cut words based on the word count of a line

UNIX for Advanced & Expert Users
1

I would like to cut words based on the word count of a line. This over here inspired me with some ideas but I wasn't able to get what I needed.

If the line has 6 words I would like to use this.

awk -F" " '{print $6}'

If the line has 4 words I would like to use this.

awk -F" " '{print $4}'

I don't want to use a shellsript. I would like this to be just one line that I can put into my shell and it work.

2 
awk '{print $NF}' myFile
3

I saw that from the link I gave. How do I cut the 4th word and 6th word based on word count.

4

This will return the last 'word' from the record/line.
Isn't that what your examples illustrate?

When you say 'cut words' what do you example mean?
Please provide a sample input a desired output.

5

I have several lines similar to this. The first one if you count it you can see has 6 words and the second one has 4 words. On the lines that have 6 words I would like to cut the 6th word and output it to my screen and on the lines that have 4 words I would like to cut the 4th word and output it to my screen.

Output would be
sat
thur

6

Then what I've posted should give you a desired result.
Does it not?

7

No. I was trying to keep my example simple. And what you gave just cuts the last word. This is the full line of text.

Output would be
sat
thur

8

This still doesn't make much sense.

It's better if you keep your example representative, than "simple"!

Is "This is a X day week" part of the line?

You can test how many "words" (fields) a line has using NF.

awk 'NF == 4 || NF == 6 { print $NF }' file

(assuming "This is a X day week" isn't part of the line!)

9

Ah, I think I see what you want...

nawk '{print $($(NF-2))}' myFile
OR
nawk '{print $(NF-6)}' myFile