I have a log file and the following commands
grep '"REFERENCE":"78","STATUS":"Buffered"' file.log | wc -l
grep '"REFERENCE":"78","STATUS":"Delivered"' file.log | wc -l
grep '"REFERENCE":"78","STATUS":"Not Delivered"' file.log | wc -l
Is there a way using a one liner to get an output like:
Buffered: 30
Delivered: 1200
Not Delivered: 589
The output has not to be exactly the same.
I just find a solution to my problem, borrowed from here
grep -oP '"REFERENCE":"79","STATUS":".*?"' file.log | sort | uniq -c
Example output:
15605 "REFERENCE":"79","STATUS":"Buffered"
24280 "REFERENCE":"79","STATUS":"Delivered"
10224 "REFERENCE":"79","STATUS":"Not Delivered"
Sure, instead of counting lines consider -c switch for grep - it returns number of matches found, and $() or SINGLEQUOTEcommandSINGLEQUOTE syntax in bash, then combine it with bash simple arithmetics $(($X+$Y)) into one command, similarly to:
OUTCOUNT=$((`grep -c aaa file1.txt` + `grep -c bbb file2.txt` + `grep -c ccc file3.txt`))
Use awk to extract the desired lines and output only the status, then use sort and uniq to obtain the counts for each unique status:
awk -F, '/"REFERENCE":"78"/ { print $2 }' | sort | uniq -c
This answer may need to be adjusted, depending on the exact format of you input lines. I assume for now that the second field in a comma-delimited line is "STATUS":"<status>".
