How to add leading zeroes

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  • lenygold via DBMonster.com

    #1

    How to add leading zeroes

    I have the following column:

    1.4.1
    1.10.1
    1.10.1.2.2
    1.22.99.1
    2
    2.8.11
    2.7.7

    I am trying to add leading zeros to every part before dots, to make them 3
    digit numbers:
    001.004.001
    001.010.001
    001.010.001
    001.022.099.001
    002
    002.008.011
    002.007.007

    My first try did't work:
    SELECT
    REPLACE ('1.4.22.34 ', S, B)
    FROM (SELECT '.34 ' , '.034 '
    FROM SYSIBM.SYSDUMMY 1
    UNION SELECT '.4.' , '.004.'
    FROM SYSIBM.SYSDUMMY 1
    UNION SELECT '.22.' , '.022.'
    FROM SYSIBM.SYSDUMMY 1) T(S,B)

    output:

    1.4.022.34
    1.4.22.034
    1.004.22.34

    Any idea how to produce requested result?
    Thank's in advance
    Lenny G.

    --
    Message posted via DBMonster.com


  • Lennart

    #2
    Re: How to add leading zeroes

    On May 5, 5:27 pm, "lenygold via DBMonster.com" <u41482@uwewrot e:
    I have the following column:
    >
    1.4.1
    1.10.1
    1.10.1.2.2
    1.22.99.1
    2
    2.8.11
    2.7.7
    >
    I am trying to add leading zeros to every part before dots, to make them 3
    digit numbers:
    001.004.001
    001.010.001
    001.010.001
    001.022.099.001
    002
    002.008.011
    002.007.007
    >
    My first try did't work:
    SELECT
    REPLACE ('1.4.22.34 ', S, B)
    FROM (SELECT '.34 ' , '.034 '
    FROM SYSIBM.SYSDUMMY 1
    UNION SELECT '.4.' , '.004.'
    FROM SYSIBM.SYSDUMMY 1
    UNION SELECT '.22.' , '.022.'
    FROM SYSIBM.SYSDUMMY 1) T(S,B)
    >
    output:
    >
    1.4.022.34
    1.4.22.034
    1.004.22.34
    >
    Any idea how to produce requested result?
    Thank's in advance
    Lenny G.
    >
    You can do it with a case expression:

    case when length(rtrim(x) ) = 1 then '00' || x ...

    but you are (IMO) going in the wrong direction. Let's look at a
    simplified example:

    [lelle@53dbd181 ~]$ db2 "select * from lateral(values( '1+21+131'))
    x(c), table(elements( x.c))"

    C ORDINAL INDEX
    -------- ----------- -----------
    1+21+131 0 0
    1+21+131 1 2
    1+21+131 2 5

    3 record(s) selected.

    index+1 will give us the startpos for each new number:

    [lelle@53dbd181 ~]$ db2 "select substr(x.c,inde x+1) from
    lateral(values( '1+21+131')) x(c), table(elements( x.c))"
    1
    --------
    1+21+131
    21+131
    131

    The stop pos for each number will be the next '+' and in case there
    are no '+' left, we can use the length of the remaining string:

    [lelle@53dbd181 ~]$ db2 "select substr(x.c,inde x
    +1,coalesce(nul lif(locate('+', substr(x.c,inde x+1)),
    0)-1,length(substr (x.c,index+1))) ) from lateral(values( '1+21+131'))
    x(c), table(elements( x.c))"

    1
    --------
    1
    21
    131

    IMO, this kind of problem is better solved in the application layer

    /Lennart

    Comment

    • Serman D.

      #3
      Re: How to add leading zeroes

      On May 5, 5:27 pm, "lenygold via DBMonster.com" <u41482@uwewrot e:
      Any idea how to produce requested result?
      I would have used perl:

      $ cat /tmp/foo.pl
      use strict;
      use warnings;

      while(my $line = <DATA>){
      print join(q{.}, map { sprintf("%03d", $_); } split(/\./,
      $line));
      }

      __DATA__
      1.4.1
      1.10.1
      1.10.1.2.2
      1.22.99.1
      2
      2.8.11
      2.7.7

      $ perl -wl /tmp/foo.pl
      001.004.001
      001.010.001
      001.010.001.002 .002
      001.022.099.001
      002
      002.008.011
      002.007.007

      --
      Serman D.

      Comment

      • Lennart

        #4
        Re: How to add leading zeroes

        On May 6, 1:48 pm, "lenygold via DBMonster.com" <u41482@uwewrot e:
        Thank' s everybody for help.
        The reason why i need this conversion is to sort the following rows:
        >
        I didn't realize that it was sorting that you where after. The
        following post by Vadim is an alternative way of attacking traversal
        of a tree, and therefor might be of interest to you.



        A total ordering is defined as:

        6. For any nodes A and B we write A B whenever
        i. B is Ancestor of A or
        ii. there exists node B' which is an ancestor of B,
        and A' which is an ancestor of A,
        and both A' and B' having the same parent,
        and A' B'

        /Lennart

        Comment

        • --CELKO--

          #5
          Re: How to add leading zeroes

          >Although I agree with Lennart, I want to try in SQL simply for my entertainment. <<

          I agree, but it is fun. Life is easier if the input string always ends
          with a period, and I think that is required in one of the ISO or US
          Government Standards for outlines.

          1) Can you nest function calls 100 levels deep? So we just nest
          REPLACE() calls that deep:

          SELECT REPLACE ( ..
          REPLACE (sourcestring, '1.', '001.'),
          ..
          '99.', '099.')
          FROM Foobar;

          Since the functions will be on the stack, this ought to run pretty
          fast. But only a LISP programmer would love it.

          2) Go back to procedural programming and write a loop?

          CREATE TABLE LeadingZeros
          (instring VARCHAR(3) NOT NULL PRIMARY KEY,
          outstring CHAR(4) NOT NULL
          CHECK (outstring SIMILAR TO '[:DIGITS:][:DIGITS:][:DIGITS:]\.');

          INSERT INTO LeadingZeros (convert_nbr, instring, outstring)
          VALUES (1, '1.', '001.'), (2, '2.', '002.'), .., (99, '99.', '099.');

          CREATE PROCEDURE PaddingZeros ()
          LANGUAGE SQL
          READS SQL DATA
          BEGIN DECLARE i INTEGER;
          SET i = 1;
          WHILE i < 100
          DO UPDATE Foobar
          SET sourcestring
          = REPLACE(sources tring,
          (SELECT instring
          FROM LeadingZeros
          WHERE i = convert_nbr),
          (SELECT outstring
          FROM LeadingZeros
          WHERE i = convert_nbr));
          SET i = i+1;
          END WHILE;
          END;

          This can be done with a recursive CTE, but that might be even worse.

          Comment

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