Hide the output of a shell command only on success?

Solution 1:

It should be easy enough to write a script for this purpose.

Something like this completely untested script.

OUTPUT=`tempfile`
program_we_want_to_capture &2>1 > $OUTPUT
[ $? -ne 0 ]; then
    cat $OUTPUT
    exit 1
fi
rm $OUTPUT

On the other hand for commands I run as part of a script I usually want something better than simply print all the output. I often limit what I see to the unknown. Here is a script I adapted from something I read over a decade ago.

#!/bin/bash

the_command 2>&1 | awk '
BEGIN \
{
  # Initialize our error-detection flag.
  ErrorDetected = 0
}
# Following are regex that will simply skip all lines
# which are good and we never want to see
/ Added UserList source/ || \
/ Added User/ || \
/ init domainlist / || \
/ init iplist / || \
/ init urllist / || \
/ loading dbfile / || \
/^$/ {next} # Uninteresting message.  Skip it.

# Following are lines that we good and we always want to see
/ INFO: ready for requests / \
{
  print "  " $0 # Expected message we want to see.
  next
}

# any remaining lines are unexpected, and probably error messages.  These will be printed out and highlighted.
{
  print "->" $0 # Unexpected message.  Print it
  ErrorDetected=1
}

END \
{
  if (ErrorDetected == 1) {
    print "Unexpected messages (\"->\") detected in execution."
    exit 2
  }
}
'
exit $?

Solution 2:

I don't think there is a clean way of doing this, the only thing I can think of is

• Capture the output of the command.
• Check the return value of the command and if it failed
  • display the captured output.

Implementing this might though be a interesting project but perhaps beyond Q&A.

---

Solution 3:

I'd setup a bash function like this:

function suppress { /bin/rm --force /tmp/suppress.out 2> /dev/null; ${1+"$@"} > /tmp/suppress.out 2>&1 || cat /tmp/suppress.out; /bin/rm /tmp/suppress.out; }

Then, you could just run the command:

suppress foo -a bar

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Solution 4:

Try so:

out=`command args...` || echo $out