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Old 09-16-2017, 11:20 AM   #1
freeroute
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awk - print lines contains only digits


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.
 
Old 09-16-2017, 11:30 AM   #2
Turbocapitalist
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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.
 
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Old 09-16-2017, 11:45 AM   #3
freeroute
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Thank you.
 
Old 09-16-2017, 12:11 PM   #4
freeroute
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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}
....
 
Old 09-16-2017, 12:16 PM   #5
Turbocapitalist
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Quote:
Originally Posted by freeroute View Post
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"
 
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Old 09-18-2017, 01:31 AM   #6
MadeInGermany
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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}'

Last edited by MadeInGermany; 09-18-2017 at 04:04 AM. Reason: ! has higher precedents than ~ brackets added
 
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Old 09-18-2017, 02:55 AM   #7
freeroute
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Thanks for the another solution...
 
  


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