Ask Ben: Cleaning Two Digit Years Using Javascript And Regular Expressions
Posted January 26, 2007 at 8:34 AM by Ben Nadel
Hey Ben,
Trying to replace a date "/" character with both the character and either a string of 19 or 20, for 2 year purposes.
For example, people are accustomed to plugging in the dates into their form with mm/dd/yy format. So when they plug in 12/23/97, I'd like to replace the second instance of / with /19 preceeding the 97.
If the last two year figures begin with 0 or 1, I'd replace the "/" with "/20" preceeding the rest of the figure. i.e. 09/17/02 would become 09/17/2002.
Is there a way to replace this with regular expressions? How would that be done?
The beauty of using Regular Expressions in Javascript is that each match made by the regular expression can be passed off to a Javascript function literal. You won't see this happening in any kind of compiled language such as Java or ColdFusion, which is part of why Javascript, as an interpreted language, is so freakin' sweet-ass!
For this solution, all we have to do is write a function that takes a given date string, checks for a two digit year, check the leading character and then returns the two digits with the proper two digits pre-pended.
- // This takes a date string that MIGHT have a two digit year
- // as the last two digits. If it does, this function replaces
- // the two digit year with what it *assumes* is the proper
- // four digit year.
- function CleanDate( strDate ){
-
- // Return the cleaned date.
- return(
- strDate.replace(
-
- // This regular expression will search for a slash
- // followed by EXACTLY two digits at the end of
- // this date string. The two digits are being
- // grouped together for future referencing.
- new RegExp( "/(\\d{2})$", "" ),
-
- // We are going to pass the match made by the
- // regular expression off to this function literal.
- // Our arguments are as follows:
- // $0 : The entire match found.
- // $1 : The first group within the match.
- function( $0, $1 ){
-
- // Check to see if our first group begins with
- // a zero or a one. If so, replace with 20 else
- // replace with 19.
- if ($1.match( new RegExp( "^[01]{1}", "" ) )){
-
- // Replace with 20.
- return( "/20" + $1 );
-
- } else {
-
- // Replace with 19.
- return( "/19" + $1 );
-
- }
-
- }
- )
- );
-
- }
Hope that helps!
Reader Comments
Having users enter a two digit year is part of the Y2K issue. A common solution for 2 year digit data was to use a pivot formula, for the below example, I set 70 as the pivot year.
Easy enough to set the pivot base, depending on the type of question being asked, such as your birth date, and then depending on the age of your user base, lower the pivot. For me, entering in 61 would result in 2061, so I'd probably lower the pivot date a bit.
<script language="javascript">
function getPivot(year) {
if (year < 70) return (2000 + year);
if (year < 1900) return (1900 + year);
return year;
}
function getDate(){
var myDate = new Date(document.getElementById('testDate').value);
var pivotDate = new Date(getPivot(myDate.getYear()),myDate.getMonth(),myDate.getDate());
alert(pivotDate);
}
</script>
<input id="testDate" value="" onchange="getDate();" />
Chris,
That's a cool little snippet of code there. Thanks!
Ben,
your example is pretty nifty. I didn't realize that you can do the equivalent of the matchEvaulator in .NET but with Java/CF Script! Nice! On a more relevant note do deal with the issue I usually just provide a calendar widget. The one at http://www.dynarch.com/projects/calendar/ is free, full of features and customizable. I use it all the time. A quick question for you, why did you have to escape the "\" in front of the special sequence "\d" as in "\\d"?
On a side note, you never got back to me about the code snippets idea.
Boyan,
"\d" matches any numeric digit (0-9) as "\\d" will match the same digits as literal text -- which is helpful when using regex to evaluate a numeric value that is of data type string.
Nice post Ben!
@Ryan,
Good point; however, it depends on the context. In Javascript strings, "\", is actually a special characters. For example, "\n" is the new line character:
"line one \n line two"
As such, if you want to create a special characters class in a Javascript RegExp, you actually have to escape the back-slash so that it doesn't try to get evaluated as a special string character.
So, to get the digit character class, \d, in a string in Javascript, you need to escape the back slash:
"\\d"
Now, that's IF you use the new RegExp() constructor. If you use the implicit regex constructor, you don't have to:
/\d/gi
In this case (which I find much less readable), since you are outside of a string literal, there's no need to escape the back-slashes.
Sorry if that was WAY more information than anyone needed.
