[SOLVED] Replacing the Last Digit

NetRock · 06-23-2010, 09:53 AM

Hi

i am trying to replace the last digit in the ip address(25) with 47 using following:

Quote:

echo 192.168.0.25|sed -r s/([0-9]*\.[0-9]*\.[0-9]*)/47/g'

but not able so far, was wondering if you can help, so i can find my mistake.

Thanks

syg00 · 06-23-2010, 09:57 AM

Seems similar to your previous thread.

NetRock · 06-23-2010, 10:06 AM

Hi

Quote:

Seems similar to your previous thread.

With all respect, Never i ask anything twice!!

Have a Nice day!!

Travis86 · 06-23-2010, 10:36 AM

echo "192.168.0.25" | sed "s/\.[0-9]*$/\.47/"

grail · 06-23-2010, 10:44 AM

Code:

echo "192.168.0.25" | awk -F. 'sub(/[0-9]+$/,"47")'

or sed:

Code:

echo "192.168.0.25" | sed -r 's/[0-9]+$/47/'

ghostdog74 · 06-23-2010, 11:16 AM

Quote:

Originally Posted by grail

Code:

echo "192.168.0.25" | awk -F. 'sub(/[0-9]+$/,"47")'

if there's already a field delimiter defined as ".", then just change field 4.

Code:

echo <ip> | awk -F"." '{$4="47"}1'  OFS="."

NetRock · 06-23-2010, 11:37 AM

Thank you so much for all your help.

since i was working under DOS (windows) needed to change two things:

Quote:

echo 192.168.0.25| sed -r 's/[0-9]+$/47/'

take the dbl quote out & NO SPACE after echo. Spent almost my whole morning to find this one out!!!

it was pretty nasty to find out this way!!

Thank you ALL ...