Wing Poon - 7 months ago 17

Java Question

I'm trying to construct a single regex (for Java) to truncate trailing zeros past the decimal point. e.g.

- 50.000 → 50
- 50.500 → 50.5
- 50.0500 → 50.05
- -5 → -5
- 50 → 50
- 5.5 → 5.5

Idea is to represent the real number (or integer) in the most compact form possible.

Here's what I've constructed:

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

I'm using

`$1`

The problem with the pattern above is that 50 gets truncated to 5. I need some way to express that the

`0+`

`.`

I've tried using negative-behind, but couldn't get any matches.

Answer

The best solution could be using built-in language-specific methods for that task.

If you cannot use them, you may use

```
^(-?\d+)(?:\.0+|(\.\d*?)0+|\.+)?$
```

And replace with `$1$2`

.

See the regex demo. Adjust the regex accordingly. Here is the explanation:

`^`

- start of string`(-?\d+)`

-Group 1 capturing 1 or 0 minus symbols and then 1 or more digits`(?:\.0+|(\.\d*?)0+|\.+)?`

- An**optional**(matches 1 or 0 times due to the trailing`?`

) non-capturing group matching 3 alternatives:`\.0+`

- a decimal point followed with 1+ zeros`(\.\d*?)0+`

- Group 2 capturing a dot with any 0+ digits but as few as possible and matching 1+ zeros`\.+`

- (**optional branch, you may remove it if not needed**) - matches the trailing dot(s)

`$`

- end of string.

```
String s = "50.000\n50\n50.100\n50.040\n50.\n50.000\n50.500\n50\n-5";
System.out.println(s.replaceAll("(?m)^(-?\\d+)(?:\\.0+|(\\.\\d*?)0+|\\.+)?$", "$1$2"));
// => [50, 50, 50.1, 50.04, 50, 50, 50.5, 50, -5]
```