如何在Java中用regex匹配嵌套重复组？

Question

String s = "The very first line\n"
        + "\n"
        + "AA (aa)\n"
        + "BB (bb)\n"
        + "CC (cc)\n"
        + "\n";

Pattern p = Pattern.compile(
        "The very first line\\s+"
        + "((?<gr1>[a-z]+)\\s+\\((?<gr2>[^)]+)\\)\\s*)+",
        Pattern.DOTALL | Pattern.CASE_INSENSITIVE);

Matcher m = p.matcher(s);

if (m.find()) {
    for (int i = 0; i <= m.groupCount(); i++) {
        System.out.println("group #" + i + ": [" + m.group(i).trim() + "]");
    }
    System.out.println("group gr1: [" + m.group("gr1").trim() + "]");
    System.out.println("group gr2: [" + m.group("gr2").trim() + "]");
}

group #0: [The very first line

AA (aa)
BB (bb)
CC (cc)]
group #1: [CC (cc)]
group #2: [CC]
group #3: [cc]
group gr1: [CC]
group gr2: [cc]

edit:第一行作为标识字符串在模式中-请参见下面对gknicker答案的注释

Answer 1

似乎您希望您的模式不匹配整个输入字符串，而只匹配单个重复部分。如果这是真的，您的模式将是：

    Pattern p = Pattern.compile(
        "((?<gr1>[a-z]+)\\s+\\((?<gr2>[^)]+)\\))",
        Pattern.CASE_INSENSITIVE);

在本例中，您将使用while循环来查找每个匹配项：

    Matcher m = p.matcher(s);

    while (m.find()) {
        System.out.println("group gr1: ["
            + m.group("gr1").trim() + "]");
        System.out.println("group gr2: ["
            + m.group("gr2").trim() + "]");
    }

但是如果您需要整个匹配，您可能必须使用以下两个模式：

    String s = "The very first line\n"
        + "\n"
        + "AA (aa)\n"
        + "BB (bb)\n"
        + "CC (cc)\n"
        + "\n";

    Pattern p = Pattern.compile(
        "The very first line\\s+(([a-z]+)\\s+\\(([^)]+)\\)\\s*)+",
        Pattern.CASE_INSENSITIVE);

    Pattern p2 = Pattern.compile(
        "((?<gr1>[a-z]+)\\s+\\((?<gr2>[^)]+)\\))",
        Pattern.CASE_INSENSITIVE);

    Matcher m = p.matcher(s);
    while (m.find()) {
        Matcher m2 = p2.matcher(m.group());
        while (m2.find()) {
            System.out.println("group gr1: ["
                + m2.group("gr1").trim() + "]");
            System.out.println("group gr2: ["
                + m2.group("gr2").trim() + "]");
        }
    }