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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

    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: , , , ,

    Concatenate sets of intervals

    Thursday, 5 May, 2011 Leave a comment
    CLEAR;
    with t as (
               select 11  AS ID, 1  AS val1, 'B'  AS val2 from dual union all
               select 12,1,'A' from dual union ALL
               select 13,1,'A' from dual union all           
               select 14,1,'B' from dual union ALL
               select 15,1,'A' from dual UNION ALL
               select 16,1,'B' from dual union ALL           
               select 17,1,'B' from dual union ALL
               select 18,1,'A' from dual union ALL           
               select 19,2,'B' from dual union all
               select 20,2,'A' from dual union ALL
               select 21,2,'B' from dual union ALL
               select 22,2,'A' from dual union ALL
               select 23,2,'B' from dual union ALL           
               select 24,2,'A' from dual union ALL           
               select 25,1,'B' from dual union ALL                      
               select 26,1,'A' from dual
               ),
    tt AS(select t.*,
            case when lag (val2) over (partition by val1 order by id) || val2 in ('A', 'BA') then 1 end as start_of_group,
            case when lead(val2) over (partition by val1 order by id) || val2 in ('B', 'AB') then 1 end as   end_of_group
          from t),
    ttt AS (SELECT tt.*, COUNT(start_of_group) over(PARTITION BY val1 ORDER BY id) AS group_no FROM tt),
    tttt AS (SELECT row_number() over(ORDER BY val1, group_no) AS rn, 
           val1, 
           t1.ID AS id1, 
           t2.ID AS id2, 
           t1.val2 AS VAL2A, 
           t2.val2 AS VAL2B
      FROM (SELECT * FROM ttt WHERE start_of_group IS NOT NULL) t1
      FULL JOIN (SELECT * FROM ttt WHERE end_of_group IS NOT NULL) t2
     USING (val1, group_no)
     ORDER BY rn)
      SELECT * FROM tttt;
    
    Categories: PL/SQL Tags:

    Oracle PL/SQL – Merge table

    Thursday, 20 January, 2011 2 comments
    clear;
    create table myTable
    (pid number, sales number, status varchar2(6));
     create table myTable2
    (pid number, sales number, status varchar2(6));
    
     insert into myTable  values(2,24,'CURR');
     insert into myTable  values(3, 0,'OBS' );
     insert into myTable  values(4,42,'CURR');
     insert into myTable  values(6,56,'C44URR');
    
    
     insert into myTable2 values(1,12,'CURR');
     insert into myTable2 values(2,13,'NEW' );
     insert into myTable2 values(3,15,'CURR');
    
    select * from myTable;
    select 't2' from dual;
    
    merge into myTable2 m
    using myTable d
    on (m.pid = d.pid)
    when matched then
      update
         set m.sales = d.sales, m.status = d.status 
    when not matched then  
       insert values (d.pid, d.sales, d.status);
      
     delete myTable2 where pid in (select m.pid from myTable2 m where m.pid not in (select pid from myTable));
    
     select * from myTable order by pid;
     select * from myTable2 order by pid;
    
     drop table myTable;
     drop table myTable2;
    
    Categories: PL/SQL Tags:

    Autoincrement primary key for Oracle

    Thursday, 20 January, 2011 1 comment

    Suppose you have a database and you want each entry to be identified by a unique number. You can do this easily in mysql by specifying “auto_increment” for your number, but Oracle makes you work a little more to get it done.

    Here is one way to do it by creating two database objects, a sequence and a trigger. I find myself wanting to do this every now and then but not often enough that I remember the syntax from time to time, so I decided it was time to write myself up a little cheat sheet. This is an extremely basic outline, so please try it first on a test table if you don’t know what you’re doing.

    1. Let’s say we have a table called “test” with two columns, id and testdata. (This is just a dumb quick example, so I won’t bother to specify any constraints on id.)

    create table test (id number, testdata varchar2(255)); 
    

    2. Next we’ll create a sequence to use for the id numbers in our test table.

    create sequence test_seq 
    start with 1 
    increment by 1 
    nomaxvalue; 
    

    You could change “start with 1” to any number you want to begin with (e.g. if you already have 213 entries in a table and you want to begin using this for your 214th entry, replace with “start with 214”). The “increment by 1” clause is the default, so you could omit it. You could also replace it with “increment by n” if you want it to skip n-1 numbers between id numbers. The “nomaxvalue” tells it to keep incrementing forever as opposed to resetting at some point. i (I’m sure Oracle has some limitation on how big it can get, but I don’t know what that limit is).

    3. Now we’re ready to create the trigger that will automatically insert the next number from the sequence into the id column.

    create trigger test_trigger
    before insert on test
    for each row
    begin
    IF :new.id IS NULL then
    select test_seq.nextval into :new.id from dual;
    end IF;
    end;
    /
    

    Obviously you would replace “test_trigger” with something a little more meaningful for the database table you want to use it with, “test” would be your table name, and the “id” in :new.id would be replaced with the name of the column. Every time a new row is inserted into test, the trigger will get the next number in the sequence from test_seq and set the “id” column for that row to whatever the sequence number is. Note that sequences sometimes appear to skip numbers because Oracle caches them to be sure that they are always unique, so this may not be your ideal solution if it’s really important that the id is exactly sequential and not just mostly sequential and always unique.

    Greg Malewski writes:

    You’ve demonstrated an implementation using triggers. This is not necessary, since instead it can be included as part of the INSERT statement. Using your example, my INSERT statement would be:

        
    insert into test values(test_seq.nextval, 'voila!');
    

    Here are a couple of questions the above might raise. This is pretty intuitive stuff, but I’m aiming it at the Oracle newbie since no expert would be reading this page anyway.

    How do you tell what sequences and triggers are already out there?

    select sequence_name from user_sequences;
    select trigger_name from user_triggers;
    

    How do you get rid of a sequence or trigger you created?

    drop sequence test_seq;
    drop trigger test_trigger;
    

    Again, replace test_seq and test_trigger with the specific names you used. You can also keep the trigger but disable it so it won’t automatically populate the id column with every insert (and enable it again later if you want):

    alter trigger test_trigger disable;
    alter trigger test_trigger enable;
    

    w3

    Categories: PL/SQL Tags: ,