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Top Forums Shell Programming and Scripting Reading in all files from parent directory (GAWK) Post 302539570 by gd9629 on Monday 18th of July 2011 06:33:59 AM
Old 07-18-2011
Reading in all files from parent directory (GAWK)

Hi all,

I'm very, very new to scripting (let alone SHELL) and was wondering if anyone could help me out as I seem to be in a spot of bother.

I collect data (.dat files) which are automatically seperated into several sub directories, so the file paths I'm reading in at the moment would be something like

'/u/Picarro/DataLog/2011/june/18/CF*nc.dat'

where I use "CF*nc.dat" to read in all files from that folder (18th June in this case). Now this is problematic as I need to read in all files from a whole year and combine them, i.e combine all the files into one .csv files for processing.

Is there any way to read in ALL *.dat files in a parent directory e.g. read in all the files for june

'/u/Picarro/DataLog/2011/june'

without having to refer to the date?

Thanks! Smilie
 

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SHELL-QUOTE(1p) 					User Contributed Perl Documentation					   SHELL-QUOTE(1p)

NAME
shell-quote - quote arguments for safe use, unmodified in a shell command SYNOPSIS
shell-quote [switch]... arg... DESCRIPTION
shell-quote lets you pass arbitrary strings through the shell so that they won't be changed by the shell. This lets you process commands or files with embedded white space or shell globbing characters safely. Here are a few examples. EXAMPLES
ssh preserving args When running a remote command with ssh, ssh doesn't preserve the separate arguments it receives. It just joins them with spaces and passes them to "$SHELL -c". This doesn't work as intended: ssh host touch 'hi there' # fails It creates 2 files, hi and there. Instead, do this: cmd=`shell-quote touch 'hi there'` ssh host "$cmd" This gives you just 1 file, hi there. process find output It's not ordinarily possible to process an arbitrary list of files output by find with a shell script. Anything you put in $IFS to split up the output could legitimately be in a file's name. Here's how you can do it using shell-quote: eval set -- `find -type f -print0 | xargs -0 shell-quote --` debug shell scripts shell-quote is better than echo for debugging shell scripts. debug() { [ -z "$debug" ] || shell-quote "debug:" "$@" } With echo you can't tell the difference between "debug 'foo bar'" and "debug foo bar", but with shell-quote you can. save a command for later shell-quote can be used to build up a shell command to run later. Say you want the user to be able to give you switches for a command you're going to run. If you don't want the switches to be re-evaluated by the shell (which is usually a good idea, else there are things the user can't pass through), you can do something like this: user_switches= while [ $# != 0 ] do case x$1 in x--pass-through) [ $# -gt 1 ] || die "need an argument for $1" user_switches="$user_switches "`shell-quote -- "$2"` shift;; # process other switches esac shift done # later eval "shell-quote some-command $user_switches my args" OPTIONS
--debug Turn debugging on. --help Show the usage message and die. --version Show the version number and exit. AVAILABILITY
The code is licensed under the GNU GPL. Check http://www.argon.org/~roderick/ or CPAN for updated versions. AUTHOR
Roderick Schertler <roderick@argon.org> perl v5.8.4 2005-05-03 SHELL-QUOTE(1p)
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