Oops. Sorry, I misread the original post. Your solution is correct. Tim Rude ----- Original Message ----- From: Michael Iannantuoni To: [email protected] Sent: Wednesday, August 26, 2009 4:03 AM Subject: Re: [Dataperf] Substring odity
Further . . . Using your idea I managed to solve this using: substring[cat.t[P1F19;"#####"];1;5] Thanks, Michael Tim Rude wrote: Try this instead: substring[cat.t["#####";P1F19];0;5] It appends five pound signs to the beginning of P1F19 then grabs the last five characters in the resulting string. Substring gets the last five rather than the first five because of using 0 as the starting position number. Notice also that cat.t is used rather than cat.c to keep trailing blanks in P1F19 from causing problems. Tim Rude ----- Original Message ----- From: "Michael Iannantuoni" <[email protected]> To: "DataPerfect Users Discussion Group" <[email protected]> Sent: Tuesday, August 25, 2009 11:43 AM Subject: [Dataperf] Substring odity I have an odd situation with the Substring function which I have solved by using an if statement to avoid the oddity: In a field formula I have: P1F19 substring["#####";1;5-length[P1F19]] which is designed to add #'s to the end of P1F19 if it is shorter than five characters in length. This works if P1F19 is shorter than five characters but if it is five characters in length so that 5-length[P1F19] is zero, a # is still added to the end. Michael _______________________________________________ Dataperf mailing list [email protected] http://lists.dataperfect.nl/mailman/listinfo/dataperf _______________________________________________ Dataperf mailing list [email protected] http://lists.dataperfect.nl/mailman/listinfo/dataperf ------------------------------------------------------------------------------ _______________________________________________ Dataperf mailing list [email protected] http://lists.dataperfect.nl/mailman/listinfo/dataperf
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