[SOLVED] regex for capturing a number with a range of digits in AWK

Issue

Im trying to capture numbers inside a file using AWK, I could capture all, but im not being able to capture those in a certain amount of digits. What im doing wrong?

echo -e "$teste" | awk '/_OA/ { match($0,/\[\([:digit:]{4,13}\]/);oa = substr($0,RSTART,RLENGTH);print oa}'

File sample:

_OA ............. [6712227000168]
_OA Tasdsd, OA .. [91][355016]
_OA Tasdsd, DA .. [91][5512987000]

Expected:

6712227000168
355016
5512987000

Hint for the regex match answers: Thanks so much for all the answers, i found link that I need to use a --posix option because of my awk version.

Solution

Your regexp \[\([:digit:]{4,13}\] says:

1. \[ = the literal character [
2. \( = the literal character (
3. [:digit:] = a bracket expression containing a character set of the characters :, d, i, g, t
4. {4,13} = a regexp interval that's 4 to 13 repetitions of the preceding bracket expression
5. \] = the literal character ]

The 2 main issues with that which are causing your regexp to be unable to match any of your input are:

1. You don't have any (s in your input (from #2 above), and
2. To match digits you need a character class [:digit:] inside a bracket expression [[:digit:]], not a character set :digit: inside a bracket expression [:digit:] (from #3 above)

You also don't actually need to escape the ] at the end of the regexp as it's only a regexp metachar (end of bracket expression) if preceded by a matching unescaped [ (start of bracket expression).

So the regexp I think you wanted to write instead would have been:

\[[[:digit:]]{4,13}]

e.g.:

$ awk '/_OA/ { match($0,/\[[[:digit:]]{4,13}]/);oa = substr($0,RSTART,RLENGTH);print oa}' file
[6712227000168]
[355016]
[5512987000]

or to only print the numbers:

$ awk '/_OA/ { match($0,/\[[[:digit:]]{4,13}]/);oa = substr($0,RSTART+1,RLENGTH-2);print oa}' file
6712227000168
355016
5512987000

Answered By - Ed Morton

Answer Checked By - Marilyn (WPSolving Volunteer)