How to negate a case pattern

Question

How add I add the condition in case whereby if it does not detect the required conditions, it will execute the command.

My code:

case $price in
[0-9] | "." | "$") echo "Numbers, . , $ Only"
            ;;
esac

This command will execute if it detects numbers, "." and "$". How do change it in a sense if it does not detect those, the command will execute. Or are there other better commands to use to do this function.

Answer 1

Add a default case:

case $price in
[0-9] | "." | "$") true
            ;;
*) 
   do-something
   ;;
esac

Answer 2

This answer does not refer to the already explained case problem but to the matching problem.

First we need a definition what the valid strings may look like. The easiest definition would, of course, allow only one structure like

1. start with one or more digits (leading zeroes allowed)
2. decimal dot
3. two trailing digits (if necessary 00)
4. Dollar sign (no leading space)

As a regular expression (for e.g. grep, see man 7 regex) this would be written as:

^[0-9]+\.[0-9][0-9]\$$

• ^ marks the beginning of the string i.e. there can be nothing before the [0-9]+
• The dot is escaped \. in order to be treated as literal dot
• The Dollar sign is escaped \$ in order to be treated as literal
• the trailing $ marks the end of the string i.e. there can be nothing after the literal $

Testing:

> echo '0123.45$' | grep -E '^[0-9]+\.[0-9][0-9]\$$'
0123.45$

If the definition is changed so that both the dot with the trailing digits and the Dollar sign are optional then the regex would change to

^[0-9]+(\.[0-9][0-9])?\$?$

Testing:

> echo '0123$' | grep -E '^[0-9]+\.[0-9][0-9]\$$'

# no match
> echo '0123$' | grep -E '^[0-9]+(\.[0-9][0-9])?\$?$'
0123$
> echo '0123.45$' | grep -E '^[0-9]+(\.[0-9][0-9])?\$?$'
0123.45$

The shell can work with regular expressions directly but not within case patterns. You need the [[ ]] structure. But as you want to know only whether a string matches or not anyway there is no reason to use case:

if [[ "$price" =~ ^[0-9]+(\.[0-9][0-9])?\$?$ ]]; then
  :
else
  :
fi

If you really need case then you can set the option extglob with the shell bulitin shopt and rewrite the regex to a "shell regex":

shopt -s extglob

^[0-9]+(\.[0-9][0-9])?\$?$ becomes then

+([0-9])?(.[0-9][0-9])?($)

leading $

If you want $0123.45 instead of 0123.45$ then you obviusly have to put the check for $ at the start:

• grep / [[ ]]: ^\$?[0-9]+(\.[0-9][0-9])?$
• shell pattern matching: ?($)+([0-9])?(.[0-9][0-9])

easier check

If you don't care about the order but just about the right chars then you can use much easier expressions:

• grep / [[ ]]: ^[0-9.$]+$
• shell pattern matching: +([0-9.$])

example

#! /bin/bash

shopt -s extglob

for value in 1234 1234EUR; do
    case "$value" in
            +([0-9]))
                    echo "value OK: '${value}'"
            ;;
            *)
                    echo "value not OK: '${value}'"
            ;;
    esac
done

Answer 3

To match a string containing only a single, positive integer with a shell case statement you can do:

case "${string:--}" in
(*[!0-9]*)  
! echo 'Invalid value for $string!';;
(*[!0]*) 
echo "$string is a positive integer greater than 0.";;
(*)
echo "$string is 0.";;
esac

In the first case match I screen for any value of the ${string:--} expansion that contains even a single character which is not among 0123456789. In the event $string is either unset or null it will expand to - and so still match this pattern. The leading ! inverts echo's return (which is always 0) to a 1 - and so the case construct returns 1 if $string contains an invalid value.

In the next I match any value for $string containing a single digit which is not 0 - and so any number of leading 0s on any valid value for $string will match as a positive number.

The last matches any which did not match the previous two - which is to say it matches only one or more 0s.