Regex issue - Replace floats, not int

Following regex should match floats for you (for the example provided):

/(-?\d*\.\d+)/

To replace:

console.log (string.replace(/(-?\d*\.\d+)/, ''));

I have used positive look behind (?<=..) in below regular expression

\.\d+(?<=\d)

Use this regular expression and replace below value with ''. The result will be 1 23 33 3

1 23 33.2000 3.4445

Hope it helps

var reg = /\d+\.\d+/g
var str = "1 south african rand is 0.11044";
var onlyFloats = str.match(reg).join(" ");
console.log(onlyFloats)

Try this

string = "1 south african rand is 0.11044"
reg = /(\d+\.\d+)/g;
string.match(reg);

/**
 * @param $string
 * @return string
 * 
 * output
 * a.a.a.a.a.a.   => 0.0
 * 1.1.1.1.1.1.   => 1.11111
 * 2a$2.45&.wer.4 => 22.454 
 */
function stringToFlout($string) {
    $parts = explode('.', $string);

    for ($i = 0; $i < count($parts); $i++) {
        $parts[$i] = preg_replace('/[^0-9]/', '', $parts[$i]);
    }

    $left = $parts[0];
    if (empty($left))
        $left = 0;

    $float[] = $left;

    unset($parts[0]);
    $right = implode('', $parts);
    if (empty($right))
        $right = 0;

    $float[] = $right;

    return implode('.', $float);
}

I would use this regex:

(0|([1-9][0-9]*))?\.(0|([1-9][0-9]*))

The (0|([1-9][0-9]*))? matches an optional integer part, the \. matches dot, and the final (0|([1-9][0-9]*)) matches a mandatory fractional part. It succeeds for:

• .5
• 0.5
• 0.0
• 0.1
• 0.5foo
• 31.41592
  

It fails for:

• 00.5
• 100
• nice
• 2.
  

Anchored version:

^(0|([1-9][0-9]*))?\.(0|([1-9][0-9]*))$