Cutting fields from lines with multiple spaces

915086731 · 09-06-2011, 10:14 PM

Please see the following code, between "status" and "OK" exists many spaces, I want to get

Code:

status                        OK

. how to ignore multi spaces? If tab exists in the spaces, how to ignore it ?
Is there other commands can replace cut?

Code:

[river@localhost CLI008003]$ echo 'drv status                        OK !!'| cut -d " " -f 2-3
status 
[river@localhost CLI008003]$ echo 'drv status                        OK !!'| cut -d " " -f 2-
status                        OK !!
[river@localhost CLI008003]$ echo 'drv status                        OK !!'| cut -d " " -f 3

grail · 09-06-2011, 10:56 PM

I presume the idea is to retain the whitespace between these two words? How about:

Code:

echo 'drv status                        OK !!'| sed -rn 's/.*(status.*OK).*/\1/p'

tbrand · 09-07-2011, 08:38 AM

A very simple solution to your problem is awk:

Code:

echo 'drv status                        OK !!'| awk '{print $2 "         " $3}'

grail · 09-07-2011, 09:22 AM

Actually tbrand, your solution is guessing how much white space is required.

tbrand · 09-07-2011, 09:31 AM

Very true, grail.

It's hard to tell from the question if it is important that the number of white spaces be preserved; I assumed that it was not important and that it was more important to duplicate the actual status. The sed solution will ignore lines that report some other status than ``OK''.

colucix · 09-07-2011, 09:39 AM

What about

Code:

echo 'drv status                        OK !!' | awk '{print substr($0,index($0,"status"))}'

This assumes there are no other fields after OK that have to be left out. Otherwise you can refine the sed solution including any number of white spaces before and after the 2nd field.

tbrand · 09-07-2011, 09:51 AM

That's a good one.

One may also add a pattern to ignore any non-status lines because presumably this is meant to filter some log file. Like this:

Code:

echo 'drv status                        OK !!' | awk '/ status /{print substr($0,index($0,"status"))}'