Training, Open Source computer languages
PerlPHPPythonMySQLApache / TomcatTclRubyJavaC and C++LinuxCSS 
Search for:
Home Accessibility Courses Diary The Mouth Forum Resources Site Map About Us Contact
For 2023 (and 2024 ...) - we are now fully retired from IT training.
We have made many, many friends over 25 years of teaching about Python, Tcl, Perl, PHP, Lua, Java, C and C++ - and MySQL, Linux and Solaris/SunOS too. Our training notes are now very much out of date, but due to upward compatability most of our examples remain operational and even relevant ad you are welcome to make us if them "as seen" and at your own risk.

Lisa and I (Graham) now live in what was our training centre in Melksham - happy to meet with former delegates here - but do check ahead before coming round. We are far from inactive - rather, enjoying the times that we are retired but still healthy enough in mind and body to be active!

I am also active in many other area and still look after a lot of web sites - you can find an index ((here))
RegEx in scalar

Posted by Akuma (Akuma), 1 August 2005
hi, everyone

I try to use replacment in PERL with the function s/$pattern1/$pattern2/gi;.
the two scalars are strings with regex pattern like :

$pattern1 = "(.*?)mysearch(.*)";
$pattern2 = "$1myreplacement$2";

When I run this script, the result is :


It's not what I want of course. It's the same thing this other special caracters like '\.' which is print '\.' and not '.'.

Help me please I don't know what to do.

The two pattern are read in a file.

Posted by admin (Graham Ellis), 1 August 2005
I think you have two  immediate problems in your sample code:

a) The running up of a variable name straight into a piece of text means that Perl sees it as a longer variable name.  In other words  $1myreplacement is seen as being a single name - not $1 followed by the text myreplacement.   This can be solved by writing ${1}myreplacement - curly braces can be used in Perl to limit the extent of a variable name

b) The $2 (and after item (a) has been corrected, $1 as well) are being interpretted at the time you assign to $pattern2, which is before they're set in the regular expression match.   If you set up $pattern2 to contain \1 and \2 for the backstring substitutions, then you should be able to get the results you want - but bear in mind that you'll need to code something like \\1 and \\2 in order to prevent the " operator acting on the \ character at the assignment.   You could use ' instead.

I'm worried that this sounds like a complicated answer and I think you're trying to do something very simple.    Why not just write:

$pattern1 = "Dover";
$replaceby = "Folkestone";
$ourjourney =~ s/$pattern1/$replaceby/gi;

Regular expressions look for a pattern *within* the source string, so the whole .* business at the start and end is usually un-necessary.

Note also that the replacement string (the output) is NOT a regular expression ... your use of the variable name $pattern2 in the original example was misleading.

This page is a thread posted to the opentalk forum at and archived here for reference. To jump to the archive index please follow this link.

You can Add a comment or ranking to this page

© WELL HOUSE CONSULTANTS LTD., 2024: Well House Manor • 48 Spa Road • Melksham, Wiltshire • United Kingdom • SN12 7NY
PH: 01144 1225 708225 • FAX: 01144 1225 793803 • EMAIL: • WEB: • SKYPE: wellho