如何计算正则表达式的匹配数?

问题描述：

假设我有一个包含以下内容的字符串:

Let's say I have a string which contains this:

HelloxxxHelloxxxHello

我编译一个模式以查找"Hello"

I compile a pattern to look for 'Hello'

Pattern pattern = Pattern.compile("Hello");
Matcher matcher = pattern.matcher("HelloxxxHelloxxxHello");

它应该找到三个匹配项.我怎样才能知道有多少场比赛?

It should find three matches. How can I get a count of how many matches there were?

我已经尝试过各种循环并使用matcher.groupCount()，但是没有用.

I've tried various loops and using the matcher.groupCount() but it didn't work.

答

matcher.find()找不到所有匹配项，只有 next 匹配项.

matcher.find() does not find all matches, only the next match.

long matches = matcher.results().count();

Java 8和更早版本的解决方案

您必须执行以下操作. (从Java 9开始，有一个更好的解决方案)

int count = 0;
while (matcher.find())
    count++;

顺便说一句，matcher.groupCount()完全不同.

Btw, matcher.groupCount() is something completely different.

完整示例:

import java.util.regex.*;

class Test {
    public static void main(String[] args) {
        String hello = "HelloxxxHelloxxxHello";
        Pattern pattern = Pattern.compile("Hello");
        Matcher matcher = pattern.matcher(hello);

        int count = 0;
        while (matcher.find())
            count++;

        System.out.println(count);    // prints 3
    }
}

处理重叠匹配

在计算aaaa中aa的匹配项时，上面的代码段将为您提供 2 .

Handling overlapping matches

When counting matches of aa in aaaa the above snippet will give you 2.

aaaa
aa
  aa

要获得3个匹配项，即此行为:

To get 3 matches, i.e. this behavior:

aaaa
aa
 aa
  aa

您必须在索引<start of last match> + 1处搜索匹配项，如下所示:

You have to search for a match at index <start of last match> + 1 as follows:

String hello = "aaaa";
Pattern pattern = Pattern.compile("aa");
Matcher matcher = pattern.matcher(hello);

int count = 0;
int i = 0;
while (matcher.find(i)) {
    count++;
    i = matcher.start() + 1;
}

System.out.println(count);    // prints 3

如何计算正则表达式的匹配数?

Java 8和更早版本的解决方案

处理重叠匹配

Handling overlapping matches

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