Bash: grep pattern from command output

Question

I\'m really new with bash, but it\'s one of the subjects on school. One of the exercises was:

Give the line number of the file \"/etc/passwd\" where the informat

Answer 1

cat /etc/passwd -n | grep `whoami` | cut -f1 

Surrounding a command in ` marks makes it execute the command and send the output into the command it's wrapped in.

Answer 2

Check command substitution in the bash man page.

You can you back ticks `` or $() , and personally I prefer the latter.

So for your question:

grep -n -e $(whoami) /etc/passwd | cut -f1 -d :

will substitute the output of whoami as the argument for the -e flag of the grep command and the output of the whole command will be line number in /etc/passwd of the running user.

Answer 3

You can do this with a single awk invocation:

awk -v me=$(whoami) -F: '$1==me{print NR}' /etc/passwd

In more detail:

• the -v creates an awk variable called me and populates it with your user name.
• the -F sets the field separator to : as befits the password file.
• the $1==me only selects lines where the first field matches your user name.
• the print outputs the record number (line).