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How would I parse following output of lein test
$ lein test
lein test nepleaks-engine.core-test
Ran 1 tests containing 1 assertions.
0 failures, 0 errors.
Where I want is get number before failures using bash script.
I only know better is grep,
$ lein test | grep 'failures' | cut -d' ' -f1
0
Suggest me the better ways of doing this.
Your solution returns the part of every line that contains "failures" up to the first space character.
If you want to be more specific, you could do:
sed -n '$s/^\([[:digit:]]\{1,\}\) failures.*/\1/p'
That is, only consider the last line ($), and only if it follows a specific pattern: any non-empty sequence of decimal digits followed by failures (and return that sequence of digits).
There is nothing wrong with the way that you are doing it, provided that failures will only occur on one line of the output. Another option might be to anchor on the fact that the number of failures is always on the 4th line of output. This is very easy with awk:
lein test | awk 'NR==4 { print $1 }'
Or to anchor on a line where failures, is the second field:
lein test | awk '$2=="failures," { print $1 }'
Or to anchor on the last line:
lein test | awk 'END { print $1 }'
Or using sed to anchor on the last line:
lein test | sed -n '$ s/^\([0-9]*\).*/\1/p'
lein test | awk '$2=="failures," { print $1 }'is the safest one.
Jun 12, 2014 at 19:11
With awk (print $1, means first column):
lein test | grep 'failures' | awk '{print $1}'
awk '/failures/{print $1}' achieves the same thing. Even better, match the output completely. awk '/^[[:digit:]]+ failures, [[:digit:]]+ errors/{print $1}'
Try this:
lein test | tail -n 1 | tr -s " " "\n" | head -n 1
How does it work?
tail -n 1 takes the last linetr -s " " "\n" replaces all spaces with the new line character. It's a nice trick to tokenize the input by some character.head -n 1 takes the first line
Through awk,
lein test | awk '/ failures,/{print $1}'
And with GNU sed,
lein test | sed -n '/ failures,/s/^\([^ ]\+\).*/\1/p'
Or portably:
lein test | sed -n '/ failures,/s/^\([^ ]\{1,\}\).*/\1/p'
\+ and \? here are GNU extensions and don't work with all versions of sed. Surely all you want is just * anyway?
sed -n '/ failures,/s/^\([^ ]\{1,\}\).*/\1/p'
Jun 12, 2014 at 12:41
s/^\([^ ]*\).*/\1/ is all you need for the replace part in this case.
Considering the output, maybe (if you have the GNU implementation of grep):
lein test | grep -m 1 -o '^[[:digit:]]*'
Safer, with the same implementation:
lein test | grep -E 'failures.*errors' | grep -o '^[[:digit:]]*'
From the manpages:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
-m NUM, --max-count=NUM
Stop reading a file after NUM matching lines.[...]
You could try using this code:
lein test | tail | tr " " "\n" | head -1
which should give you what you want.
