UNIX for Dummies Questions & Answers

You don't really want grep.

Code:

case $input in *[!0-9]*) echo Need a number >&2; exit 1;  esac

fabtagon's solution does not check the number of digits at all, it just checks that there is a number in there somewhere. But the thing you really want to check is that there is nothing there which is not a digit. You could do that with grep, but case is absolutely a better solution.

that code is really helpfull

i am just wondering if i want to check to number, is it possible to combine the code

example
case [ $input1 -a $input2 ] in
*[!1-9]*) echo not number
esac

I wrote a utility menu to do this. Try some of this code:

# find file names containing text

1) echo ""
echo 'which dir or .?'
read dir
echo 'what text to find?'
read text
echo " do you want to put the findings to a file?"
read answer
if [ "$answer" = "y" ]
then
echo "outputting to text_found"
find $dir -exec grep -l $text {} \; > text_found;
echo "findings have been output to `pwd`/text_found"
echo "would you like to view? y/n"
read answer
if [ "$answer" = "y" ]
then
more text_found
fi
else
echo ""
find $dir -exec grep -l $text {} \; 2>/dev/null|more
fi
;;

that is so helpfull, thanks for the help

Yes, the syntax you want is quite different though.

[ is a separate command (believe it or not!) and is customarily only used as an argument to if and perhaps occasionally while; case simply looks at the next token and compares it. But you can do two comparisons:

Code:

case $input1 in *[!1-9]*) echo not number;; esac
case $input2 in *[!1-9]*) echo not number;; esac

In fact, you could check them both in one go, if you don't care which one is not correct:

Code:

case "$input1$input2" in *[!1-9]*) echo not number;; esac

Sorry for taking so long to respond, I had this open in my browser but got interrupted ...

The earlier example I posted lacked the double semicolons; sorry for omitting them (though some shells can apparently cope if it's in the last branch in a case structure).