To cut a string based on last index of identifier

Shell Programming and Scripting
1

here below is sample string

null pointer dereference of 'resourceList' where null is returned from a method/opt/bld/fetch/ds/interzone/notification/LocalLineStatusNotificationListener.java:79
null pointer dereference of 'reList' where null is returned from a method/opt/bld/fetch/dunkindev6jbe4/src/com/ip/interzone/notification/LineStatusListener.java:45

i want to cut based on last index of '/' identifier.

i tried

 
cat fileName.txt | awk '.java' '{print $1}' | cut -d'/' f5

but that would return only for partial since there are different number of '/' for each line.. so i need something to find last index of '/' and cut it and provide me only java file name

2

This will get you the filename:

sed -e 's#.*/##' -e 's/:.*//' inputfile

Or simply use grep:

grep -o [^/]*java inputfile
3 
awk -F'[/:]' '{print $(NF-1)}' file
4 
awk '{gsub(/.*\/|:.*/,"")}1' file