Java匹配正则表达式并提取组oneliner

Question

I have a simple regex like

%(\d+)\$@[_a-zA-Z0-9]+@

I don't want to write

Matcher m = Pattern.compile(myRegex).matcher(myText);
if (m.matches())
   // do something with m.group(1);

What I would really like to do is to have a one liner like

// do something with
Pattern.compile(myRegex).matcher(myText).match().group(1);

Do you know a good way to do that in Java (I'm using Java 7 but perhaps something changed in 8)?

Answer 1

As of Java 9, you can stream match results and grab the first group of the first result:

String result = Pattern.compile(myRegex)
        .matcher(myText)
        .results()
        .map(m -> m.group(1))
        .findFirst()
        .orElse(null);

Answer 2

Here is one :

Integer.valueOf(Pattern.compile(myRegex).matcher(myText).matches() ? 
        Pattern.compile(myRegex).matcher(myText).group(1) : "0");
//-----------------------------------------Not match---------^

Edit

You ca also use :

Matcher m;
Integer.valueOf((m = Pattern.compile(myRegex).matcher(myText)).matches() ? 
        m.group(1) : "0");

Answer 3

create a static matcher:

private static Matcher matcher = Pattern.compile(myRegex).matcher("");

and use it this way:

public String match(String myText, String defaultValue) {
  matcher.reset(myText);
  return matcher.matches() ? matcher.group(1) : defaultValue;
}

This is the most efficient way of using regexs (as far as I know).