can anyone here tell me why the * is not passed to the bash shell script?
My script named as top40 is
#!/bin/bash sudo du -shx $1 | sort -rh | head -n 40
when I try to run it as
top40 /var/* the * is ignored. It is like
top40 /var but I want to see the top 40`s size of the directories.
when I do it whithout script and type it on the prompt, it works fine.
I really can not find the reason for this. Thanks for opening my eyes. I use Ubuntu 18.04 LTS