Need to parse big log file in one run and print id, address and service_name of found requests. The problem is that service_name is in request body that is quite big.
If I list all patterns with -e option -
grep -e 'ID: [0-9]\+' -e 'Address: .*' -e ':Body><[^ ]*'
the full request body will be printed.
What is needed is
grep -e 'ID: [0-9]\+' -e 'Address: .*' -o ':Body><[^ ]*'
or
grep -o 'ID: [0-9]\+' -o 'Address: .*' -o ':Body><[^ ]*'
to print only first word from request body that is name of the service;
but in this case grep: :Body><[^ ]*: No such file or directory error received
UPD: solution with -oe and regex works, but as it turned out -o significantly slows the operation
If you wish to print only bits of a file that match these 3 regular expressions and those three are never on the same line, you may use \|, which is grep's logical or :
grep -o 'ID: [0-9]\+\|Address: .*\|:Body><[^ ]*' my.log