Java正则表达式模式 [英] Java regex patterns

问题描述

我需要帮助解决这个问题。请看以下正则表达式：

I need help with this matter. Look at the following regex:

Pattern pattern = Pattern.compile("[A-Za-z]+(\\-[A-Za-z]+)");
Matcher matcher = pattern.matcher(s1);

我想找这样的字：自制，aaaa-bbb和不是aaa - bbb，而是不
aaa - aa - aaa。基本上，我想要以下内容：

I want to look for words like this: "home-made", "aaaa-bbb" and not "aaa - bbb", but not "aaa--aa--aaa". Basically, I want the following:

word - hyphen - word。

word - hyphen - word.

它适用于所有事情，除了这种模式将通过：aaa - aaa - aaa而不应该。什么正则表达式适用于这种模式？

It is working for everything, except this pattern will pass: "aaa--aaa--aaa" and shouldn't. What regex will work for this pattern?

推荐答案

可以从表达式中删除反斜杠：

Can can remove the backslash from your expression:

"[A-Za-z]+-[A-Za-z]+"

以下代码应该可以使用

Pattern pattern = Pattern.compile("[A-Za-z]+-[A-Za-z]+");
Matcher matcher = pattern.matcher("aaa-bbb");
match = matcher.matches();

请注意，您可以使用 Matcher.matches（） 而不是 Matcher.find（）以检查匹配的完整字符串。

Note that you can use Matcher.matches() instead of Matcher.find() in order to check the complete string for a match.

如果相反，你想要使用 Matcher.find（）查看字符串内部你可以使用表达式

If instead you want to look inside a string using Matcher.find() you can use the expression

"(^|\\s)[A-Za-z]+-[A-Za-z]+(\\s|$)"

但请注意，只会找到以空格分隔的单词（即没有像 aaa-bbb这样的单词。）。为了捕捉这种情况，你可以使用lookbehinds和lookaheads：

but note that then only words separated by whitespace will be found (i.e. no words like aaa-bbb.). To capture also this case you can then use lookbehinds and lookaheads:

"(?<![A-Za-z-])[A-Za-z]+-[A-Za-z]+(?![A-Za-z-])"

将会读取

(?<![A-Za-z-])        // before the match there must not be and A-Z or -
[A-Za-z]+             // the match itself consists of one or more A-Z
-                     // followed by a -
[A-Za-z]+             // followed by one or more A-Z
(?![A-Za-z-])         // but afterwards not by any A-Z or -

一个例子：

Pattern pattern = Pattern.compile("(?<![A-Za-z-])[A-Za-z]+-[A-Za-z]+(?![A-Za-z-])");
Matcher matcher = pattern.matcher("It is home-made.");
if (matcher.find()) {
    System.out.println(matcher.group());    // => home-made
}

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