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I've asked this question on many other forums and no one seems to know, so that's why I posted in the experts forume. So I just got a new sip phone. It is connected via usb to box A in my bedroom....

You can try something like this:

awk 'BEGIN {FS="[, /]"} {gsub (/"/, "")} {FS="/" ; print $6"/"$5"/"$4}' file

I was attempting to upgrade SSL on my netbsd machine and after a 'make install' I was logged out(I was logged in via ssh), and when I attempted to log back in(via ssh) I was given this message:...

Thank you.

Yes you're correct, sorry I must've pasted the first bit of text after I removed the excess whitespace with sed. Anyway, would you mind explaining your solution.

It does get it all on one line, but results look like this:

1600-1700 NA U. S. A. WYFR 11565, 11830, 13695,, 17760...

They're shortwave radio broadcasts.

They should look something like this:

UTC Time Notes Country Station Frequencies
0000-0015 Am, Su U. S. A. WRMI ...

I have some text:


0400-0427 NA Czech Republic R. Prague 5990ca, 6200, 7345
0400-0456 NA, As Romania R. Romania Int'l 6115, 9515, 9690,
11895
0400-0500 NA ...

Thank you for the replies. I believe I've got it now.

Hrm, I'm not quite sure I understand. If I just put the value returned from st_mode(33152)through like this:

%04o\n", statbuf.st_mode

I get 100600. If it's not too much trouble could you...

I'm new to C, and I'm attempting to write a script similar to the stat command for practice. I only had a problem converting st_mode to an octal permission format. I remembered I had littleutils...

You're welcome.

Try this:

sed "s/=.*/=$(date +%F)/g"

Assumeing your new password file is formated with one password per line, something like this could work:


#!/bin/bash -
#

i=1
lines=$(cat passfile | wc -l)

while [ $i -le $lines ]...

I'm just wondering if there's a simple way (that could be used in a shell script) to tell if the sound card is in use (as in audio is being played)?

I'm sure someone will be able to get your desired output into a single field, but here it is in two:

awk 'BEGIN { RS = "/^~/" ; FS = "\n" ; OFS = "~" } {gsub(/ +/, "") ; print $1,$2 }'

Could someone explain this? I understand that "/^[^ ]/ " grabs all the usernames. and I think I understand the ternary operator; found equals true (1) if name (the username) is found in $1, or found...

Could someone explain this? I understand that "/^[^ ]/ " grabs all the usernames. and I think I understand the ternary operator; found equals true (1) if name (the username) is found in $1, or found...

Well, either of the two commands I gave will will work on the entire file. I don't see any reason why you'd need a loop just to print your chosen field.

Lots of ways:

awk -F":" '{print $1}' file

cut -d':' -f1 file

Just pick the field you'd like to print ( i.e. $1, $2, or -f1 -f2).

You're welcome.

This seems to work:


#!/bin/bash -

if [ $# -lt 1 ]; then
echo "Usage: `basename $0` <user>"
exit
fi

USER=$1
LOGINS=`who | grep $USER | wc -l`
TIMES=`who | awk '{print...

With bash version 3.0 or higher you can use
${BASH_ARGV[0]}

This little explanation is of courtesy Dark_Helmet over at linxquestions.org