CODEWITHSUNDEEP
awksedcut
user3721640
user3721640

Reputation: 927

print last two words of last line

I have a script which returns few lines of output and I am trying to print the last two words of the last line (irrespective of number of lines in the output)

$ ./test.sh

service is running..
check are getting done
status is now open..
the test is passed

I tried running as below but it prints last word of each line.

$ ./test.sh |  awk '{ print $NF }'

running..
done
open..
passed

how do I print the last two words "is passed" using awk or sed?

Upvotes: 1

Views: 7073

Answers (3)

Ed Morton
Ed Morton

Reputation: 203229

For robustness if your last line can only contain 1 field and your awk doesn't retain the field values in the END section:

awk '{split($0,a)} END{print (NF>1?a[NF-1]OFS:"") a[NF]}'

Upvotes: 1

potong
potong

Reputation: 58371

This might work for you (GNU sed):

sed '$s/.*\(\<..*\<.*\)/\1/p;d' file

This deletes all lines in the file but on the last line it replaces all words by the last two words and prints them if successful.

Upvotes: 0

fedorqui
fedorqui

Reputation: 289535

Just say:

awk 'END {print $(NF-1), $NF}'

"normal" awks store the last line (but not all of them!), so that it is still accessible by the time you reach the END block.

Then, it is a matter of printing the penultimate and the last one. This can be done using the NF-1 and NF trick.

Upvotes: 4

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