java正则表达式捕获组在OR运算符后返回null

Question

Matcher matcher = Pattern.compile("\\bwidth\\s*:\\s*(\\d+)px|\\bbackground\\s*:\\s*#([0-9A-Fa-f]+)").matcher(myString);
if (matcher.find()) {
    System.out.println(matcher.group(2));
}

示例数据： myString = width:17px;background:#555;float:left;将产生null。 我想要的是：

matcher.group(1) = 17
matcher.group(2) = 555

我刚开始在Java上使用正则表达式，有什么帮助吗

Answer 1

我建议把事情分开一点

而不是构建一个大型正则表达式（也许你想在字符串中添加更多规则？）您应该将字符串拆分为多个部分：

String myString = "width:17px;background:#555;float:left;";
String[] sections = myString.split(";"); // split string in multiple sections
for (String section : sections) {

  // check if this section contains a width definition
  if (section.matches("width\\s*:\\s*(\\d+)px.*")) {
    System.out.println("width: " + section.split(":")[1].trim());
  }

  // check if this section contains a background definition
  if (section.matches("background\\s*:\\s*#[0-9A-Fa-f]+.*")) {
    System.out.println("background: " + section.split(":")[1].trim());
  }

  ...
}

Answer 2

下面是一个有效的例子。在regexp中包含|（or）通常会让人感到困惑，因此我又添加了两个匹配器，以展示我将如何做到这一点

public static void main(String[] args) {
    String myString = "width:17px;background:#555;float:left";

    int matcherOffset = 1;
    Matcher matcher = Pattern.compile("\\bwidth\\s*:\\s*(\\d+)px|\\bbackground\\s*:\\s*#([0-9A-Fa-f]+)").matcher(myString);
    while (matcher.find()) {
        System.out.println("found something: " + matcher.group(matcherOffset++));
    }

    matcher = Pattern.compile("width:(\\d+)px").matcher(myString);
    if (matcher.find()) {
        System.out.println("found width: " + matcher.group(1));
    }

    matcher = Pattern.compile("background:#(\\d+)").matcher(myString);
    if (matcher.find()) {
        System.out.println("found background: " + matcher.group(1));
    }
}