[SOLVED] awk

freeroute · 09-16-2017, 11:20 AM

I would like to print lines that contains only digits and the length is equal 5.

I can do that with sed with this command:

Code:

sed -E -n '/^[0-9]{5}$/p' /usr/share/wordlists/rockyou.txt | less

This awk command print lines, length only 5 chars:

Code:

awk 'length($0) ==5' /usr/share/wordlists/rockyou.txt | less

How could I print lines with length only 5 chars and contains only digits?

Thanks your help.

Turbocapitalist · 09-16-2017, 11:30 AM

You'd do it about the same way.

If you want the whole line to be just five digits, no more, no less, and nothing else, then:

Code:

awk '/^[0-9]{5}$/' /usr/share/wordlists/rockyou.txt

If you want the same but limited only to the first column, then:

Code:

awk '$1 ~ /^[0-9]{5}$/' /usr/share/wordlists/rockyou.txt

See "man 7 regex". Of course you might also use [[:digit:]] instead of [0-9] but that depends on your locale a little.

freeroute · 09-16-2017, 11:45 AM

Thank you.

freeroute · 09-16-2017, 12:11 PM

I tried this command, also working.

Quote:

awk '/^[0-9]{5}$/ {print $0}' /usr/share/wordlists/rockyou.txt | less

The print command optional in this case?
I just learning awk and lots of time I use "print" command

Quote:

{print $0}

....

Turbocapitalist · 09-16-2017, 12:16 PM

Quote:

Originally Posted by freeroute

The print command optional in this case?

Yes. There are some shortcuts in awk. If you have nothing after a selector then a { print; } is implied. And { print; } is short for { print $0; }.

Take a peek at the awk manual pages periodically. It's a reference not a tutorial and looking up things in it will make the language more familiar.

Code:

man awk

Scroll down to the section on "PATTERNS AND ACTIONS"

MadeInGermany · 09-18-2017, 01:31 AM

and of course you can use the logical AND and NOT operators

Code:

awk 'length==5 && ! /[^0-9]/'

Length is 5 and there is not a character that is not 0-9.
Even some builtin functions have a default of $0 (analogue to $_ in perl).
The long form is

Code:

awk 'length($0)==5 && !($0~/[^0-9]/) {print $0}'

freeroute · 09-18-2017, 02:55 AM

Thanks for the another solution...