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Shell Programming and Scripting
How can I ignore only the lines which have # at the begining?
Post 97399 by vino on Monday 30th of January 2006 02:35:55 AM
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Is there a standard way to make a shell script read a file, or list, and skip each line that contains # at the begining, or ignores the content starting after a # in line?
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I am trying to print those line which has no # in the begining of the line.
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I have a shell script, which reads a *.txt file - line by line. In this text file, I have some lines beginning with "#" that I want to ignore :
MY_FILE
#blah blah blah 1
blah blah blah 2
blah blah blah 3
#blah blah blah 4
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now i want to append $var value in the begining of all the lines present in the file. Can u please suggest?
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LEARN ABOUT ULTRIX
zgrep
ZGREP(1) General Commands Manual ZGREP(1) NAME
zgrep - search possibly compressed files for a regular expression SYNOPSIS
zgrep [ grep_options ] [ -e ] pattern filename... DESCRIPTION
Zgrep invokes grep on compressed or gzipped files. These grep options will cause zgrep to terminate with an error code: (-[drRzZ]|--di*|--exc*|--inc*|--rec*|--nu*). All other options specified are passed directly to grep. If no file is specified, then the standard input is decompressed if necessary and fed to grep. Otherwise the given files are uncompressed if necessary and fed to grep. If the GREP environment variable is set, zgrep uses it as the grep program to be invoked. EXIT CODE
2 - An option that is not supported was specified. AUTHOR
Charles Levert (charles@comm.polymtl.ca) SEE ALSO
grep(1), gzexe(1), gzip(1), zdiff(1), zforce(1), zmore(1), znew(1) ZGREP(1)