[SOLVED] Bash conditional help

the_gripmaster · 09-07-2010, 11:17 PM

Code:

if  [ ${METAL} = "Au" -o ${METAL} = "Ag" ]; then
  echo
else
  echo "This utility is only for Ag and Au."
  exit 1
fi

How can you negate this Bash condition so that no else is required? I am trying to get rid of the else.

In C++, what I am trying to do would be:

Code:

if (!(metal == "Au" || metal == "Ag")) {
  cout << "This utility is only for Ag and Au.";
  exit(1);
}

smoker · 09-07-2010, 11:48 PM

Code:

if  [ ${METAL} != "Au" -o ${METAL} != "Ag" ]; then
    echo "This utility is only for Ag and Au."
  exit 1
fi

Code:

if  [[ ${METAL} != "Au" || ${METAL} != "Ag" ]]; then
    echo "This utility is only for Ag and Au."
  exit 1
fi

the_gripmaster · 09-08-2010, 12:01 AM

Quote:

Originally Posted by smoker

Code:

if  [ ${METAL} != "Au" -o ${METAL} != "Ag" ]; then
    echo "This utility is only for Ag and Au."
  exit 1
fi

Code:

if  [[ ${METAL} != "Au" || ${METAL} != "Ag" ]]; then
    echo "This utility is only for Ag and Au."
  exit 1
fi

I did try code1 before. But unfortunately none of these works.

the_gripmaster · 09-08-2010, 12:08 AM

Found the fix!

Code:

if ! [ ${METAL} = "Au" -o ${METAL} = "Ag" ]; then
  echo "This utility is only for Ag and Au."
  exit 1
fi

ghostdog74 · 09-08-2010, 01:01 AM

you can do this with case/esac

Code:

case "$METAL" in
  Au|Ag ) echo "This utitlity..... ";;
  *) echo "....";;
esac

slakmagik · 09-08-2010, 01:33 AM

I hate posting on 'solved' threads, but...

If you have to use old style tests, you can also do it '[ foo ] || [ bar ]' (or &&) since '[ foo -o bar ]' (or -a) can be kinda funky.

Code:

if [ ${METAL} != "Au" ] && [ ${METAL} != "Ag" ]; then

or

Code:

if ! [ ${METAL} = "Au" ] && ! [ ${METAL} = "Ag" ]; then

<pedantic>If using the old-style test, quoting "$METAL" is a good idea and it doesn't need the braces.</pedantic>

But tweaking smoker's line

Code:

[[ ${METAL} != "Au" && ${METAL} != "Ag" ]]; then

or using ghostdog74's is probably about the best.