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Top Forums UNIX for Dummies Questions & Answers Du command and modified date in ssh Post 302929440 by junior-helper on Sunday 21st of December 2014 02:38:48 PM
Old 12-21-2014
Quote:
What should be modified to get this corrected?
Not sure if I understand correctly, but try
Code:
<(ls -lh | sed 1d)

The sed command will remove the first line of the output produced by ls -lh

---------- Post updated at 09:38 PM ---------- Previous update was at 08:09 PM ----------

Quote:
Originally Posted by Newuzer
The best way is your previous method. I just need help on the justification part now for the date/time and name columns. - thanks
Try this:
Code:
paste -d' ' \
 <(du -hs * | awk -F'\t' '{printf "%8s\n", $1}') \
 <(ls -lh | awk 'NR>1{for (i=6;i<=NF;i++)
    if ( ((i==7) && (length($i)==1)) || ((i==8) && (length($i)==4)) ) printf " %s ", $i; else printf "%s ", $i;
    print ""}')

The long if line means following: if the field 7 (day of month) is a single digit (field length = 1) OR the field 8 (time) is a year (field length = 4), "space-pad" the particular value (first print the space, then the value + a trailing space), else print it "as is" + add a trailing space.
 

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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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