Java Regular Expression Matcher 没有找到所有可能的匹配项

Question

我在 TutorialsPoint 上查看一段代码，从那以后有些事情一直困扰着我……看看这段代码:

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

public class RegexMatches
{
    public static void main( String args[] ){

      // String to be scanned to find the pattern.
      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);
      while(m.find( )) {
         System.out.println("Found value: " + m.group(1));
         System.out.println("Found value: " + m.group(2));
         System.out.println("Found value: " + m.group(3));
      }
   }
}

此代码成功打印:

Found value: This was placed for QT300 
Found value: 0
Found value: ! OK?

但是根据正则表达式 "(.*)(\\d+)(.*)"，为什么它不返回其他可能的结果，例如:

Found value: This was placed for QT30 
Found value: 00
Found value: ! OK?

或

Found value: This was placed for QT 
Found value: 3000
Found value: ! OK?

如果此代码不适合这样做，那么我如何编写一个可以找到所有可能匹配项的代码？

Answer 1

这是因为 greediness的 *然后是backtracking .

字符串:

This order was placed for QT3000! OK?

正则表达式:

(.*)(\\d+)(.*)

我们都知道.*是贪心的，尽可能匹配所有字符。所以第一个.*匹配所有字符直到最后一个字符 ?然后它回溯以提供匹配。我们正则表达式中的下一个模式是 \d+ ，所以它回溯到一个数字。一旦它找到一个数字，\d+匹配该数字，因为此处满足条件( \d+ 匹配一个或多个数字)。现在第一个(.*)捕获 This order was placed for QT300和以下 (\\d+)捕获数字 0位于 ! 之前符号。

现在下一个模式(.*)捕获所有剩余字符 !<space>OK? . m.group(1)指的是存在于组索引 1 和 m.group(2) 中的字符指的是索引 2，就这样继续下去。

查看演示 here .

得到你想要的输出。

String line = "This order was placed for QT3000! OK?";
  String pattern = "(.*)(\\d{2})(.*)";

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

  // Now create matcher object.
  Matcher m = r.matcher(line);
  while(m.find( )) {
     System.out.println("Found value: " + m.group(1));
     System.out.println("Found value: " + m.group(2));
     System.out.println("Found value: " + m.group(3));
  }

输出:

Found value: This order was placed for QT30
Found value: 00
Found value: ! OK?

(.*)(\\d{2}) , 回溯最多两位数以提供匹配。

把你的模式改成这个，

String pattern = "(.*?)(\\d+)(.*)";

要得到这样的输出，

Found value: This order was placed for QT
Found value: 3000
Found value: ! OK?

?在*之后强制 *进行非贪婪匹配。

使用额外的捕获组从单个程序中获取输出。

String line = "This order was placed for QT3000! OK?";
String pattern = "((.*?)(\\d{2}))(?:(\\d{2})(.*))";
Pattern r = Pattern.compile(pattern);
      Matcher m = r.matcher(line);
      while(m.find( )) {
         System.out.println("Found value: " + m.group(1));
         System.out.println("Found value: " + m.group(4));
         System.out.println("Found value: " + m.group(5));
         System.out.println("Found value: " + m.group(2));
         System.out.println("Found value: " + m.group(3) + m.group(4));
         System.out.println("Found value: " + m.group(5));
     }

输出:

Found value: This order was placed for QT30
Found value: 00
Found value: ! OK?
Found value: This order was placed for QT
Found value: 3000
Found value: ! OK?