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Top Forums UNIX for Dummies Questions & Answers Using a list menu as a variable Post 302277075 by Great Uncle Kip on Thursday 15th of January 2009 10:35:44 AM
Old 01-15-2009
Using a list menu as a variable

Hi again

I have the follwing - cat ~/ABCFILE | grep "$SYSTEM" | grep "$USERNAME"

What I'm looking to do is have the variable for $SYSTEM determined by the user making a selection from a numbered list. So, input 1 would be system ABC etc.

I'm very puzzled as to how to go about this?

Any advice at all would be hugely appreciated

Regards
Kip
 

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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)
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