CODEWITHSUNDEEP
regexawk
fedorqui
fedorqui

Reputation: 290025

Print lines from match until end of file with awk

In Print lines in file from the match line until end of file I learnt that you can use 0 in range expressions in awk to match until the end of the file:

awk '/matched/,0' file

However, today I started playing around this 0 and I found that this also works:

awk '/matched/,/!./' file

I guess 0 is working because in the range expression /regex1/,/regex2/ the lines are printed until the regex2 evaluates as True. And since 0 evaluates as False, this never happens and it ends up printing all the lines. Is this right?

However, with /!./ I would imagine this to work until a line without any character is found, like if we said awk '/matched/,!NF'. But it does not: instead, it always evaluates to False. What is the reasons for this?

Examples

$ cat a
1
matched
2

3
4
end

Both awk '/matched/,/!./' a and awk '/matched/,0' a return:

matched
2

3
4
end

Whereas awk '/matched/,!NF' a returns:

matched
2

Upvotes: 5

Views: 3993

Answers (1)

anubhava
anubhava

Reputation: 785491

You have exclamation at wrong place. Keep it outside the regex like this to negate the match:

awk '/matched/,!/./' a

matched
2

Your regex /!./ is matching a literal ! followed by any character.

If your file is this:

cat a
1
matched
2

3
!4
end

Then 

awk '/matched/,/!./' a
matched
2

3
!4

And:

awk '/matched/,!/./' a

matched
2

Upvotes: 4

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