Script as login shell (passing args to login shell)

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# 1  
Old 03-31-2014
Script as login shell (passing args to login shell)

Hello all,

for security reasons my compagny imposes that my script be launch remotly via ssh under the users login shell.

So serverA launches the ssh command to serverB which has a local user with my script as a login shell.

Local script works like a charm on his own.

serverB$ grep user1 /etc/passwd

serverA$ ssh -t user1@serverB

Everything works fine until i bring up arguments

For testing purpose i've replaced the 3 first lines of my script with:
echo $#
echo $*
sleep 3

and i get this:
 ssh user1@serverB -- "-i -n WYWYOU"
-c -i -n WYWYOU

Which is not good cause the script needs at least 4 args to run:
# Check number of arguments
case $# in
         4|5) ;;
         *) usage; exit 1 ;;

Any way too make him pass more than 2 arguments?... and he's probably counting the -c as one.
# 2  
Old 03-31-2014
It would help if you show the script. How are you executing this? What is the command you send over to the server. Since you didn't supply the script I created that just outputs argument count, arguments...
echo $# 
echo $* 

I didn't have problems with argument counting. This works....
[josephgr@oc0887178221 ~]$ ssh josephgr@freezer1 ./  "-i -n WYWYOU"
josephgr@freezer1's password: 
-i -n WYWYOU

This User Gave Thanks to blackrageous For This Post:
# 3  
Old 03-31-2014
Thanks for the anwser but the real problem is the script is actually the remote user login shell which is in the home of the remote user.

- serverA ssh serverB with user1
- user1 login shell is
- user1's home on serverB contains which gets exec upond login.
- can't pass args too that exec via ssh

Keys are handle and scripts exec is handle too. Only params poses a problem. Hope it is clearer (been banging my head on that since Friday).
# 4  
Old 03-31-2014
What are the contents of If it's mangling arguments, it will have to be worked around, and one can hardly tell that without knowing what it is.
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# 5  
Old 03-31-2014
Originally Posted by maverick72
Any way too make him pass more than 2 arguments?... and he's probably counting the -c as one.
Yes, "-c" is one argument and the rest is the second. Your problem is not the passing of the parameters but the splitting of them.

You could emulate the splitting along IFS characters inside your shell this way:

#! /bin/ksh
IFS=" "
while : ; do
     print - "next: ${args%%${IFS}*}"
     if [ "$args" != "${args#*${IFS}}" ] ; then
          print - "rest: $args"
          print - "rest:"

When run it looks like this:

# ./split.ksh 'a     b c def g h'
next: a
rest:     b c def g h
rest:    b c def g h
rest:   b c def g h
rest:  b c def g h
rest: b c def g h
next: b
rest: c def g h
next: c
rest: def g h
next: def
rest: g h
next: g
rest: h
next: h

Notice that you will have to throw out empty "args" in the loop to compensate for multiple consecutive IFS chars.

I hope this helps.

This User Gave Thanks to bakunin For This Post:
# 6  
Old 03-31-2014
Ohhhh i get it now... Yeah well like you guys said the problem will be the parsing.

For the script it is a script that edit's a dns file and relaunches the process.

$ insert_host_ip.bash -i <ip> -n <name record> [-d] (to delete the record)

So basicly the args are "-i" "-n" "ip address" "name_record" "-D"

Nothing with spaces or weird chars but i do parse my options with a basic getops

# Parse the arguments
while getopts i:n:R OPTION; do
  case "$OPTION" in
        D) REMOVE=Y
        :) usage ; exit 5
        \?) usage ; exit 5

shift $((OPTIND-1))

# 7  
Old 03-31-2014
No, I mean -- what are the contents of The shell script that gets executed on login?
This User Gave Thanks to Corona688 For This Post:
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