Javascript Regexp for all numeric and decimal point format

i'd like to make a javascript validation that will accept all numeric and decimal point format.

For example :

1,000,000.00 is OK

1.000.000,00 is OK

1.000,000.00 is not OK

1.000,000,00 is not OK

1,000.000,00 is not OK

1,000.开发者_StackOverflow中文版000.00 is not OK

Based on what i got here is :

/^[1-9][0-9]{0,2}(,[0-9]{3})*(\.[0-9]{2})?$/ is only valid for 1,000,000.00 not for 1.000.000,00

How can i validate both format ?

Updated :

What if the thousand points are not compulsory such as :

1000000.00 is OK or

1000000,00 is OK

Assuming that the decimal part and thousands separators are mandatory, not optional, and that 0 is not an allowed value (as suggested by your examples and your regex):

^[1-9]\d{0,2}(?:(?:,\d{3})*\.\d{2}|(?:\.\d{3})*,\d{2})$

As a verbose regex:

^              # Start of string
[1-9]\d{0,2}   # 1-3 digits, no leading zero
(?:            # Match either...
 (?:,\d{3})*   # comma-separated triple digits
 \.\d{2}       # plus dot-separated decimals
|              # or...
 (?:\.\d{3})*  # dot-separated triple digits
 ,\d{2}        # plus comma-separated decimals
)              # End of alternation
$              # End of string

Here is the regex that you want..

^(([1-9]\d{0,2}(((\.\d{3})*(,\d{2})?)|((,\d{3})*(\.\d{2})?)))|(0(\.|,)\d{1,2})|([1-9]\d+((\.|,)\d{1,2})?))$

This is the link that proves that it can handles all cases

http://regexr.com?2tves

The best way to look at a regular expression this big is to blow it up to a very large font and split it on the alternatives (|)

var s='1,000,000.00';// tests

var result= /(^\d+([,.]\d+)?$)/.test(s) || // no thousand separator

/((^\d{1,3}(,\d{3})+(\.\d+)?)$)/.test(s) || // comma thousand separator

/((^\d{1,3}(\.\d{3})+(,\d+)?)$)/.test(s); // dot thousand separator

alert(result)

Put together its a brute-

function validDelimNum2(s){
var rx=/(^\d+([,.]\d+)?$)|((^\d{1,3}(,\d{3})+(\.\d+)?)$)|((^\d{1,3}(\.\d{3})+(,\d+)?)$)/;
return rx.test(s);
}

//tests

var N= [
'10000000',
    '1,000,000.00',
    '1.000.000,00',
    '1000000.00',
    '1000000,00',
    '1.00.00',
    '1,00,00',
    '1.000,00',
    '1000,000.00'
]
var A= [];
for(var i= 0, L= N.length; i<L; i++){
    A.push(N[i]+'='+validDelimNum2(N[i]));
}
A.join('\n')

/*  returned values
10000000=true
1,000,000.00=true
1.000.000,00=true
1000000.00=true
1000000,00=true
1.00.00=false
1,00,00=false
1.000,00=true
1000,000.00=false
*/

The simplest (though not most elegant by far) method would be to write analogous RE for another case and join them with 'OR', like this:

/^(([1-9][0-9]{0,2}(,[0-9]{3})*(\.[0-9]{2})?)|([1-9][0-9]{0,2}(\.[0-9]{3})*(,[0-9]{2})?))$/

UPDATE

A little cleaned up version

/^[1-9]\d{0,2}(((,\d{3})*(\.\d{2})?)|((\.\d{3})*(,\d{2})?))$/

You can replace the , and \. with [,.] to accept either in either location. It would also make 1,000.000.00 OK though.

Its harder to make the regexp behave like that in JavaScript because you can't use lookbehinds

/^(0|0[.,]\d{2}|[1-9]\d{0,2}((,(\d{3}))*(\.\d{2})?|(\.(\d{3}))*(,\d{2})?))$/

/^      #anchor to the first char of string
(       #start group
  0       # 0
|       # or
  0[.,]   # 0 or 0 followed by a . or ,
  \d{2}   # 2 digits
|       # or
  [1-9]   #match 1-9
  \d{0,2} #0-2 additional digits
  (       #start group
    (,(\d{3}))* # match , and 3 digits zero or more times
    (\.\d{2})?  # match . and 2 digits zero or one
  |       # or
    (\.(\d{3})* # match . and 3 digits zero or more times
    (,\d{2})?   # match , and 2 digits zero or one time
  )       #end group
)       #end group
$/      #anchor to end of string

http://jsfiddle.net/AC3Bm/