Bash RegEx to check floating point numbers from user input

Question

I'm relatively new to bash programming and i am currently creating a simple calculator.

It needs to use floating point numbers and check they are so.

I have a checkNumbers function:

function checkNumber {

    regExp=[0-9]

    if [ $testNo =~ $regExp ]
    then
        echo "That is a number!"
        let check=1
    else
        echo "Damn! Not A Number!"
    fi
}

where i get the user to input a number like this:

while [ $check -eq 0]
do
    echo "Please enter a number
    read testNo
    echo ""
    checkNumber
done

This does not work, i get this error:

./calculator.sh: line 39: [: =~: binary operator expected

line 39 is:

if [ $testNo =~ $regExp ] 

I have tried lots of Regular Expressions like:

^*[0-9]*$

and

^*[0-9]\.[0.9]*$

etc etc.

also, i have tied other ways of performing the check:

case $testNo
in
    ^*[0-9]*$) echo "That is a number!";;
    *) echo "Error! Not a number!";;
esac

and

if [ $testNo = ^*[0-9]*$ ]
then
    echo "etc etc"
else
    echo "oops"
fi

I also need this to work with floating point numbers.

could someone please let me know how i go about this?

Answer 1

This regex ^[-+]?[0-9]+\.?[0-9]*$ will match only digits with an optional .:

$ echo 30 | grep -Eq '^[-+]?[0-9]+\.?[0-9]*$' && echo Match
Match

$ echo 30.10 | grep -Eq '^[-+]?[0-9]+\.?[0-9]*$' && echo Match
Match

$ echo 30. | grep -Eq '^[-+]?[0-9]+\.?[0-9]*$' && echo Match
Match

$ echo +30 | grep -Eq '^[-+]?[0-9]+\.?[0-9]*$' && echo Match
Match

$ echo -30 | grep -Eq '^[-+]?[0-9]+\.?[0-9]*$' && echo Match
Match

I think when you tried ^*[0-9] you wanted ^[0-9]*

Rexeplanation:

^       # Match start of string
[-+]?   # Match a leading + or - (optional)
[0-9]+  # Match one or more digit
\.?     # Match a literal . (optional, escaped)
[0-9]*  # Match zero or more digits
$       # Match the end of the string

Note: this matches numbers followed by a . like 30., not sure if this is acceptable for you.

Edit: Don't quote the regex

testNo=30.00

if [[ $testNo =~ ^[+-]?[0-9]+\.?[0-9]*$ ]]; then 
    echo Match
fi

>>> Match

Answer 2

To use that type of feature, you need the [[ ... ]] version of the conditional. [ is the "old" test command and doesn't handle regular expressions at all.

#! /bin/bash
function checkNumber {
    regExp='^[+-]?([0-9]+\.?|[0-9]*\.[0-9]+)$'
    if [[ $testNo =~ $regExp ]]
    then
        echo "That is a number!"
        let check=1
    else
        echo "Damn! Not A Number!"
    fi
}
testNo=1
checkNumber
testNo=-1.2
checkNumber
testNo=+.2
checkNumber
testNo=+0.
checkNumber
testNo=a
checkNumber
testNo=hello2you
checkNumber

$ ./t.sh
That is a number!
That is a number!
That is a number!
That is a number!
Damn! Not A Number!
Damn! Not A Number!

See What is the difference between test, [ and [[ ?.

An explanation on the regex:

^ Anchor at start of string
$ Anchor at end of string

These two make the regex match the whole string passed, partial matches are not allowed.

[+-]

matches either + or -.

[+-]?

makes that part optional, so the above matches exactly +, - or nothing at all.

Then there's an alternation (part1|part2) which will match if part1 or part2 matches.

Part one is:

[0-9]+\.?

which matches one or more (+) digits (but not zero digits/empty set) and an optional .. This handles numbers of the form 123 and 534.. But not just ..

Part two is:

[0-9]*\.[0-9]+

This matches zero or more (*) digits, followed by a ., followed by one or more digits. This matches all other floats like 1.3 or .543 (without exponent notation), but still excludes just ..