Java: Find index of first Regex

Question

I need to change a piece of code which includes this:

string.indexOf(\"bc\")

How can this be changed by a solution that skips the occurrenc

Answer 1

As requested a more complete solution:

    /** @return index of pattern in s or -1, if not found */
public static int indexOf(Pattern pattern, String s) {
    Matcher matcher = pattern.matcher(s);
    return matcher.find() ? matcher.start() : -1;
}

call:

int index = indexOf(Pattern.compile("(?<!a)bc"), "abc xbc");

Answer 2

Use regex to find the String that matches your criteria, and then find the index of that String.

int index = -1;
Pattern p = Pattern.compile("[^Aa]?bc");
Matcher m = p.matcher(string);
if (m.find()) {
    index = m.start();
}

Something like this. Where 'string' is the text you're searching and 'index' holds the location of the string that is found. (index will be -1 if not found.) Also note that the pattern is case sensitive unless you set a flag.

Answer 3

To add to Arne's answer - if you wanted to also add indexing:

public static int indexOf(String regex, String s, int index)
{
    Pattern pattern = Pattern.compile(regex);
    Matcher matcher = pattern.matcher(s);
    return matcher.find(index) ? matcher.start() : -1;
}

call:

int index = indexOf("(?<!a)bc", "abc xbc", 2);

Answer 4

You could use a regex with a negative lookbehind:

(?<!a)bc

Unfortunately to reproduce .indexOf with Regex in Java is still a mess:

Pattern pattern = Pattern.compile("(?!a)bc");
Matcher matcher = pattern.matcher("abc xbc");
if (matcher.find()) {
    return matcher.start();
}