STDOUT and STDERR redirection fails for forked process

[Peter]

Hi,

Read the manpages, googled around, found some samples and followed
those to get it to work... but still fails..

What do i want? Simply, fork a command using the OPEN instruction,
redirect STDERR to STDOUT and capture the returncode of the executed
command.
I wrote the code wich is printed below, which works fine. Only output
from STDERR is not redirected to STDOUT but printed to the terminal
and the returncode is not captured. (So somehow redirected to STDOUT
but outside the open statement).

e.g. when the routine executes 'ls -Q' which sends the errormessage to
STDERR and a RC=2, the output from STDERR is printed to the terminal
and the returncode ($?) is 0.

Any ideas? suggestions?

Peter

Jul19 10:01:40 INF Data received from client "05794A6E5CD1122F ls -Q"
Jul19 10:01:40 DBG subroutine loccmd (15) (ls -Q)
Jul19 10:01:40 DBG Issuing command 'ls -Q'
ls: Not a recognized flag: Q
Usage: ls [-1ACFLNRabcdefgilmnopqrstuxE] [File...]
Jul19 10:01:40 DBG leaving subroutine loccmd passing values '0' '' ''


eval {
local $SIG{ALRM} = sub { die "Timed Out!\n" } ;
# set the alarm
alarm($timeout);

## Issue the command by forking the process and monitoring the
output
if (open STATUS, "$cmdstr 2>&1 |") {
while (<STATUS>) {
$result .= $_ ;
}
$errmsg = $! ;
$rcverror = $? ;

close STATUS ;
}
else {
$errmsg = "Fork failed for \'$cmdstr\'" ;
$rcverror = -2 ;
}

# reset the alarm
alarm(0) ;
$rcvtimeout = 0 ;
} ;

 

[anno4000]

Hi,

Read the manpages, googled around, found some samples and followed
those to get it to work... but still fails..

What do i want? Simply, fork a command using the OPEN instruction,
redirect STDERR to STDOUT and capture the returncode of the executed
command.
I wrote the code wich is printed below, which works fine. Only output
from STDERR is not redirected to STDOUT but printed to the terminal
and the returncode is not captured. (So somehow redirected to STDOUT
but outside the open statement).

e.g. when the routine executes 'ls -Q' which sends the errormessage to
STDERR and a RC=2, the output from STDERR is printed to the terminal
and the returncode ($?) is 0.

Any ideas? suggestions?

Peter

Jul19 10:01:40 INF Data received from client "05794A6E5CD1122F ls -Q"
Jul19 10:01:40 DBG subroutine loccmd (15) (ls -Q)
Jul19 10:01:40 DBG Issuing command 'ls -Q'
ls: Not a recognized flag: Q
Usage: ls [-1ACFLNRabcdefgilmnopqrstuxE] [File...]
Jul19 10:01:40 DBG leaving subroutine loccmd passing values '0' '' ''

This is not the output of the code you showed. How can we help
you when you don't give us consistent information?

eval {
local $SIG{ALRM} = sub { die "Timed Out!\n" } ;
# set the alarm
alarm($timeout);

## Issue the command by forking the process and monitoring the
output
if (open STATUS, "$cmdstr 2>&1 |") {
while (<STATUS>) {
$result .= $_ ;
}
$errmsg = $! ;
$rcverror = $? ;

close STATUS ;
}
else {
$errmsg = "Fork failed for \'$cmdstr\'" ;
$rcverror = -2 ;
}

# reset the alarm
alarm(0) ;
$rcvtimeout = 0 ;
} ;

That code is a mess. It isn't strict-safe. Why is everything wrapped
in an eval block? What is the value of $timeout? Or is it $rcvtimeout?
You are accessing $? too early. The return code is only put there after
close().

Reducing your code to the essentials (something *you* should have done),
this is what remains:

my $cmdstr = 'ls -9'; # -Q is a valid flag in some systems
open STATUS, '-|', "$cmdstr 2>&1";
print ">>> $_" while <STATUS>;
close STATUS;
print "retcode: $?\n";
exit;

This prints:
retcode: 256

Note the ">>>" in front of the error message. It shows that the error
is not printed to stdout by the command but is caught and printed by
the perl program. The effect you complain about simply isn't there.

So what is this all about?

Anno