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Full Discussion: Parsing file list in variable
Top Forums Shell Programming and Scripting Parsing file list in variable Post 302456502 by prasbala on Friday 24th of September 2010 11:26:09 AM
Old 09-24-2010
Parsing file list in variable
Hello,
somewhere in a shell script, i am storing the output of "ls" into a variable. My question is how can i parse this variable to get each filepath. I don't want to create a temporary file to write down all the filenames and then parse it..

is there a easy way out..

here is what i am doing

Code:
tempVar=`find {some files according to pattern} | grep "some pattern"`

So tempVar has a list of filepaths, i would like to parse this. May be this is simple, but since i am not so well versed with shell scripting, please help me.

rgds
Prash
 

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LEARN ABOUT LINUX

zegrep

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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