AWK end of line sign in regular expressions

Question

I have a simple awk script named "script.awk" that contains:

/\/some_simple_string/ { print $0;}

I'm using it to parse some file that contains: (by using: cat file | awk -f script.awk)

14 catcat one_two/some_thing
15 catcat one_three/one_more_some_simple_string
16 dogdog one_two/some_simple_string_again
17 dogdog one_four/some_simple_string
18 qweqwe firefire/ppp

I want the script to only print the stroke that fully reflect "/some_simple_string[END_OF_LINE]" but not 2 or 3. Is there any simple way to do it?

I think, the most appropriate way is to add end-of-line sigh to the regular expression. So it will parse only strokes that starting with "/some.." and have a new line at the end of "..string[END_OF_LINE]"

Desired output:

17 dogdog one_four/some_simple_string

Sorry for confusion, I was asking for END OF LINE sign in regular expressions. The correct answer is:

/\/some_simple_string$/ { print $0;}

Solution 1

You can always use:

/\/some_simple_string$/ { print $0 }

I.e. match not only "some_simple_string" but match "/some_simple_string" followed by the end of the line ($ is end of line in regex)

Solution 2

grep '\some_simple_string$' file | tail -n 1 should do the trick.

Or if you really want to use awk do awk '/\/some_simple_string/{x = $0}END{print x}'

Solution 3

To return just the last of a group of matches, ...

Store the line in a variable and print it in the END block.

/some_simple_string/ { x = $0 }
END{ print x }

Solution 4

To print all the matches that end with the string /some_simple_string using regular expression you need to anchor to the the end of the line using $. The most suitable tool for this job is grep:

$ grep '/some_simple_string$' file

In awk the command is much the same:

$ awk '/[/]some_simple_string$/' file

To print all lines after the matching you would do:

$ awk 'print_flag{print;f=0} /[/]some_simple_string$/{print_flag=1}' file

Or just combine grep and tail if it makes it clearer using context option -A to print the following lines:

$ grep -A1 '/some_simple_string$' file | tail -n 1

View more solutions

Comments

[lc2817]

why not END{print} using the fact that $0 is set at each line and that print implicitly prints $0?

[luser droog]

I forgot about that. Haven't really used awk in a while. :)

[Chris Seymour]

Both of these just print the last line the file not the last line matching the regular expression

[Mikhail Kalashnikov]

That is probably what I've been looking for. I used to use: \\<some_simple_string\>\, but unfortunately it works only in gawk but not in awk.

[Chris Seymour]

@MikhailKalashnikov this answer won't print the last match only, I thought that is what you wanted? Also this will work with any POSIX compliant awk not just gawk.

[Mikhail Kalashnikov]

Unfortunately, It didn't work in awk v1.6, and I wanted it to be as portable as possible.

[Jotne]

END {print} is NOT a good solution. It will print any last line, why do the search then?? File may contain other string at last line than some_simple_string. Do use END{ print x }

[luser droog]

@sudo_O Thanks! Twice!

[Chris Seymour]

@luserdroog it seems this is not what the OP wants after all.

[starbeamrainbowlabs]

Warning: If your file has Windows line endings (\r\n), this might not work as expected. I suggest adding a \s* or similar (it depends on your dataset of course) to the end before the $.