Regex for remove unnecessary zeros after decimal point - regex

Question

I need to cut some numbers:
21250000.022000 -> 21250000.022
20.00 -> 20
200 -> 200
20.50 -> 20.5
UPDATE:
This is what i try:
https://regex101.com/r/Sdfe5D/1
In this regex the problem is with the numnbers '200' and '20'.

Answer 1

You don't need a regex for this. Just cast your string to double:
$arr = array('21250000.022000', '20.00', '200', '20.50');
foreach ($arr as $n)
echo (double) $n. "\n";
Output:
21250000.022
20
200
20.5
Code Demo
Update: If you're looking for a regex solution then use:
Search regex:
(?:(\.\d*?[1-9]+)|\.)0*$
replacement:
$1
RegEx Demo

Answer 2

You can use the following regular expression. I used a positive look-ahead to only select the relevant numbers: (?=)
^[^\.]+?(?=\.0*$)|^[^\.]+?\..*?(?=0*$)|^[^\.]*$
^[^\.]+?(?=\.0*$): The first part looks for a number which is not a dot ([^\.]), followed by a dot and a random number of zeros at, which continue until the end ($). This matches numbers like 20.000 -> 20 or 25.0 -> 25. It won't match any dots at the end of the line.
^[^\.]+?\..*?(?=0*$): The second part looks is pretty similar. It looks for a number which is not a dot, followed by a dot, any characters and trailing zeros. The difference is in the brackets of the positive look-ahead: This means, it matches also characters after the dot (\.), but only if they are non-trailing zeros. This regex alone would match 20.0 -> 20.. Examples for matches here are 20.04020000 -> 20.0402 or 24.120300 -> 24.1203
^[^\.]*$: matches full numbers without any dot in it. This one is necessary for numbers like 200 or 21.
This regular expression is based on the beginning (^) and the end of a string ($). It won't match multiple numbers in one string. When using multiple numbers in one string, line breaks need to be used to match the numbers.
You can test the regular expression with the following snippet:
// expected results:
// 21250000.022000 -> 21250000.022
// 20.00 -> 20
// 200 -> 200
// 20.50 -> 20.5
var regex = /^[^\.]+?(?=\.0*$)|^[^\.]+?\..*?(?=0*$)|^[^\.]*$/g;
var texts = [
'21250000.022000',
'20.00',
'200',
'20.50'
];
for(var i = 0; i < texts.length; i++) {
var text = texts[i];
console.log(text, '->', text.match(regex)[0]);
}

Answer 3

This regex has better performance:
\.0+$|(\.\d*[1-9])(0+)$
Replace with $1
Here's an explanation, if you're interested:
First, we deal with any whole numbers, since they're easier:
\.0+$
Then, we match the important numbers (That we want to keep) after the decimal point in a capturing group:
(\.\d*[1-9])
Finally, we match all trailling zeroes, and the end of the string (replace $ with \b if needed)
(0+)$
See it live

Answer 4

If you really need to use regex for this, following would be one way:
preg_replace("/(\.[0-9]*[1-9]+)[0]+/", "$1", "21250000.022000");
However, the trailing zeroes can be removed simply by casting the string to a double:
(double) "21250000.022000";