How can I get exit status from a called shell.

dthacker · 06-26-2008, 04:37 PM

I'm working on some really old boxes with no bash, perl, or other scripting language. The task is to ftp several files to a long set of machines.

I've set this up as two shells. ftp-send.sh builds the list and calls ftpit.sh, which does the actual ftp command. Here's ftp-send:

Code:

RUNDATE=`date +%Y%m%d%H%M`
echo $RUNDATE
LOGFILE=ftp-send$RUNDATE.log
#start the logfile
echo "ftp-send-update started at `date`">>$LOGFILE

DESTLIST=ftp-destination-list.txt
for i in `cat $DESTLIST`
    do
    echo "Now sending to $i" >> $LOGFILE
    if /upgrade_work/ftpit.sh $i; then
        echo "Transfer to $i successful"
    else
        echo "Transfer to $i failed"
    fi
    done

and here's ftpit.sh

Code:

ftp -n $1 <<EOF
    user root foopass
    binary
    put foo1.txt
    bye
    EOF
print $?

The problem I'm having is that ftpit.sh is always returning zero. I've tried putting ftpit into the main shell as a function, but that doesn't work either. Can someone recommend a way to catch and pass a non-zero exit code when ftpit isn't able to connect?

TIA Dave

jailbait · 06-26-2008, 04:51 PM

Quote:

Originally Posted by dthacker

Can someone recommend a way to catch and pass a non-zero exit code when ftpit isn't able to connect?

The exit code is passed by the exit command or equivalent. In ftp.sh you are taking the default exit which is exit 0. So you need to have exit pass the return code returned from the ftp command. Without testing this is how I would change your script:

Code:

ftprc=ftp -n $1 <<EOF
    user root foopass
    binary
    put foo1.txt
    bye
    EOF
print $?
exit ftprc

------------------
Steve Stites

chrism01 · 06-26-2008, 06:23 PM

The default exit status of a script is the status of the last cmd executed, so you don't need anything except the ftp cmd in the 2nd script.
In your case you are receiving the exit status of the 'print' cmd, which always works, so is always zero.
If you really wanted to be explicit about it you could say

exit $?

but really that's redundant, as I said.

AnanthaP · 06-27-2008, 08:06 AM

I think that the exit code is to be trapped by the calling program.

ie. you should replace

Code:

if /upgrade_work/ftpit.sh $i; then

with

/upgrade_work/ftpit.sh $i
if [ $? -eq 0 ]; then

... and so on

dthacker · 06-27-2008, 09:51 AM

Quote:

Originally Posted by jailbait

The exit code is passed by the exit command or equivalent. In ftp.sh you are taking the default exit which is exit 0. So you need to have exit pass the return code returned from the ftp command. Without testing this is how I would change your script:

Code:

ftprc=ftp -n $1 <<EOF
    user root foopass
    binary
    put foo1.txt
    bye
    EOF
print $?
exit ftprc

------------------
Steve Stites

Putting the ftp command into a shell variable seems to prevent the -n argument from getting read properly.

Code:

# ./ftp-send-update.sh
200806270944
/upgrade_work/ftpit.sh: -n:  not found
Transfer to 10.27.48.9 failed
/upgrade_work/ftpit.sh: -n:  not found
Transfer to 172.16.110.39 failed
/upgrade_work/ftpit.sh: -n:  not found
Transfer to 172.16.110.37 failed

I've tried a couple of different quoting methods on that first line, but I haven't found anything that works. Thanks for the suggestion though. I'm continuing to experiment with it.

Dave