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Top Forums Shell Programming and Scripting Find not finding stuff if run remotely Post 302895388 by bitlord on Monday 31st of March 2014 03:06:08 PM
Old 03-31-2014
Find not finding stuff if run remotely

Hello
I'm working on script to find tomcat on all my servers. Then find out what version of tomcat is installed. Basically I want to check and see if the latest version is installed. I'm testing the script on Solaris 10. I'm also going to need it to work on RHEL and SLES.

If I run the following find command remotely, It doesn't find the file. If I run it on the server locally it it finds this file: /usr/serena/vm/common/tomcat/bin/version.sh

Local command
Code:
sudo find / \( -name 10_Recommended* -o -name share \) -prune -o -type f -name version.sh -print 2>/dev/null


The script

Code:
for s in `cat sol`
do echo $s
ssh -q  $s sudo find / \( -name 10_Recommended* -o -name share \) -prune -o -type f -name version.sh -print 2>/dev/null
done

Can someone tell me why this is happening?
 

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

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 v5.16.3 2010-06-11 SHELL-QUOTE(1)
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