Hello All,

#!/depot/local/bin/perl5.8 -w

my $stat;
$stat=`rsh linx01 'ls ;echo \$status'`;
print $stat;


$stat variable contains the ls command returned output (i mean current directory ls execution output present in the $stat). $stat is being initialized from "echo $status".

I need only the execution status not the ls command output in the $status shell variable.

can anyone help me to get execution status of the ls command in the $status variable and ls command output should display in the screen.

Thanks
dayalan

Interesting that your title references $?, but your script doesn't.

Use:

$stat=`rsh linx01 'ls >/dev/null;echo $?'`;

I'm not sure what your overall point is. But if you don't want the output of ls, then route it to /dev/null; and, if you what the status of the command, it is $?, not $status. Also, with the single quotes, you don't want to escape the $.

security note: consider using ssh instead of rsh.

Hi,

i need "ls" command output should display in STDOUT. the execution status should return in to $? or $status. now the issue is "ls" command output and return status both are appended in $? or $status.

how can i extract the execution status from $? or $status?

Please let me know.

I think (if I understand you correctly) you want something like:

Please go through the mail communication, have a look at that perl script. echo $status returns exectuion STDOUT as well as command execution status. i wanted to extract the return status from $status variable.

please let me know.

OK, you're getting tangled up because you are running a shell command from inside of a perl script and you want to capture both the output and the status from the shell back to perl. Even within a pure shell script that can sometimes get complicated.

Just to give a somewhat messy example, I have a script in which I want to run a ufsdump and capture the status code from that. But I want to pipe the output of the ufsdump through ssh as a different user to a tape drive on a remote machine. Just a bit messy. I'm going to simplify it by pulling a lot of variables and options that were in my original.

( ufsdump <options-etc>; STATUS=$?; echo $STATUS > $FILE ) | su - backuser \
-c "ssh $TAPESERVER \"dd obs=64b of=/dev/rmt/0n\"";

Then I can do

STATUS=`cat $FILE`;

So, basically, I put the ufsdump into a subprocess within which I captured the status code. The standard out of that entire subprocess was piped to the su, which issued the ssh and ran it through to dd on the remote server to write the tape.

In the next step in the script I could then test the success of the ufsdump.

In your case you have (quoting exactly without subsequent comments or suggestons):

$stat=`rsh linx01 'ls ;echo \$status'`;
print $stat;

You can change this to send the $? somewhere that you can pick it up from, or you can change it to send the ls output somewhere that you can pick it up from.

You might decide that having the ls output go to a file to allow further manipulation might be the way to go. Then you could do something like:

$status=`rsh linx01 'ls > $file; echo $?'`;

following which you can immediately use the return value of $status to decide what your program does next. If it was successful, it can process the output of the ls command from the file -- copying it out, or whatever you want.