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;
}
Pattern pattern = Pattern.compile(regexPattern);
List<String> list = new ArrayList<String>();
Matcher m = pattern.matcher(input);
while (m.find()) {
list.add(m.group());
}
for (MatchResult match : allMatches(pattern, input)) {
// Use match, and maybe break without doing the work to find all possible matches.
}
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());
}
6 回答
来自Official Regex Java Trails:
使用
find
并将结果group
插入您的数组/列表/等等 .这是一个简单的例子:
(如果你有更多的捕获组,你可以通过它们的索引引用它们作为组方法的参数 . 如果你需要一个数组,那么使用
list.toArray()
)Java使得regex过于复杂,并且它不遵循perl风格 . 看看MentaRegex,了解如何在一行Java代码中实现这一目标:
在Java 9中,您现在可以使用Matcher#results()来获取Stream<MatchResult>,您可以使用它来获取匹配列表/数组 .
(如果您可以假设Java> = 9,则4castle's answer优于以下内容)
您需要创建一个匹配器并使用它来迭代查找匹配项 .
在此之后,
allMatches
包含匹配项,如果您确实需要,可以使用allMatches.toArray(new String[0])
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,您可以编写一个惰性迭代器来执行此操作
做这样的事情:
有了这个,
产量