Parsing Karma Coverage Output in Bash for a Jenkins Job (Scripting) - regex

Question

I'm working with the following output:
=============================== Coverage summary ===============================
Statements : 26.16% ( 1681/6425 )
Branches : 6.89% ( 119/1727 )
Functions : 23.82% ( 390/1637 )
Lines : 26.17% ( 1680/6420 )
================================================================================
I would like to parse the 4 coverage percentage numbers without the percent via REGEX, into a comma separated list.
Any suggestions for a good regex expression for this? Or another good option?

Answer 1

The sed command:
sed -n '/ .*% /{s/.* \(.*\)% .*/\1/;p;}' input.txt | sed ':a;N;$!ba;s/\n/,/g'
gives the output:
26.16,6.89,23.82,26.17
Edit: A better answer, with only a single sed, would be:
sed -n '/ .*% /{s/.* \(.*\)% .*/\1/;H;};${g;s/\n/,/g;s/,//;p;}' input.txt
Explanation:
/ .*% / search for lines with a percentage value (note spaces)
s/.* \(.*\)% .*/\1/ and delete everything except the percentage value
H and then append it to the hold space, prefixed with a newline
$ then for the last line
g get the hold space
s/\n/,/g replace all the newlines with commas
s/,// and delete the initial comma
p and then finally output the result
To harden the regex, you could replace the search for the percentage value .*% with for example [0-9.]*%.

Answer 2

I think this is a grep job. This should help:
$ grep -oE "[0-9]{1,2}\.[0-9]{2}" input.txt | xargs | tr " " ","
Output:
26.16,6.89,23.82,26.17
The input file just contains what you have shown above. Obviously, there are other ways like cat to feed the input to the command.
Explanation:
grep -oE: only show matches using extended regex
xargs: put all results onto a single line
tr " " ",": translate the spaces into commas:
This is actually a nice shell tool belt example, I would say.
Including the consideration of Joseph Quinsey, the regex can be made more robust with a lookahead to assert a % sign after then numeric value using a Perl-compatible RE pattern:
grep -oP "[0-9]{1,2}\.[0-9]{2}(?=%)" input.txt | xargs | tr " " ","

Answer 3

Would you consider to use awk? Here's the command you may try,
$ awk 'match($0,/[0-9.]*%/){s=(s=="")?"":s",";s=s substr($0,RSTART,RLENGTH-1)}END{print s}' file
26.16,6.89,23.82,26.17
Brief explanation,
match($0,/[0-9.]*%/): find the record matched with regex [0-9.]*%
s=(s=="")?"":s",": since comma separated is required, we just need print commas before each matched except the first one.
s=s substr($0,RSTART,RLENGTH-1): print the matched part appended to s

Answer 4

Assuming the item names (Statements, Branches, ...) do not contain whitespaces, how about:
#!/bin/bash
declare -a keys
declare -a vaues
while read -r line; do
if [[ "$line" =~ ^([^\ ]+)\ *:\ *([0-9.]+)% ]]; then
keys+=(${BASH_REMATCH[1]})
values+=(${BASH_REMATCH[2]})
fi
done < output.txt
ifsback=$IFS # backup IFS
IFS=,
echo "${keys[*]}"
echo "${values[*]}"
IFS=$ifsback # restore IFS
which yields:
Statements,Branches,Functions,Lines
26.16,6.89,23.82,26.17

Answer 5

Yet another option, with perl:
cat the_file | perl -e 'while(<>){/(\d+\.\d+)%/ and $x.="$1,"}chop $x; print $x;'
The code, unrolled and explained:
while(<>){ # Read line by line. Put lines into $_
/(\d+\.\d+)%/ and $x.="$1,"
# Equivalent to:
# if ($_ =~ /(\d+\.\d+)%/) {$x.="$1,"}
# The regex matches "numbers", "dot", "numbers" and "%",
# stores just numbers on $1 (first capturing group)
}
chop $x; # Remove extra ',' and print result
print $x;
Somewhat shorter with an extra sed
cat the_file | perl -ne '/(\d+\.\d+)%/ and print "$1,"'|sed 's/.$//'
Uses "n" parameter which implies while(<>){}. For removing the last ',' we use sed.