I am writing a shell script to search for previous versions of an application...the application is called TAU and basically i want to search the users home directory and /Applications for any instances of a "TAU" folder.. If found i want to give the user the option to remove the old folders and if nothing is found i just want to continue with the script...this is what i have come up with so far...

echo "Searching for previous versions of TAU..."
found=$(find ~ /Applications -type d -maxdepth 2 -name "TAU")
if [ ! $found ]; then
**commands**
else
other commands
esac

the problem is that if more than one TAU folder is found the expression doesn't evaluate correctly...i get a unary or binary operator expected error...any help would be appreciated

Thanks,
meskue

How about:

if [ -n "$found" ]; then
**commands**
else
**Other-Commands**
fi

Would that work for you ? Since the find command does return strings, just checking for a "non-zero" string length (-n) should
help you make your "decision".

An addition.

Would all the directories HAVE to be named "TAU" ? Would checking for -name '*TAU*' find all possible versions ?
[Like TAU.old , old_TAU , etc..]

#!/bin/sh

find ~ /Applications -type d -maxdepth 2 -name "TAU" 2>/dev/null | while IFS= read vo
do
# file "$vo" was found, do what you want with it, e.g.
echo "$vo"
done

great thanks a lot for your help works perfect