The following Perl one-liner will do what you need
perl -n -e 's/(.{8,8}).(.*)/$1$2/;print ;'
The first set of parentheses strips out the first 8 characters, the second set
strips out everything after the 9th character which is lost.
My test line was:-
(echo "1234567890";echo "abcdefghijkl")|perl -n -e 's/(.{8,8}).(.*)/$1$2/;print
;'
The two echo statements are in parentheses to keep them within the same
sub-shell.
I'm assuming all along you are running a *nix type system.
--
Andrew in Edinburgh,Scotland
On Tue, 5 Aug 2008, Chris Charley wrote:
----- Original Message ----- From: "Chris Charley" <[EMAIL PROTECTED]>
To: <beginners@perl.org>
Sent: Tuesday, August 05, 2008 10:45 PM
Subject: Re: question about text operation using regex.
----- Original Message ----- From: ""Remy Guo"" <[EMAIL PROTECTED]>
Newsgroups: perl.beginners
To: "Perl Beginners" <beginners@perl.org>
Sent: Tuesday, August 05, 2008 10:14 PM
Subject: question about text operation using regex.
hi all,
i have a txt log file and i want to delete the 9th character of each line.
how can i do it using regex?...thanks!
-Remy
Ooops. Sent it to the poster by mistake.
You could do it from the command line
perl -pe "8 < length && substr $_,8, 1, ''" logfile.txt > altered_file.txt
Chris
Just to clarify the substr function here. Following the 1, are 2 *single*
quotes. This assigns the empty dtring to the 9th position of the string,
effectively eliminating the 9th character.
Chris
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