Ruby Regular expression to get percentage from specific line

Question

I have the following output:

Name                              Stmts   Miss  Cover   Missing
---------------------------------------------------------------
src/global_information.py             8      1    88%   6
src/settings.py                      38      0   100%
src/storage_backends.py               4      4     0%   1-5
src/urls.py                           8      0   100%
users/admin.py                        1      0   100%
users/apps.py                         3      3     0%   1-5
users/forms.py                        5      0   100%
users/models.py                       1      0   100%
users/tests/tests_views_urls.py       5      0   100%
users/urls.py                         5      0   100%
users/views.py                        1      1     0%   1
---------------------------------------------------------------
TOTAL                                79      9    89%

I need to get the TOTAL percentage, which is 89%. I try the following two regex:

TOTAL\s+\d+\s+\d+\s+\d+\%

and

(?<=TOTAL\s).*

I can get the correct line but not sure how to extract the percentage part of it. This needs to be achieved in a regular expression due that I don’t have access to any tool

Thanks

what tools are you using for the job? extract in regex usually use capture group. — delta, May 09 '18 at 01:01

score 2 · Accepted Answer · answered May 09 '18 at 01:04

2

You can use a regex like this:

TOTAL.*?(\d+)%

Working demo

Or if you want to capture the % then

TOTAL.*?(\d+%)

Then grab the content from the capturing group $1

answered May 09 '18 at 01:04

Federico Piazza

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@MarcoHerrarte, have you tried the regex without the parentheses? `TOTAL.*?\d+%` – Federico Piazza May 09 '18 at 03:43

Cary Swoveland · Answer 2 · 2018-05-09T06:20:26.163

str=<<_
Name                              Stmts   Miss  Cover   Missing
---------------------------------------------------------------
src/global_information.py             8      1    88%   6
src/settings.py                      38      0   100%
src/storage_backends.py               4      4     0%   1-5
src/urls.py                           8      0   100%
users/admin.py                        1      0   100%
users/apps.py                         3      3     0%   1-5
users/forms.py                        5      0   100%
users/models.py                       1      0   100%
users/tests/tests_views_urls.py       5      0   100%
users/urls.py                         5      0   100%
users/views.py                        1      1     0%   1
---------------------------------------------------------------
TOTAL                                79      9    89%
_

str[/^TOTAL.*?\K\d+%/] #=> "89%

\K means discard everything matched so far. The non-greedy modifier ? in .*? is needed. Without it the match before \K would end with the next-to-last digit in the total percentage (here the "8" in "89%", the "3" in "1234%").

score 0 · Answer 3 · answered May 09 '18 at 00:58

0

awk '/TOTAL/{ print $4}' <INPUT_FILE>

answered May 09 '18 at 00:58

delta

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15
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This certainly provides an `awk` solution but OP needs Ruby's regex solution. – builder-7000 May 09 '18 at 03:47
Please add some explanatory text. – CJ Dennis May 09 '18 at 03:58
If you surround your expression with a pair of backticks you can execute it within a Ruby program: `\`awk '/TOTAL/{ print $4}' test\` #=> "89%\n"`. To get rid of the newline add `.chomp` after the ending backtick. The various ways of invoking system programs is discussed in answers to [this SO question](https://stackoverflow.com/questions/6338908/ruby-difference-between-exec-system-and-x-or-backticks). – Cary Swoveland May 09 '18 at 05:42

Ruby Regular expression to get percentage from specific line

3 Answers3