index of a regex in a String

[Guest]

How can I implement this?
(Without checking characters one-by-one with my code)

String a = " abc";
int b = a.indexOfRegex("\s", 0);
// b is 2 (index of first non-whitespace character)

 

[John McGrath]

How can I implement this?
(Without checking characters one-by-one with my code)

String a = " abc";
int b = a.indexOfRegex("\s", 0);
// b is 2 (index of first non-whitespace character)

Note that you need to escape the backslash: "\\s", not "\s".

Pattern pattern = Pattern.compile( "\\s" );
Matcher matcher = pattern.matcher( a );
if ( matcher.find() ) {
int start = matcher.start();
} else {
// not found
}

 

[Alan Moore]

Note that you need to escape the backslash: "\\s", not "\s".

Pattern pattern = Pattern.compile( "\\s" );
Matcher matcher = pattern.matcher( a );
if ( matcher.find() ) {
int start = matcher.start();
} else {
// not found
}

That will return the index of the first *whitespace* character, zero.
If you want the index of the first non-whitespace character, either
change the regex to "\\S" or use matcher.end().

 

[John McGrath]

That will return the index of the first whitespace character, zero.
If you want the index of the first non-whitespace character, either
change the regex to "\\S" or use matcher.end().

Yes, that is true. I think the OP was looking for how to find the
location that an arbitrary pattern matches - which is what the code I
posted does. I was not paying attention to the actual pattern he provided.