CODEWITHSUNDEEP
awk
akaur
akaur

Reputation: 415

print "m"th column of "n"th line after a match if found in a file using awk

I need to print 6th line after a particular pattern is found in a file. In that 6th line, I want to print only the 5th column. I can do the first part of this problem using the following command,

awk 'c&&!--c;/pattern/{c=6}' file

but I can't find a way to modify it to print just the 5th column of this 6th line instead. Any help will be greatly appreciated.

Upvotes: 0

Views: 295

Answers (2)

Ed Morton
Ed Morton

Reputation: 204064

It's important to really think about and ask questions if you don't understand a script you're using.

awk 'c&&!--c;/pattern/{c=6}' file

is just using default behavior as shorthand for

awk 'c&&!--c{print $0} /pattern/{c=6}' file

which you can trivially tweak to:

awk 'c&&!--c{print $5} /pattern/{c=6}' file

Upvotes: 0

Matthew Story
Matthew Story

Reputation: 3783

You can use the built-in NR variable for this

awk '/pattern/ { nrs[NR + 6] = 1; } NR in nrs { print $5; delete nrs[NR] }'

This will test for pattern and make an entry in an array of the it's line number plus six (NR + 6). We then do a simple lookup on that array to see if our current line-number is one we want to print (nrs[NR] == 1) and then print the 5th column (print $5) and then clean up the array.

This solution accounts for the fact that a pattern might occur multiple times within any given 6 line range.

Upvotes: 1

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