On Jan 2, 2008 10:48 AM, Chas. Owens <[EMAIL PROTECTED]> wrote:
> On Jan 1, 2008 4:20 PM, <[EMAIL PROTECTED]> wrote:
> > Hi there,
> >
> > I'm quite new to perl, and now having problem with using parentheses
> > in translation strings. For example, I want to replace the 4 digit
> > year code 'yyyy' with '(\d\d\d\d)', so it can be used for grouping in
> > string matching later. I expected the code would be like:
> >
> > $org_str =~ tr/yyyy/\(\d\d\d\d\)/;
> >
> > Then the result is '(((('. It seems the parens were not escaped
> > properly, shouldn't the backslash escape the round brackets? If I
> > remove the backslashes before brackets, the result is the same. If I
> > use double backslashes, that is, \\(\d\d\d\d\\), the result is '\\\\'.
> > Can someone inform the right way to do this? Many thanks in
> > advance! :-)
> snip
>
> You want substitute (s///) not transliterate (tr///). The s///
> operator takes a regex in the first part and replaces what martches in
> the target string it with the string in the second part. The tr///
> operator replaces the any character in the first argument with the
> corresponding character from the second argument in the target string.
> So you should be using
>
> $org_str =~ s/yyyy/(\\d\\d\\d\\d)/;
>
I just thought of something. If it is your intent to use $org_str
later in a regex you probably want the replacement string to be
([0-9]{4}) not (\\d\\d\\d\\d). The \d character class matches any
numeric character. This includes characters like U+1811 (the
Mongolian digit one). If you want only the characters 0 through 9,
then you need to make that character class yourself: [0-9]. The {4}
modifier say that there must be 4 of the last pattern (so it is that
same thing as [0-9][0-9][0-9][0-9]).
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