GID;
    
    Categories: PL/SQL Tags: , , ,

    Summarizes the start and end dates (or how to eleminate overlaps in dates)

    Thursday, 18 August, 2011 1 comment

    TimeSheet table
    StartDate EndDate
    ---------- ----------
    2005-01-01 2005-01-03
    2005-01-02 2005-01-04
    2005-01-04 2005-01-05
    2005-01-06 2005-01-09
    2005-01-09 2005-01-09
    2005-01-12 2005-01-15
    2005-01-13 2005-01-14
    2005-01-14 2005-01-14
    2005-01-17 2005-01-17

    result
    StartDate EndDate
    ---------- ----------
    2005-01-01 2005-01-05
    2005-01-06 2005-01-09
    2005-01-12 2005-01-15
    2005-01-17 2005-01-17

    Data creation script

    create table TimeSheets(
    StartDate Date,
    EndDate   Date,
    primary key (StartDate));
    
    insert into TimeSheets values(to_date('20050101','yyyymmdd'),to_date('20050103','yyyymmdd'));
    insert into TimeSheets values(to_date('20050102','yyyymmdd'),to_date('20050104','yyyymmdd'));
    insert into TimeSheets values(to_date('20050104','yyyymmdd'),to_date('20050105','yyyymmdd'));
    insert into TimeSheets values(to_date('20050106','yyyymmdd'),to_date('20050109','yyyymmdd'));
    insert into TimeSheets values(to_date('20050109','yyyymmdd'),to_date('20050109','yyyymmdd'));
    insert into TimeSheets values(to_date('20050112','yyyymmdd'),to_date('20050115','yyyymmdd'));
    insert into TimeSheets values(to_date('20050113','yyyymmdd'),to_date('20050114','yyyymmdd'));
    insert into TimeSheets values(to_date('20050114','yyyymmdd'),to_date('20050114','yyyymmdd'));
    insert into TimeSheets values(to_date('20050117','yyyymmdd'),to_date('20050117','yyyymmdd'));
    commit;
    

    SQL

    Solution 1
    How to use an inline view

    select StartDate,min(EndDate) as EndDate
    from
    (select a.StartDate,b.EndDate
       from TimeSheets a,TimeSheets b
      where a.EndDate     and not exists(select 1 from TimeSheets c
                        where (c.StartDate <  a.StartDate and a.StartDate                        or (c.StartDate group by StartDate
    order by StartDate;
    

    Solution 2
    How to use an inline view

    select a.StartDate,min(b.EndDate) as EndDate
       from TimeSheets a,TimeSheets b
      where a.EndDate     and not exists(select 1 from TimeSheets c
                        where (c.StartDate <  a.StartDate and a.StartDate                        or (c.StartDate group by a.StartDate
    order by a.StartDate;
    

    Solution 3
    How to use an inline view

    SELECT startdate, MIN(enddate) AS enddate
      FROM (SELECT T1.startdate, T2.enddate
              FROM Timesheets T1, Timesheets T2
             WHERE T1.enddate  T3.startdate AND T1.startdate = T3.startdate AND T2.enddate < T3.enddate)))
     GROUP BY startdate;
    

    Solution 4
    How to use an inline view

    SELECT startdate, MIN(enddate) AS enddate
      FROM (SELECT T1.startdate AS startdate, T2.enddate AS enddate
              FROM Timesheets T1, Timesheets T2, Timesheets T3
             WHERE T1.enddate  T3.startdate AND T1.startdate = T3.startdate AND T2.enddate < T3.enddate) THEN 1 END) = 0)
     GROUP BY startdate;
    

    Solution 5
    How to use a hierarchical queries (> 10g)

    select min(connect_by_root StartDate) as StartDate,EndDate
      from TimeSheets
     where connect_by_IsLeaf = 1
    connect by nocycle prior EndDate between StartDate and EndDate
    group by EndDate
    order by StartDate;
    

    Solution 6
    How to use hierarchical queries (> 10g)

    select connect_by_root StartDate as StartDate,EndDate
      from TimeSheets a
     where connect_by_IsLeaf = 1
    start with not exists(select 1 from TimeSheets b
                           where a.RowID != b.RowID
                             and a.StartDate between b.StartDate and b.EndDate)
    connect by nocycle prior EndDate between StartDate and EndDate
    order by StartDate;
    

    Solution 7
    How to use analysis of the relevant number range

    select min(StartDate) as StartDate,max(EndDate) as EndDate
    from (select StartDate,EndDate,
          sum(willSum) over(order by StartDate) as GID
          from (select StartDate,EndDate,
                case when StartDate
                                                     range between unbounded preceding
                                                 and 1 preceding)
                     then 0 else 1 end as willSum
                  from TimeSheets))
    group by GID
    order by StartDate;
    

    from

    Categories: PL/SQL Tags: , , ,

    How does one get the time difference between two date columns?

    Friday, 12 August, 2011 Leave a comment

    Oracle allows two date values to be subtracted from each other returning a numeric value indicating the number of days between the two dates (may be a fraction). This example will show how to relate it back to a time value.

    Let’s investigate some solutions. Test data:

    CREATE TABLE dates (date1 DATE, date2 DATE);
    INSERT INTO dates VALUES (SYSDATE, SYSDATE-1);
    INSERT INTO dates VALUES (SYSDATE, SYSDATE-1/24);
    INSERT INTO dates VALUES (SYSDATE, SYSDATE-1/60/24);
    SELECT (date1 - date2) FROM dates;
    
    DATE1-DATE2
    -----------
    1
    .041666667
    .000694444
    

    Solution 1

    SELECT floor(((date1-date2)*24*60*60)/3600)
    || ' HOURS ' ||
    floor((((date1-date2)*24*60*60) -
    floor(((date1-date2)*24*60*60)/3600)*3600)/60)
    || ' MINUTES ' ||
    round((((date1-date2)*24*60*60) -
    floor(((date1-date2)*24*60*60)/3600)*3600 -
    (floor((((date1-date2)*24*60*60) -
    floor(((date1-date2)*24*60*60)/3600)*3600)/60)*60) ))
    || ' SECS ' time_difference
    FROM dates;
    
    TIME_DIFFERENCE
    --------------------------------------------------------------------------------
    24 HOURS 0 MINUTES 0 SECS
    1 HOURS 0 MINUTES 0 SECS
    0 HOURS 1 MINUTES 0 SECS
    

    An enhancement to solution 1

    SELECT floor((date1-date2)*24)
    || ' HOURS ' ||
    mod(floor((date1-date2)*24*60),60)
    || ' MINUTES ' ||
    mod(floor((date1-date2)*24*60*60),60)
    || ' SECS ' time_difference
    FROM dates;</code>
    
    TIME_DIFFERENCE
    --------------------------------------------------------------------------------
    24 HOURS 0 MINUTES 0 SECS
    1 HOURS 0 MINUTES 0 SECS
    0 HOURS 1 MINUTES 0 SECS
    

    Solution 2

    If you don’t want to go through the floor and ceiling math, try this method (contributed by Erik Wile):

    SELECT to_number( to_char(to_date('1','J') +
    (date1 - date2), 'J') - 1) days,
    to_char(to_date('00:00:00','HH24:MI:SS') +
    (date1 - date2), 'HH24:MI:SS') time
    FROM dates;
    
    DAYS TIME
    ---------- --------
    1 00:00:00
    0 01:00:00
    0 00:01:00
    

    Solution 3: If u want an easier method, use numtodsinterval()

  • Submitted by Shilpa Petrim
  • NUMTODSINTERVAL: This function is new to Oracle 9i. It takes two arguments numtodsinterval(x,c) where x is a number and c is a character string denoting the units of x. Valid units are ‘DAY’, ‘HOUR’, ‘MINUTE’ and ‘SECOND’.

    This function converts the number x into an INTERVAL DAY TO SECOND datatype.

    select numtodsinterval(date1-date2,'day') time_difference from dates;
    TIME_DIFFERENCE
    ----------------------------------------------------------------
    +000000001 00:00:00.000000000
    +000000000 01:00:00.000000000
    +000000000 00:01:00.000000000
    

    from

    Categories: PL/SQL Tags: , , , ,