Print out the directory size with condition [duplicate]

I want to check the directory size:

• Low if the directory uses less than 3 MB of disk space

• Medium if the directory uses more than 3 MB and less than 8 MB of disk space

• High if the directory uses equal to or more than 8 MB disk space

For example:

> ./size.sh /usr/bin High > ./size.sh /tmp Low 

Here is the code that I try even though the folder contains this directory I still cannot do it.

#!/bin/bash dir="breakfast" DIRSIZE=`du -h "$dir"` a=3145728 b=8388608 if (DIRSIZE -le "$a"); then echo "Low" else if (DIRSIZE -gt "$a" && DIRSIZE -lt "$b") echo "Medium" else if (DIRSIZE -ge "$b") echo "High" fi 

Output:

sh quota.sh : not found: du: cannot access 'breakfast'$'\r': No such file or directory quota.sh: 9: Syntax error: word unexpected (expecting "then")