On 6/6/05, Nischi wrote:
> This works in ASCII but my question is why is it not working in EBCIDIC. why
> $1 is not having \xa0\xa0\xa0.
As I said, I don't know. I don't have access to such a system. (It's
called EBCDIC, by the way - no "I" between the "C" and "D")
Maybe it's your Perl version.
On 6/6/05, Nischi wrote:
> Hello
> i am having something like this ...
>
> $b = "\xa0\xa0\xa0\x{100}" =~ /(\xa0+)/;
Try not to use "$b" (or "$a") as variable names. "$a" and "$b" are the
special Perl sort variables and using them can cause strange bugs, for
example because even under "use strict"
Hello
i am having something like this ...
$b = "\xa0\xa0\xa0\x{100}" =~ /(\xa0+)/;
$b =~ \(\xa0+)\;
print "Valid" if ($1 eq "\xa0\xa0\xa0";
print " $b $1";
Here according to the regular expression the $1 should contain the matching
string. so according $1 should have \xa0\xa0\xa0. This is happ
ay, June 02, 2005 10:16 PM
Subject: problem with $1
> Hello
>
> The matched string which is stored in $1 is getting corrupted.
>
> the regular expression i have is
>
> "\xa0\xa0\xa0\x{100}" =~ /(\xa0+)/;
>
> when i say ... if($1 eq "\xa0\xa0\xa0") this
Hello
The matched string which is stored in $1 is getting corrupted.
the regular expression i have is
"\xa0\xa0\xa0\x{100}" =~ /(\xa0+)/;
when i say ... if($1 eq "\xa0\xa0\xa0") this is true. It is working fine in
Linux but not working (ASCII platform) but it is not working in EBCIDIC
pl
