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Top Forums Shell Programming and Scripting Capturing shell script command output Post 100734 by designflaw on Wednesday 1st of March 2006 02:53:27 PM
Old 03-01-2006
Network Capturing shell script command output

I am trying to check to see if a file exists on a ftp server, well, I know that cant be done, atleast directly, So I came up with this small script

Quote:
check_file
Code:
ftp -n $HOST <<END_SCRIPT
quote USER $USER
quote PASS $PASSWD
cd public_html/crap
dir $FILE
quit
END_SCRIPT

Where the $ variable have corresponding values, so if I run this on the promt as check_file > check.txt, I can then just check the file size for the check.txt file and know if the file exists or not. Well good deal, But I want to do the above check in the same script, for example, something like

Quote:
check_file_improved
Code:
{
ftp -n $HOST <<END_SCRIPT
quote USER $USER
quote PASS $PASSWD
cd public_html/crap
dir $FILE
quit
END_SCRIPT
} > $CHECK_FILE
if $CHECK_FILE = 0 {
  # file does not exist, lets upload the file to the server
  }
else {
  # file exists on the server, rename the file on server before uploading
  }

but somehow, I cannot capture any information into the $CHECK_FILE. Anyone ?
 

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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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