Question

我无法理解组的输出，它是如何考虑模式中的每个paranthesis并将其与变量'line'等同于此处。请解释

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegexTut3 {

  public static void main(String args[]) {
    String line = "This order was placed for QT3000! OK?"; 
    String pattern = "(.*)(\\d+)(.*)";

    // Create a Pattern object
    Pattern r = Pattern.compile(pattern);

    // Now create matcher object.
    Matcher m = r.matcher(line);

    if (m.find()) {
      System.out.println("Found value: " + m.group(0));
      System.out.println("Found value: " + m.group(1));
      System.out.println("Found value: " + m.group(2));
    } else {
      System.out.println("NO MATCH");
    }
  }
}

Answer 1

模式"(.*)(\\d+)(.*)"表示：

(.*) - Consume anything and store it in group 0
(\\d+) - Consume a number and store it in the next group, i.e. group 2
(.*) - Consume anything and store it in the next group

通过这种方式，模式将在字符串中找到一个数字，并将其中的所有内容存储在组0中，组1中的数字以及组2中其余字符串的数字

Answer 2

m.group（0）始终匹配整个模式。你的比赛实际上从1开始，而不是0。

m.group(0): matches the whole pattern
m.group(1):  first parenthesis
m.group(2):  second set of parenthesis
m.group(3): third set

编辑：纠正

Java Regex组

2 个答案: