Subject AW: AW: [firebird-support] string difficulty
Author checkmail

.. and if the ID has the length of 15 chars and only the las * is for filling out to 16, i get a wrong result too.

 

Can I compact my code?

 

  if(char_length(str)>16) then

  begin

    str = left(:str,16);

    while (k = 1) do

    begin

      if(right(str, 1) = '*') then

      begin

        str = left(str,char_length(str)-1);

      end

      else

      begin

       teil = str;

       k = 0;

      end

    end

  end

 

(exept the one begin-end (only one command can be without..)

 

Von: firebird-support@yahoogroups.com [mailto:firebird-support@yahoogroups.com]
Gesendet: Montag, 7. Dezember 2015 11:32
An: firebird-support@yahoogroups.com
Betreff: Re: AW: [firebird-support] string difficulty

 

 

select replace('ABC*DEFG********01','**','') from RDB$DATABASE

 

 

On Monday, December 7, 2015 12:26 PM, "Virna Constantin costel_61@... [firebird-support]" <firebird-support@yahoogroups.com> wrote:

 

 

sorry, wrong response :(

 

 

 

On Monday, December 7, 2015 12:22 PM, "Virna Constantin costel_61@... [firebird-support]" <firebird-support@yahoogroups.com> wrote:

 

 

a quick answer :

m_char='**'

 

 

 

On Monday, December 7, 2015 12:12 PM, "'checkmail' check_mail@... [firebird-support]" <firebird-support@yahoogroups.com> wrote:

 

But If the ID is ABC*DEFG, I get ABC*DEFG********01 and in this case only all * from right should be deleted.

 

I have create this, is there a simpler way?