Regex: How to match a decimal '.' if it's follow by or preceded by (or both) a '\d' decimal number

Here's what I currently have, but it only works if the decimal is preceded and followed by a decimal.

^\$?(\d*?(?<=\d)\.?(?=\d)\d*?)$

So the string: '$50.00' matches, but '$.50', and '$50.' don't match (I want them to)

I want to retrie开发者_如何学Pythonve the matched decimal like '50.00' in a single group if possible so that I can grab the normalized value if there is a match.

---

Try this:

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

It will match:

^\$?           # an optional $ at the begin
  (\d+         # one or more digits
    (?:\.\d*)? # followed by an optional decimal part
  |\.\d+       # or just decimal places without any leading digits
  )$

---

(?=^.*\d.*$)^(?:\$\s*)?(?:\d|,)*?\.?(?:\d?)*$

disallows:  
$  
.  
$.
<empty>  
<whitespace>

allows:
$50,000
$500
$0
50,000
500
0
.0
.00000000000
$50,000.000000
$ 5.

---

^\$?\s*((?=\d*\.|\d)\d*(?:\.\d*)?)$

---

Err, what about

^\$?(\d*\.?\d*)$