Shell Programming and Scripting

I have a one-line command,

Code:

lsusb | awk '{ $1=""; $2=""; $3=""; $4=""; $5=""; $6=""; print $0 }'

It works, and gives the results I expect, I was just wondering if I am missing some easier way to nullify the first 6 column variables?

Something like,

Code:

lsusb | awk '{ $(1-6)=""; print $0 }'

But of course, the above doesn't work, and gives the error:

Code:

awk: (FILENAME=- FNR=1) fatal: attempt to access field -5

Any hints?

Well, you could use a for loop:[code]awk ' { for(i=1;i

@ Chubler_XL: I got this to work, also.

Code:

lsusb | awk '{ for (i=1; i<7; i++) $i=""} { print $0 }'

Or:

Code:

awk '{sub(".*" $6 FS,x)}1' file

I like that just a little better, because I get less whitespace in front of the output. Although, (at least for me, it is harder to understand the syntax)

Could you explain?

I would like to keep the ability to modify the number of columns removed from 6, to anything I need.

Hmmm. In my opinion, that's a dangerous construct and a few caveats are in order.

Since the first argument to sub() is an extended regular expression, any regular expression metacharacters in $6 could yield an unexpected result.

Also, if one of the fields which follows $6 is identical to $6 or contains $6 at its end, the greedy .* will extend beyond $6.

Just a heads up in case either of those conditions could arise.

Regards,
Alister

OK, I noticed, although slightly different from my OP, that using Franklin52's code, this works:

Code:

lsusb | awk '{sub(".*" $6 FS,x)}1'

But,

Code:

ls *.txt | awk '{sub(".*" $6 FS,x)}1'

does NOT work if there are files that have spaces in the filename.

EDIT: Nevermind I solved it with:

Code:

ls -l *.txt | awk '{sub(".*" $8 FS,x)}1'