Shell Programming and Scripting

It would have worked if you had done this:

awk '/'$(hostname)'/{print $1}' /etc/hosts

or ...

Code:

awk "/$(hostname)/ {print \$1}" /etc/hosts

... awk needs the single quotes normally but you used the double quotes which made $1 susceptible to the shell substitution ... you needed to escape $1 to get it to work with the double quotes you needed to get $(hostname) ...

Thanks ..
That Was A Good Explanation


Esham

I have a doubt in awk, the below two commands displays only directories available in the current directory, leaving regular & hidden files.

Code:

  ls -l | awk '$0~/^d/ {print $9}'
  ls -l | awk '$1~/^d/ {print $9}' 

I know $0 displays the entire record and $1 displays the very first field in the record. Though I would like to know what is the significance of $0 and $1 here and how it differs.

you should create a new thread. anyway..
The regular expression ^ says to match the letter 'd' as the first character of the line(record). Because of this, using $0 and $1 would get the same result. However if you remove the ^, and it becomes like this : $0 ~ /d/ and $1 ~ /d/, they would have different meaning. $0 ~ /d/ would match when d is anywhere in the record. $1 ~ /d/ would match only when the first field contains 'd'.