Hi,
I want to print my log separately for each day it contains;
-1st solutions was:
grep "^\[*" file1.log | awk '{print $1}' >> $1day.log
but I noticed that awk doesn't
eat records that have more than 99 fields, so I get the error "... too many fields; Broken pipe. "
-2nd one:
grep "^\[*" file1.log | sed -n <pattern> >> day.log
problem: how to identify each day in sed separately ?
file1.log comes from IBM WebSphere and looks like:
[20/07/05 mm:dd hh:MM] ...
..
many thanks from Rome.
Carmen
Hi,
I want to print my log separately for each day it contains;
-1st solutions was:
grep "^\[*" file1.log | awk '{print $1}' >> $1day.log
but I noticed that awk doesn't
eat records that have more than 99 fields, so I get the error "... too many fields; Broken pipe. "
-2nd one:
grep "^\[*" file1.log | sed -n <pattern> >> day.log
problem: how to identify each day in sed separately ?
file1.log comes from IBM WebSphere and looks like:
[20/07/05 mm:dd hh:MM] ...
..
many thanks from Rome.
Carmen