Lindsay,

You can not do a simple LEN check, otherwise 

X = " 1"
Y = "1 "
X = Y would be true

The "problem" I believe is that the runtime engine tries for a numeric match first, 
rather than strict character match.

So the "trick" is to force a character match, so

X:"X" = Y:"X" is false, which is what you "really" want I think



Ross Ferris
Stamina Software
Visage  an Evolution in Software Development


>-----Original Message-----
>From: [EMAIL PROTECTED] [mailto:owner-u2-
>[EMAIL PROTECTED] On Behalf Of Lindsay Gregory
>Sent: Thursday, 13 May 2004 12:29 PM
>To: [EMAIL PROTECTED]
>Subject: [U2] SPACE character ignored
>
>Hi
>
>
>
>I am sure this has been covered somewhere before but I could not find under
>oliver.com or u2ug.org.
>
>
>
>Can anyone tell me why: if
>
>
>
>X = '1'
>
>Y = '1 '    ;* note the trailing space character
>
>
>
>X = Y is true
>
>
>
>But if
>
>
>
>X = ''
>
>Y = ' '
>
>X = Y is false
>
>
>
>So adding a space character to a text string is ignored but adding a space
>character to a null is not ignored.
>
>I can understand why X = ''  is not equal to Y = ' ' because X is not a
>text
>string but if in the first case - the text string Y has two characters - a
>one and a space and the text string X has only one character how can they
>be
>equivalent.
>
>
>
>My guess is that by the time they come to be tested for equivalence they
>have been converted to numeric values and are being tested for numeric
>equivalence.
>
>
>
>How can I force them to be tested for TXT equivalence?
>
>
>
>Do I have to say:
>
>
>
>IF X = Y AND LEN(X) = LEN(Y) THEN true
>
>
>
>Except  if X = ' 1' and Y = '1 '  then LEN(X) = LEN(Y) ...
>
>
>
>It seems that leading or trailing spaces are ignored (but not embedded
>spaces) so I have to say
>
>
>
>IF LEN(X) = LEN(TRIM(X)) AND LEN(Y) = LEN(TRIM(Y)) AND X = Y THEN true
>
>
>
>There has got to be an easier way !!
>
>
>
>Lindsay Gregory
>
>Axys (NZ) Ltd
>
>www.axys.co.nz
>
>Ph: +64 9 5200264
>
>Fax +64 9 5200266
>
>Mob +64 21 770 462
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