I need help creating a script that will print user info. the script has to have: - named "printpasswd" - it must take 1 arguement from the command line. the argument will point to a file that follows the /etc/passwd format - it will parse the file and output each user's information in the following format: user-name,uid,gid,home,shell and notes

What do you have so far? This looks like homework, but I'll give you a hint. This loop reads the /etc/passwd file and parses the seven fields: #!/bin/ksh while IFS=":" read f1 f2 f3 f4 f5 f6 f7 do echo "$f1" echo "$f2" echo "$f3" echo "$f4" echo "$f5" echo "$f6" echo "$f7" done < /etc/passwd

Here's what I have so far, but I don't know if its correct. #!/bin/sh if [ $# != 1 ]; then echo "Usage: printpasswd [file]" exit fi cat $1 | awk '/^[^#]/{print $0}' | while read Line do IFS=":" read UserName Bleh Uid Gid Bleh2 Home Shell printf "$UserName, $Uid, $Gid, $Home, $Shell\n" done

You can do most of it with one line: awk -F: '!/^ *#/ {print $1,$3,$4,$6,$7}' testpasswd although if this is a shell exercise it won't be what they want.