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Hi Jeff,The parms for %replace are: replacement string, source string, start position, length. The %scan you have will find the comma in position 2, but you add 1 to that, so the length parm in your %replace is actually 3, i.e.
midrange.com
ChrOut = %replace ('':ChrIN:1:3)
which will replace positions 1-3 of ChrIN with a zero-length string,
which leaves you with 00.70, which is what you got.
Another point about %replace is that it only does one replacement. If there were more than one comma in ChrIn, it would only replace the first one.
If you had reversed the 3rd & 4th parms of your %replace, you'd've run into another problem -- ChrOut is too small to hold the result. "1,100.70" is 8 characters. Remove the comma and you have 7 characters, but ChrOut is only 5. Better make it the same as ChrIn in case there are no commas in ChrIn.
So what you probably want is something like:
D CHRIn s 17a INZ('1,100.70')
D CHROut s 17a
D x s 5i 0
/free
dow ChrIn <> 'QUIT';
dou x = *zero;
x = %scan(',' : ChrOut);
if x > *zero;
ChrOut = %replace('' : ChrOut : x : 1);
endif;
enddo;
DSPLY ChrOUT ' ' ChrIN;
EndDo;
*INLR = *On;
/end-free
*Peter Dow* /
Dow Software Services, Inc.
909 793-9050
pdow@xxxxxxxxxxxxxxx <mailto:pdow@xxxxxxxxxxxxxxx> /
jstevens@xxxxxxxxxxxxxx wrote:
I'm trying to remove the commas in a field using the folling code... D CHRIn s 17a INZ('1,100.70') D CHROut s 5a /free DoW ChrIn <> 'QUIT'; ChrOut = %replace('':ChrIN:1:%scan(',':ChrIN)+1 ); DSPLY ChrOUT ' ' ChrIN; EndDo; *INLR = *On;/end-freeAnd it Produces the following result...DSPLY 00.70 1,100.70But I want it to produce DSPLY 1100.70 1,100.70 Can anyone see where I'm going wrong? Thanks! Jeff Stevens Mize, Houser & Co. P.A. 913 451 1882 JStevens@xxxxxxxxxxxxxx
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