Awk to extract phrase between two words on a line?

grob115 · 05-25-2010, 06:48 AM

Hi, am trying to find a way to extract the phrase between the words Connection and is (ie the underlined words below). Can we use awk to do this? How? Is it the best command to use?

Code:

[06:25:00][i] Connection at Plant A is live
[06:25:00][i] Connection at Building_C is not live
[07:25:00][i] Connection at Terminal D is down

grail · 05-25-2010, 06:54 AM

Code:

awk '{print $4,$5}' file

Edit: sorry just noticed the underscore in second line, try:

Code:

sed -r 's/.*Connection(.*)is.*/\1/' file

grob115 · 05-25-2010, 07:00 AM

Sorry wasn't aware the underline masked the space/underscore. Here's a better representation.

[06:25:00][i] Connection at Plant A is live
[06:25:00][i] Connection at Building_C is not live
[07:25:00][i] Connection at Terminal D is down

Notice the 2nd line has an underscore between Building and C. So it's not always fixed in terms of the column numbers.

grail · 05-25-2010, 07:04 AM

Also here is a better awk:

Code:

awk 'gsub(/.*Connection|is.*/,"")' file

grob115 · 05-25-2010, 07:14 AM

grail, thanks. You are the master. Will try it out.

grob115 · 05-25-2010, 08:22 AM

BTW, sorry one more question. If the command returns many lines and some of the lines repeat, how do I select a distinct set of lines only?

grail · 05-25-2010, 08:46 AM

Well here is one a little trickier to work out

Code:

awk 'gsub(/.*Connection|is.*/,"") && !_[$0]++' file

grob115 · 05-25-2010, 10:12 AM

Um... thanks. Haven't seen this !_[$0]++ before. I'll dig around for some explanation. But yeah thanks!

catkin · 05-25-2010, 10:35 AM

Quote:

Originally Posted by grob115

Haven't seen this !_[$0]++ before. I'll dig around for some explanation.

Not easy to find so I'll help some. "!" is a logical negation. "_" is an array name. awk arrays are "content addressable" -- you can use strings as their indexes. In awk the number zero is logical "false" and any other number is logical "true". Uninitialised variables when referenced as numbers are given value 0. "++" in that position is a post-reference increment operator. String those concepts together!

grail · 05-26-2010, 12:07 AM

@catkin - thanks, very comprehensive explanation

Jerry Mcguire · 05-26-2010, 08:53 PM

Hi, I walked by and saw this interesting thread about awk. I did the following to test:

Code:

$ awk '!_[$0]++' file

and it miraculously prints the first occurence of $0 in the file. Why does it 'print'? Is it a default behaviour if a condition evaluates true?

Code:

$ awk '!_[$0]++ {print}' file

Thanks.

grail · 05-26-2010, 09:35 PM

Quote:

Is it a default behaviour if a condition evaluates true?

Yes

Explanation found here: http://www.gnu.org/manual/gawk/html_...ml#Very-Simple

Quote:

In an awk rule, either the pattern or the action can be omitted, but not both. If the pattern is omitted, then the action is performed for every input line. If the action is omitted, the default action is to print all lines that match the pattern.

NetRock · 05-26-2010, 09:46 PM

Hi grail.....
YOU ROCK..!! Keep doing the good job!! share your knowledge & THE UNIVERSE will return to you MUCH MORE than you share....