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Top Forums Shell Programming and Scripting Assign command (with pipe) output to a variable Post 302372597 by cfajohnson on Wednesday 18th of November 2009 10:05:36 AM
Old 11-18-2009
Quote:
Originally Posted by jeff_cen
Hi ,

I would like to assign command (with pipe) output to a variable. The code is as follows. The goal of the code is to get the last folder folder with a particular name pattern.

myDate=`ls | grep 2009 | tail -1`
echo "myDate=" $myDate

However, in the presence of the pipe, the code doesn't work.

What does "doesn't work" mean?
What does happen?


You don't need ls or grep or tail:

Code:
set -- *2009*
shift $(( $# - 1 ))
myDate=$1

 

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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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