Problem with simple regular expression

[haiaw]

String stringTest="555";
Pattern p = Pattern.compile("^[0-9]");
Matcher m=p.matcher(stringTest);
while(m.find()){

throw new RuntimeException("FOUND BAD THING");
}

I want to throw exception if stringTest has not only numbers, but this code always throws RuntimeException("FOUND BAD THING");

[haiaw]

thanks, i found the solution
But what is the difference between [^0-9] and ^[0-9] ?

[destin]

thanks, i found the solution
But what is the difference between [^0-9] and ^[0-9] ?

One works and one doesn't. Look at the Pattern class.

Character classes
[abc] a, b, or c (simple class)
[^abc] Any character except a, b, or c (negation)
[a-zA-Z] a through z or A through Z, inclusive (range)
[a-d[m-p]] a through d, or m through p: [a-dm-p] (union)
[a-z&&[def]] d, e, or f (intersection)
[a-z&&[^bc]] a through z, except for b and c: [ad-z] (subtraction)
[a-z&&[^m-p]] a through z, and not m through p: [a-lq-z](subtraction)

There is no ^[abc].

[graviton]

particullarly there is a ^[0-9].
^ not within squared brackets and at the start of a regex means to match from beginning of input.
while the regex "abc" matches in "xxabc"
the regex "^abc" doesn't, since the pattern doesn't appear at the beginning of the input.
in square brackets ^ means negotiation.