Regex for decimal numbers - regex

Question

Could anybody provide a regular expression for a number that has to be between 1 and 17 in length, and could optionally contain a mantissa of up to 4 places? The length of 17 includes both the characteristic and the mantissa.
Edit:
The length of 17 excludes the decimal point.
Valid examples:
12345678901234567
1234567890123.4567
123456789012345.67
12.34
Invalid:
12345678901234.5678 (Length of numerals = 18)
Thanks.

Answer 1

^\d{17}$|^\d{13}(?=.{5}$)\d*\.\d*\d$
Regex explained:
^\d{17}$ //A string of 17 digits
| //or
^\d{13} //13 digits followed by
(?=.{5}$) //5 characters, of which
\d*\.\d* //one is a decimal point and others are digits
\d$ //and the last one is a digit

Answer 2

OK, this is the best I could do:
/^\d{1,17}$|(?=^.{1,18}$)^\d+\.\d{1,4}$/
Basically, match 1-17 digits, or strings of length 1-18 which consist of two sets of digits separated by a period. The right set can only contain between 1-4 digits.

Answer 3

Don't do this completely in regex. The problem becomes nearly trivial in most programming languages, and that way will be easier for you to write, verify, test, and maintain. You can still use regex for part of the solution, of course, but you don't have to. Pseudocode:
m = re.match(r"(?P<before>[0-9]+)(?P<after>\.[0-9]{1,4})?$", input_string)
if not m:
return "no match"
before, after = m.group("before", "after")
after = after[1:] if after else "" # remove period or set to empty string
if len(before) + len(after) > 17:
return "incorrect length"
return "valid"

Answer 4

It's not particularly pretty, but with so few possibilities (0,1,2,3,4 length mantissa) I would probably just list them all:
\d{17}|\d{16}\.\d{1}|\d{15}\.\d{2}|\d{14}\.\d{3}|\d{13}\.\d{4}

Answer 5

in your favourite language, you can do a couple of logical checks, eg Python
num="1234567890133.3456"
if "." in num and len(num)==17 :
n=num.split(".")
if len(n[1])>4:
print "cannot have more than 4 decimal places"
elif len(n)==2 and n[0].isdigit() and n[1].isdigit():
print "yes, decimal"
elif len(num)==17 and num.isdigit():
print "%s is number with no decimal and is exactly 17 digits." % num
else:
print "%s not ok, check length is 17" % num

Answer 6

I have created this regex from above great solutions. may it help any one. Please let me know if you find any bug in it.
String decimalRegex =""+
"^(?!0[\d,])\+?" + // ^ Start of Number
"(\d{0,"+size+"}|" + // Numeric value without group symbol | (OR)
"(\d{0,"+rem(size,groupSize)+"},)?"+
"(\d{0,"+groupSize+"},) {0,"+div(size,groupSize)+"}\d{"+groupSize+"})" + // Numeric value with group symbol
"((([a-zA-Z]{0,2}|\"|\')\s?\+?)|\.)"+
"(\d{0,"+scale+"})?" + // Decimal value without group symbol
"(\s?([a-zA-Z]{0,2}|\"|\'))$"; // Ends with
private int rem(int size,int groupSize ){
int rem = (size - groupSize)%groupSize;
return rem;
}
private int div(int size,int groupSize ){
int div = (size - groupSize)/groupSize;
return div;
}