I am writing simple born shell script.

How can I extract 'firefox' from '/usr/local/bin/firefox' string ??
I don't care to use any tools just like 'gawk', 'sed' and etc.

thanks!

basename /usr/local/bin/firefox

Prior poster was correct - the basename command gives you the end of a fully qualified path (the files base name without its path). Similarly there is a dirname command that would give you the path without the base name.

However if you'd wanted to extract something out of the path you would have need something like awk. In awk you can change the default field separator (which is white space) with the -F flag/ so you could do:

echo /usr/local/bin/firefox |awk -F/ '{print $NF}'


NF is a predefined value meaning number of fields - by putting a $ in front of it you tell the print to print the last field. $1 would be the first (null as there is nothing in front of the first separator), $2 the second (usr), $3 the third (local), $4 the fourth (bin) and $5 the fifth (firefox). In this case NF = 5 so $NF is the same as $5.

Or sed:
Or bash :
Yves.

BTW, it's a "Bourne" shell, after the original author.
The improved ver. we use now, "bash", is the "Bourne again shell"

Why not sed or awk? They are very useful & almost essential to anything beyond elementary in shell scripting.