Regex greater than zero with 2 decimal places

Question

I need a RegEx for a numeric value with up to two decimal places greater than zero and may or may not have a zero in the ones column. I should also add....whole numbers are

Answer 1

/^[0-9]*(\.{1})?([0-91-9][1-9])?$/

try this it passes all your cases

Answer 2

How about this:

^\s*(?=.*[1-9])\d*(?:\.\d{1,2})?\s*$

Explanation:

^            # Start of string
\s*          # Optional whitespace
(?=.*[1-9])  # Assert that at least one digit > 0 is present in the string
\d*          # integer part (optional)
(?:          # decimal part:
 \.          # dot
 \d{1,2}     # plus one or two decimal digits
)?           # (optional)
\s*          # Optional whitespace
$            # End of string

Test in Python:

>>> import re
>>> test = [".1", "0.1", "1.12", "123.12", "92", "092", "092.13", "0", "0.0", "0.00", "00", "1.234", "-1", "-1.2"]
>>> r = re.compile(r"^\s*(?=.*[1-9])\d*(?:\.\d{1,2})?\s*$")
>>> for item in test:
...     print(item, "matches" if r.match(item) else "doesn't match")
...
.1 matches
0.1 matches
1.12 matches
123.12 matches
92 matches
092 matches
092.13 matches
0 doesn't match
0.0 doesn't match
0.00 doesn't match
00 doesn't match
1.234 doesn't match
-1 doesn't match
-1.2 doesn't match

Answer 3

This expression will not allow any white spaces in beginning and last

/^\d*(?:\.\d{1,2})*$/

Answer 4

[0-9]+\.[0-9]{1,2}

That will find:

• At least one number
• A decimal point
• One or two digits after the decimal point.

Answer 5

The code below allows both , and ..

^(?=.*[1-9])[0-9]*[.,]?[0-9]{1,2}$