Java regex not compiling

Question

I have the code

private String regexHHMM = "^([0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]$";

The code does not compile with the error "Unclosed Character class"

This must be something very basic, is there any particular escaping I should be using, also I am mainly interested on WHY it cannot be accepted by the java compiler.

UPDATE: I have tried

Pattern.quote("^([0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]$");

And now it compiles, but the regex does not match the HH:MM format now...

Pattern.quote is for escaping all special character in the String, so that when it is compiled, the regex will match the characters in the original String literally. — nhahtdh
– nhahtdh, Commented Nov 26, 2014 at 5:08

hwnd · Accepted Answer · 2014-11-26 04:48:04Z

4

You need to remove the initial open square bracket inside of your pattern.

^([0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]$
  ^

Should be:

^(0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]$

answered Nov 26, 2014 at 4:48

hwnd

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5 Comments

Juan Antonio Gomez Moriano Over a year ago

Good one, and I guess the reason it does not compile is because the java compiler parses every string?

nhahtdh Over a year ago

@Gary: The character class can be nested arbitrarily deep in Java. It's called character class union.

ajb Over a year ago

@JuanAntonioGomezMoriano Are you sure it "does not compile"? Your declaration of regexHHMM is just a String variable initialized with a string literal. As far as I know, it isn't parsed as a regex until runtime when you call Pattern.compile or some other methods. What exactly is the "compiler"'s behavior?

nhahtdh Over a year ago

@ajb: The regex is compiled at run time by Pattern class (implicitly via some method in String class or explicitly), and the OP's regex return a PatternSyntaxException.

ajb Over a year ago

@nhahtdh Right, "at run time" ... but OP is saying it's the compiler doing it, and asking if the "java compiler parses every string"...

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