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Top Forums Shell Programming and Scripting In Sed how can I replace starting from the 7th character to the 15th character. Post 302455851 by alister on Wednesday 22nd of September 2010 01:50:21 PM
Old 09-22-2010
One more (minor variation of Franklin52's second option):
Code:
sed "s/\(.\{6\}\).\{9\}/\1$VAR1/"

Regards,
Alister
This User Gave Thanks to alister For This Post:
 

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REPLACE(1)						      General Commands Manual							REPLACE(1)

NAME
replace - A utility program that is used by msql2mysql, but that has more general applicability as well. replace changes strings in place in files or on the standard input. Uses a finite state machine to match longer strings first. Can be used to swap strings. USAGE
replace [-?svIV] from to from to ... -- [files] or replace [-?svIV] from to from to ... < fromfile > tofile SYNOPSIS
replace [-?|-I] [-s] [-v] DESCRIPTION
replace -?|-I info -s silent -v verbose EXTRA INFO
Special characters in from string: ^ Match start of line. $ Match end of line.  Match space-character, start of line or end of line. For a end  the next replace starts locking at the end space-character. A  alone in a string matches only a space-character. EXAMPLE
this command swaps a and b in the given files: shell> replace a b b a -- file1 file2 ... SEE ALSO
isamchk (1), isamlog (1), mysqlaccess (1), mysqladmin (1), mysqlbug (1), mysqld (1), mysqldump (1), mysqlshow (1), msql2mysql (1), perror (1), replace (1), safe_mysqld (1), which1 (1), zap (1), AUTHOR
Ver 1.0, distribution 3.23.29a Michael (Monty) Widenius (monty@tcx.se), TCX Datakonsult AB (http://www.tcx.se). This software comes with no warranty. Manual page by L. (Kill-9) Pedersen (kill-9@kill-9.dk), Mercurmedia Data Model Architect / system developer (http://www.mer- curmedia.com) 20 December 2000 REPLACE(1)
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