Regular Expression Escaped Digit "\d" Illegal Escape Character

Hello,
I'm trying to write a regular expression to determine if a String matches a date format that is defined as YYYYMMDD. For example, March 11, 2009 would be "20090311"
For the time being I don't care if an invalid month or day is entered. I've attempted both of the following
if (date.matches("(19|20)\d{4}")) {
// warn the user
}and
if (java.util.regex.Pattern.matches("(19|20)\d{4}"), date)) {
// warn the user
}And both yield Illegal Escape Character compilation errors, for the "\d" part of the regular expression.
http://java.sun.com/j2se/1.4.2/docs/api/java/util/regex/Pattern.html#sum
Says that "\d" is the predefined digit character class. So at this point, I don't know what I'm doing wrong. I realize I could just define the character class myself, and use a pattern like "(19|20)[0-9]{4}", but I would like to know why "\d" isn't being recognized by the compiler.
Thanks,
Paul

paulwooten wrote:
Can someone give me an explanation of heuristics, as they might apply to SimpleDateFormat? Does this mean that if the format was similar the parser might figure it out? Say, if instead of "yyyyMMdd", it was "yyyyddMM", or "yyMMdd"?No. Since all of these are valid formats, there's no way for the parser to distinguish this.
Or does this have to do with rejecting February 29, and other dates like that.That's the one. When setLenient(false) is called, then the 29th February is only accepted in leap years.
It will also reject the 57th January when lenient is set to false (try parsing that with lenient=true, you'll be surprised).
I've read some of the wikipedia article about heuristics, but I'm confused as to how it would apply to this example.Don't concentrate to much on the term heuristics. Just remember: lenient=true means that not-really-correct dates will be accepted, lenient=false means more strict checks.

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