ksh validating string with logical OR

Question

I am writing a script in ksh which needs to check a variable is set to one of the two accepted values. I have got the following code but when I enter a valid value e.g. "C" I get the error message (Not a valid value):

if  [[ "$Mode" != "C" ]] || [[ "$Mode" != "M" ]] 
then

   echo ""
   echo "*Not a valid value*"

   exit 2;

fi

I am not sure why it is failing on the string comparison, I have tried varations on syntax but still not getting anywhere, any help would be much appreciated.

Solution 1

if ! [[  "$Mode" == "C" || "$Mode" == "M" ]] ; then
   echo ""
   echo "*Not a valid value*"
   exit 2;
fi

I think this is the easier way to test multiple values. I like to think of it 'if Mode is NOT in the list of C, or M, or ....; then ....' I see beginners post code like yours (and I did it myself when I was a beginner), because then your mental model is 'if Mode is NOT C OR Mode is NOT M then ...', but you you really need the 'AND' instead (as illustrated by jlliagre).

I hope this helps.

Solution 2

Your logic is flawed.

if  [[ "$Mode" != "C" ]] && [[ "$Mode" != "M" ]] 
then

   echo ""
   echo "*Not a valid value*"

   exit 2;

fi

Solution 3

Years later...

For those who may follow, the reason amer's logic is flawed is, $Mode is always either not "C" OR not "M", because even a value like "C" is not "M", which makes the 2nd condition true.

Here are 2 one-liner alternatives:

[[ $Mode != "C" && $Mode != "M" ]] && echo -e "\n*Not a valid value*" && exit 2

This bash alterative allows for easy and concise adding of more valid values:

[[ `echo $Mode | egrep "C|M"` ]] || echo -e "\n*Not a valid value*" && exit 2

Why is everyone writing "$Mode" instead of just $Mode? Try it before you answer, please.