Regex for validating zero and string [duplicate]
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 fine. See somments below but there could be leading or trailing white spaces
Good values:
.1
0.1
1.12
123.12
92
092
092.13
Error values:
0
0.0
0.00
00
1.234
-1
-1.2
Anything less than zero
Solution 1:
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*$")
>>> for item in test:
... print(item, "matches" if r.match(item) else "doesn't match")
...
.1 matches
0.1 matches
1.12 matches
123.12 matches
92 matches
092 matches
092.13 matches
0 doesn't match
0.0 doesn't match
0.00 doesn't match
00 doesn't match
1.234 doesn't match
-1 doesn't match
-1.2 doesn't match
Solution 2:
The code below allows both ,
and .
.
^(?=.*[1-9])[0-9]*[.,]?[0-9]{1,2}$
Solution 3:
[0-9]+\.[0-9]{1,2}
That will find:
- At least one number
- A decimal point
- One or two digits after the decimal point.