Get last field using awk substr

awk, substr, unix

Solution

Use the fact that `awk` splits the lines in fields based on a field separator, that you can define. Hence, defining the field separator to `/` you can say:

awk -F "/" '{print $NF}' input

as `NF` refers to the number of fields of the current record, printing `$NF` means printing the last one.

So given a file like this:

/home/parent/child1/child2/child3/filename
/home/parent/child1/child2/filename
/home/parent/child1/filename

This would be the output:

$ awk -F"/" '{print $NF}' file
filename
filename
filename

Problem

I am trying to use `awk` to get the name of a file given the absolute path to the file. For example, when given the input path `/home/parent/child/filename` I would like to get `filename` I have tried: ``` awk -F "/" '{print $5}' input ``` which works perfectly. However, I am hard coding `$5` which would be incorrect if my input has the following structure: ``` /home/parent/child1/child2/filename ``` So a generic solution requires always taking the last field (which will be the filename). Is there a simple way to do this with the awk substr function?

Original source