正则表达式第一个字符不匹配

Question

I am having some Java Pattern problems. This is my pattern:

"^[\\p{L}\\p{Digit}~._-]+$"

It matches any letter of the US-ASCII, numerals, some special characters, basically anything that wouldn't scramble an URL.

What I would like is to find the first letter in a word that does not match this pattern. Basically the user sends a text as an input and I have to validate it and to throw an exception if I find an illegal character.

I tried negating this pattern, but it wouldn't compile properly. Also find() didn't help out much.

A legal input would be hello while ?hello should not be, and my exception should point out that ? is not proper.

I would prefer a suggestion using Java's Matcher, Pattern or something using util.regex . Its not a necessity, but checking each character in the string individually is not a solution.

Edit: I came up with a better regex to match unreserved URI characters

Answer 1

Try this :

^[\\p{L}\\p{Digit}.'-.'_]*([^\\p{L}\\p{Digit}.'-.'_]).*$

The first character non matching is the group n°1

I made a few try here : http://fiddle.re/gkkzm6 1

Explanation :

I negate your pattern, so i built this :

[^\\p{L}\\p{Digit}.'-.'_]      [^...] means every character except for
^                       ^             the following ones.
|  your pattern inside  |

The pattern has 3 parts :

^[\\p{L}\\p{Digit}.'-.'_]*

Checks the regex from the first character until he meets a non matching character

([^\\p{L}\\p{Digit}.'-.'_]) 

The non-matching character (negation) inside a capturing group

.*$

Any character until the end of the string.

Hope it helps you

EDIT :

The correct regex shoud be :

^[\\p{L}\\p{Digit}~._-]*([^\\p{L}\\p{Digit}~._-]).*$

It is the same method, i only change the contents of the first and second part.

I tried and it seems to work.

Answer 2

Try out this one to find the first non valid char:

Pattern negPattern = Pattern.compile(".*?([^\\p{L}^\\p{Digit}^.^'-.'^_]+).*");
Matcher matcher = negPattern.matcher("hel?lo");
if (matcher.matches())
{
    System.out.println("'" + matcher.group(1).charAt(0) + "'");
}

Answer 3

The "^[\\\\p{L}\\\\p{Digit}.'-.'_]+$" pattern matches any string containing 1+ characters defined inside the character class. Note that double ' and . are suspicious and you might be unaware of the fact that '-. creates a range and matches '()*+,-. . If it is not on purpose, I think you meant to use .'_- .

To check if a string starts with a character other than the one defined in the character class, you can negated the character class, and check the first character in the string only:

if (str.matches("[^\\p{L}\\p{Digit}.'_-].*")) {
   /* String starts with the disallowed character */ 
}

I also think you can shorten the regex to "(?U)[^\\\\w.'-].*" . At any rate, \\\\p{Digit} can be replaced with \\\\d .