Hi
I'm trying to split a dir listing
eg
/home/foo1/foo2
I'm using ksh

I've tried
dir=/home/foo1/foo2
splitit=`echo $dir | awk -F '\/' '{print $1}'`
echo $splitit

nothing is output!
I have checked the escape character. The only one I have found is \

BTW `pwd` | awk -F \/ '{print $1}' works from cmdline


Why is this not working?

I want to split the string and then put it into an array, get the size of it and then get the right most directory

no need to 'escape'.
Given a sample input, if you need to echo 'foo2':
the same could be accomplished easier using 'basename' - 'man basename'

thank you for reply vgersh99 I am on Linux but '/' returns blank in inline awk

I look at basename, but is it only for directory name that you know already as argument passed?

thanks

I'm not following what's being said here - could you provide a sample input AND a desired output, pls?

Sorry
I have used basename foo2 and it returns foo2

I did not see the NF on the end - now I get 6 (yes 6!) returned to this script
dir=/home/foo1/foo2
splitit=`echo $dir | awk -F '/' '{print $NF}'`
echo $splitit

when i use the script below blank is returned
dir=/home/foo1/foo2
splitit=`echo $dir | awk -F '/' '{print $1}'`
echo $splitit

desired output is foo2.
last folder name is only known when the script is run (its run from another script)

thanks

Hm...... I don't see where '6' would be returned...

a 'blank' is what I would expect.
you're outputting '$1' - FIRST field. You have an ABSOLUTE path with the LEADING '/'.
What would be the FIRST field in a string '/a/b/c/d'?
It would be the string BEFORE the FIRST fieldSeparator. What is it in this case? It's the 'blank' character.
field1 -> 'blank'
field2 -> a
field3 -> b
field4 -> c
field5 -> d