gensub on multiple lines

Question

I have a file which has many random lines like

aaa bbb
ccc ddd
eee mark: 98 fff
ggg ggg jjjj iii
jjj kkkk

I want to use awk AND only gensub to match the number "98" above. So far I have this code below, I think it does not work cause I need to make gensub treat "\n" as any other character.

cat file.txt | awk 'printf(gensub(/^.*mark: ([0-9]+).*$/,"\\1","g"))}'

I need the output of the code above to be only "98". How do I do that?

EDIT

even when I use the s or m modifier it does not work as it should cause as far as I know the "s" modifier should make regex treat . as any character including \n.

Answer 1

You seem to think that awk treats its input as a multiline string. It doesn't. When you run an awk script on a file, the script is applied to each line of the file separately. So, your gensub was run once per line. You can actually do what you want with awk but it really isn't the best tool for the job.

As far as I can tell, you have a large file and only want to print a number that comes after mark: and whitespace. If so, all of these approaches are simpler than fooling around with gensub:

1. Use grep with Perl Compatible Regular Expressions (-P)

   $ grep -oP 'mark:\s*\K\d+' file 
   98
   

   The -o makes grep only print the matching portion of the line. The \K is a PCRE construct which means "ignore anything matched before this point".

2. sed

   $ sed -n 's/.*mark:\s*\([0-9]\+\).*/\1/p' file
   98
   

   The -n suppresses normal output. The p at the end makes sed print only if the substitution was successful. The regex itself captures a string of numbers following mark: and 0 or more whitespace characters and replaces the whole line with what was captured.

3. Perl

   $ perl -ne 'print if s/.*mark:\s*(\d+).*/$1/' file
   98
   

   The -n tells perl to read an input file line by line and apply the script given by -e. The script will print any lines where the substitution was successful.

If you really, really want to use gensub, you could do something like:

$ awk '/mark:/{print gensub(/.*mark:\s*([0-9]+).*/,"\\1","g")}' file
98

Personally, I would do it this way in awk:

$ awk '/mark:/{gsub(/[^0-9]/,"");print}' file
98

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Since you seemed to be trying to get awk to receive multiline input, this is how you can do that (assuming there are no NULL characters in your file):

$ awk '{print(gensub(/^.*mark: ([0-9]+).*$/,"\\1","g"))}' RS='\0' file
98

The RS='\0' sets the input record separator (that's what defines a "line" for awk) to \0. Since there are no such characters in your file, this results in awk reading the whole thing at once.

Answer 2

The smallest change to get it working will be:

cat file | awk '/mark:/{printf( "%s\n",gensub(/^.*mark: ([0-9]+).*$/,"\\1","g"))}'

The /mark:/ is to select a line that contain "mark:".
But, then, why is a printf needed? This will also work:

cat file | awk '/mark:/{print(gensub(/^.*mark: ([0-9]+).*$/,"\\1","g"))}'

But that would be a "useless use of cat", as awk could directly read from a file:

awk '/mark:/{print(gensub(/^.*mark: ([0-9]+).*$/,"\\1","g"))}' file

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Edit:

On user request: How to use the regex on file and string.

Well, with the rules you set: awk with only gensub is not possible.
Also, the idea of matching with .*mark: ([0-9]+).* to replace all of that with the match inside the parenthesis will mean that it is needed to match the whole file to extract a part. That is one reason why grep was created.

Just use:

grep -oP "mark: \K([0-9]+)" file

or:

echo "$string" | grep -oP "mark: \K([0-9]+)"

And you will get the result.