Remove everything up to the last numbers of a string w/ sed or awk

OutThere · 04-23-2009, 04:54 PM

I know that the expression for the last numbers in a line is "[0-9]*$" if that makes any more sense. Ive got somthing that looks like "hey1234" and I need "1234" to be output. Of course, it could also read "hey123", or "hey there 12345" It could even read "hey123 there1234" and I would still need "1234", so I cant just remove all the numbers. Thanks.

custangro · 04-23-2009, 05:17 PM

Quote:

Originally Posted by OutThere

I know that the expression for the last numbers in a line is "[0-9]*$" if that makes any more sense. Ive got somthing that looks like "hey1234" and I need "1234" to be output. Of course, it could also read "hey123", or "hey there 12345" It could even read "hey123 there1234" and I would still need "1234", so I cant just remove all the numbers. Thanks.

Try tr....

Code:

root@host:~<1>$ echo "hey12345" | tr -d '[:alpha:]'
12345
root@host:~<2>$ echo "hey12345 there" | tr -d '[:alpha:]'
12345 
root@host:~<3>$ echo "heythere12345 " | tr -d '[:alpha:]'
12345 
root@host:~<4>$ echo "hey there12345 " | tr -d '[:alpha:]'
 12345

arckane · 04-23-2009, 05:21 PM

echo "hey12345" | tr -d [:alpha:] [:space:]
12345

Just to be tidy

OutThere · 04-23-2009, 05:26 PM

yeesh- sorry- I should have mentioned that I also have non-alphanumeric characters. I'm actually going through Apache logs that have the miliseconds that the request took at the end of each line.

Kenhelm · 04-23-2009, 07:01 PM

Try

Code:

echo "hey123 there1234" | grep -o '[0-9]*$'
1234

string="hey123 there1234"
echo ${string##*[^0-9]}     # bash parameter substitution
1234