Regular expression not working despite testing

Question

I'm trying to enforce validation of an ID that includes the first two letters being letters and the next four being numbers, there can be one 0 i.e. 0333 but can never be full zeroes with 0000 therefore something like ID0000 is not allowed. The expression I came up with seems to check out when testing it online but doesn't seem to work when trying to enforce it in the program:

\b(?![A-Z]{2}[0]{4})[A-Z]{2}[0-9]{4}\b

and heres the code I'm currently using to implement it:

String pattern = "/\b(?![A-Z]{2}[0]{4})[A-Z]{2}[0-9]{4}\b/";
Pattern regEx = Pattern.compile(pattern);
String ingID = ingredID.getText().toString();
Matcher m = regEx.matcher(ingID);

if (m.matches()) {
    ingredID.setError("Please enter a valid Ingrediant ID");
}

For some reason it doesn't seem to validate correctly with accepting ids like ID0000 when it shouldn't be. Any thoughts folks ?

The `/` is a regex string delimiter. So you shouldn't have both `/` and `"`. Past that, it depends on what language you're using--what is this? — Kyle Strand, Apr 29 '14 at 19:48

score 0 · Accepted Answer · answered Apr 29 '14 at 19:49

0

Change your regex pattern to "\\b(?![A-Z]{2}[0]{4})[A-Z]{2}[0-9]{4}\\b"

answered Apr 29 '14 at 19:49

Karakuri

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score 0 · Answer 2 · edited May 23 '17 at 12:05

Your problem is essentially that Java isn't all that Regex-friendly; you need to deal with the limitations of Java strings in order to create a string that can be used as a Regex pattern. Since \ is the escape character in Regex and the escape character in Java strings (and since there's no such thing as a raw string literal in Java), you must double-escape anything that must be escaped in the Regex in order to create a literal \ character within the Java string, which, when parsed as a Regex pattern, will be correctly treated as the escape character.

So, for instance, the Regex pattern /\b/ (where /, as mentioned in my comment, delimits the pattern itself) would be represented in Java as the string "\\b".

Regular expression not working despite testing

2 Answers2