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Full Discussion: Passing a command in a variable
Top Forums UNIX for Dummies Questions & Answers Passing a command in a variable Post 302394073 by Corona688 on Wednesday 10th of February 2010 11:36:25 AM
Old 02-10-2010
---------- Post updated at 10:36 AM ---------- Previous update was at 10:26 AM ----------

[/COLOR]And I just realized one really brain-dead simple solution: Let the server do the parsing. I do this over ssh sometimes.

Code:
 vpass $account bash < commands

This also prevents people from inserting clever things in the 'commands' file that end up running locally.

[rest of the post]

Your method is a little naive in a couple ways. For one thing, it's a useless use of backticks:

Code:
# bad -- a Useless Use of Cat and Useless Use of Backticks in one.
# It not only makes an extra process and wastes memory,  but
# might prevent you reading all of a file >64KB since it won't
# fit in one shell variable.
for LINE in `cat file`
do
        ...
done

# better.  Reads one line at a time with no extra process at all.
# Works on files of any size.
while read LINE
do
        ...
done < file

Also, your xargs solution could be done with one fewer process like
Code:
xargs vpass < command

Basically, whenever you do 'cat foo' you can replace that with a shell redirection in nearly all circumstances. This is quite a bit more efficient.

As for why the command isn't doing what you expect, the shell doesn't substitue strings more than once -- it won't check inside things it substitutes for things like quoted strings or variables. So embedding strings and variables in a string won't work unless you explicitly tell the shell to re-evaluate it, with eval. Note that eval supports all valid syntax so someone could inject variables or local commands where you didn't expect them by escaping $ and so forth.
Code:
#!/bin/bash
STR="\"this is a string\""
STR2='$HOSTNAME'

function args
{
  echo "Has ${#} arguments"
}

# it LOOKS right...
echo $STR
# But it's being split into "\"this" "is" "a" "string"!
args $STR
# This also looks right...
eval echo $STR
# eval tells it to re-parse the line after substitution.
eval args $STR

# This will not substitute in variables.
echo $STR2
# This will!
eval echo "$STR2"

[COLOR="#738fbf"]
Last edited by Corona688; 02-10-2010 at 12:48 PM..
 

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LEARN ABOUT CENTOS

lppasswd

lppasswd(1)							    Apple Inc.							       lppasswd(1)

NAME

       lppasswd - add, change, or delete digest passwords.

SYNOPSIS

       lppasswd [ username ]
       lppasswd -a [ -g groupname ] username
       lppasswd -x username

DESCRIPTION

       lppasswd  adds, changes, or deletes passwords in the CUPS digest password file, passwd.md5. When run by a normal user, lppasswd will prompt
       for the old and new passwords.  When run by the super-user, lppasswd can add new accounts (-a username), change	existing  accounts  (user-
       name), or delete accounts (-x username) in the digest password file. Digest usernames do not have to match local UNIX usernames.

OPTIONS

       lppasswd supports the following options:

       -g groupname
	    Specifies a group other than the default system group.

SECURITY ISSUES

       By  default,  the lppasswd program is not installed to allow ordinary users to change their passwords. To enable this, the lppasswd command
       must be made setuid to root with the command:
       chmod u+s lppasswd

       While every attempt has been made to make lppasswd secure against exploits that could grant super-user privileges  to  unprivileged  users,
       paranoid system administrators may wish to use Basic authentication with accounts managed by PAM instead.

SEE ALSO

       lp(1), lpr(1),
       http://localhost:631/help

COPYRIGHT

       Copyright 2007-2013 by Apple Inc.

22 February 2008						       CUPS							       lppasswd(1)
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