创建正则表达式匹配数组

创建正则表达式匹配数组

IT小君   2021-09-15T00:08:51

在 Java 中,我试图将所有正则表达式匹配项返回到一个数组,但似乎您只能检查该模式是否匹配某些内容(布尔值)。

如何使用正则表达式匹配来形成与给定字符串中的正则表达式匹配的所有字符串的数组?

服务器费用不足...
评论(6)
IT小君

如果您可以假设 Java >= 9,则4castle 的答案比下面的要好)

您需要创建一个匹配器并使用它来迭代查找匹配项。

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

 ...

 List<String> allMatches = new ArrayList<String>();
 Matcher m = Pattern.compile("your regular expression here")
     .matcher(yourStringHere);
 while (m.find()) {
   allMatches.add(m.group());
 }

在此之后,allMatches包含匹配项,allMatches.toArray(new String[0])如果您确实需要一个数组,您可以使用它来获取数组。


您还可以使用MatchResult编写辅助函数来循环匹配,因为Matcher.toMatchResult()返回当前组状态的快照。

例如你可以写一个懒惰的迭代器来让你做

for (MatchResult match : allMatches(pattern, input)) {
  // Use match, and maybe break without doing the work to find all possible matches.
}

通过做这样的事情:

public static Iterable<MatchResult> allMatches(
      final Pattern p, final CharSequence input) {
  return new Iterable<MatchResult>() {
    public Iterator<MatchResult> iterator() {
      return new Iterator<MatchResult>() {
        // Use a matcher internally.
        final Matcher matcher = p.matcher(input);
        // Keep a match around that supports any interleaving of hasNext/next calls.
        MatchResult pending;

        public boolean hasNext() {
          // Lazily fill pending, and avoid calling find() multiple times if the
          // clients call hasNext() repeatedly before sampling via next().
          if (pending == null && matcher.find()) {
            pending = matcher.toMatchResult();
          }
          return pending != null;
        }

        public MatchResult next() {
          // Fill pending if necessary (as when clients call next() without
          // checking hasNext()), throw if not possible.
          if (!hasNext()) { throw new NoSuchElementException(); }
          // Consume pending so next call to hasNext() does a find().
          MatchResult next = pending;
          pending = null;
          return next;
        }

        /** Required to satisfy the interface, but unsupported. */
        public void remove() { throw new UnsupportedOperationException(); }
      };
    }
  };
}

有了这个,

for (MatchResult match : allMatches(Pattern.compile("[abc]"), "abracadabra")) {
  System.out.println(match.group() + " at " + match.start());
}

产量

a at 0
b at 1
a at 3
c at 4
a at 5
a at 7
b at 8
a at 10
2021-09-15T00:08:52   回复
IT小君

在Java中9,你现在可以用Matcher#results()得到Stream<MatchResult>,你可以用它来获得匹配的列表/阵列。

import java.util.regex.Pattern;
import java.util.regex.MatchResult;
String[] matches = Pattern.compile("your regex here")
                          .matcher("string to search from here")
                          .results()
                          .map(MatchResult::group)
                          .toArray(String[]::new);
                    // or .collect(Collectors.toList())
2021-09-15T00:08:52   回复
IT小君

Java 使正则表达式过于复杂,并且不遵循 perl 风格。看看MentaRegex,看看如何在一行 Java 代码中完成它:

String[] matches = match("aa11bb22", "/(\\d+)/g" ); // => ["11", "22"]
2021-09-15T00:08:52   回复
IT小君

这是一个简单的例子:

Pattern pattern = Pattern.compile(regexPattern);
List<String> list = new ArrayList<String>();
Matcher m = pattern.matcher(input);
while (m.find()) {
    list.add(m.group());
}

(如果你有更多的捕获组,你可以通过它们的索引作为 group 方法的参数来引用它们。如果你需要一个数组,那么使用list.toArray()

2021-09-15T00:08:52   回复
IT小君

来自官方正则表达式 Java Trails

        Pattern pattern = 
        Pattern.compile(console.readLine("%nEnter your regex: "));

        Matcher matcher = 
        pattern.matcher(console.readLine("Enter input string to search: "));

        boolean found = false;
        while (matcher.find()) {
            console.format("I found the text \"%s\" starting at " +
               "index %d and ending at index %d.%n",
                matcher.group(), matcher.start(), matcher.end());
            found = true;
        }

使用find并将结果插入group您的数组/列表/任何内容。

2021-09-15T00:08:53   回复
IT小君
        Set<String> keyList = new HashSet();
        Pattern regex = Pattern.compile("#\\{(.*?)\\}");
        Matcher matcher = regex.matcher("Content goes here");
        while(matcher.find()) {
            keyList.add(matcher.group(1)); 
        }
        return keyList;
2021-09-15T00:08:53   回复