Grep/Awk functionalty request

[Dave Walley]

Hi.

I am looking for your assistance once again.

This time I need to extract from a file the last field regardless of whether it is the fourth, fifth or sixth fields, for example

1679 <****** DWDW COPY VWC87.
1788 CR362 * HGHT HTYG COPY VW4B0.
1793 <** JHGF DERE HGFDE GGG COPY VW4C8.

in this example I would need VWC87., VW4B0. and VW4C8. returned.

Can anybody help please.

Thanking you.

Dave

[Rainer_1 · Accepted solution]

here the solution:
awk '{print $NF}'

[Andreas Voss]

Hi,

simply use:
awk '{print $NF}' file

Regards

[Andreas Voss]

Hi,

simply use:
awk '{print $NF}' file

Regards

[Stefan Schulz]

Try awk '{print $NF}' yourfile. $NF is the last field of every line in yourfile. The whitspace is the default field seperator. If you want to use another field seperator use awk -F: .... (to use ':' as field seperator).

Hpe this helps.

Regards

Stefan

[federico_3]

Try:


cat your_file | awk '{print $NF}'


federico

[Ralph Grothe]

Maybe a superfluous addition,

you can do integer arithmetic with the awk qualifiers like NR, NF.
Sometimes some output produces inconsistent (or rather unpredictable) numbers of fields, but you know that the wanted field is always say n-2

awk '{print $(NF-2)}' /your/file/to/parse