Regex to truncate trailing zeroes

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 → 5

Answer 1

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.

Java demo:

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]

Answer 2

For a general regex which should do the trick:

^\d+?0*\.??\d*?(?=0*?[^\d]*$)

You can replace the caret and dollar sign with whatever your boundaries should be. Those could be replaced by whatever you would expect around your number.

basically:

/d+? non-greedy match for any number (needs at least 1 number to start the match)

\.*?? optional match for a decimal. Prefers to match 0 occurrences

\d*? (?=0*?[^\d]*$) - non-greedy match for a number, but would stop at the 0 which is proceeded by a non-number

EDIT: I just realized the original expression also trimmed the last zero on integers, this should work. I added the option 0 match to catch that