Java 如何使用正则表达式匹配括号内的文本？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/1337289/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How do I match text within parentheses using regex?
提问by northpole
I have the following pattern:
我有以下模式：
(COMPANY) -277.9887 (ASP,) -277.9887 (INC.)
I want the final output to be:
我希望最终输出是：
COMPANY ASP, INC.
公司 ASP, INC.
Currently I have the following code and it keeps returning the original pattern ( I assume because the group all falls between the first '(' and last ')'
目前我有以下代码，它不断返回原始模式（我假设是因为该组都落在第一个 '(' 和最后一个 ')' 之间
Pattern p = Pattern.compile("((.*))",Pattern.DOTALL);
Matcher matcher = p.matcher(eName);
while(matcher.find())
{
System.out.println("found match:"+matcher.group(1));
}
I am struggling to get the results I need and appreciate any help. I am not worried about concatenating the results after I get each group, just need to get each group.
我正在努力获得我需要的结果并感谢任何帮助。我不担心在我得到每个组后连接结果，只需要得到每个组。
采纳答案by chaos
Pattern p = Pattern.compile("\((.*?)\)",Pattern.DOTALL);
回答by Oliver
Not a direct answer to your question but I recommend you use RegxTesterto get to the answer and any future question quickly. It allows you to test in realtime.
不是您问题的直接答案，但我建议您使用RegxTester快速获得答案和任何未来的问题。它允许您进行实时测试。
回答by ptomli
Your .* quantifier is 'greedy', so yes, it's grabbing everything between the first and last available parenthesis. As chaos says, tersely :), using the .*? is a non-greedy quantifier, so it will grab as little as possible while still maintaining the match.
您的 .* 量词是“贪婪的”，所以是的，它抓取了第一个和最后一个可用括号之间的所有内容。正如混乱所说，简洁:)，使用 .*? 是一个非贪婪的量词，所以它会在保持匹配的同时尽可能少地抓取。
And you need to escape the parenthesis within the regex, otherwise it becomes another group. That's assuming there are literal parenthesis in your string. I suspect what you referred to in the initial question as your pattern is in fact your string.
并且您需要对正则表达式中的括号进行转义，否则它将成为另一组。那是假设您的字符串中有文字括号。我怀疑您在最初的问题中提到的内容，因为您的模式实际上是您的字符串。
Query: are "COMPANY", "ASP," and "INC." required?
查询：是“COMPANY”、“ASP”和“INC”。需要吗？
If you must have values for them, then you want to use + instead of *, the + is 1-or-more, the * is zero-or-more, so a * would match the literal string "()"
如果您必须为它们指定值，那么您想使用 + 而不是 *，+ 是 1 个或多个，* 是零个或多个，因此 * 将匹配文字字符串“()”
eg: "((.+?))"
例如：“((.+?))”
回答by Brent Writes Code
If your strings are always going to look like that, you could get away with just using a couple calls to replaceAll instead. This seems to work for me:
如果您的字符串总是看起来像那样，您可以只使用几个调用来代替 replaceAll。这似乎对我有用：
String eName = "(COMPANY) -277.9887 (ASP,) -277.9887 (INC.)";
String eNameEdited = eName.replaceAll("\).*?\("," ").replaceAll("\(|\)","");
System.out.println(eNameEdited);
Probably not the most efficient thing in the world, but fairly simple.
可能不是世界上最有效的东西，但相当简单。
回答by Chetan Laddha
Tested with Java 8: /** * Below Pattern returns the string inside Parenthesis.
用 Java 8 测试：/** * 下面的模式返回括号内的字符串。
* Description about casting regular expression: \(+\s*([^\s)]+)\s*\)+
* \(+ : Exactly matches character "(" at least once
* \s* : matches zero to any number white character.
* ( : Start of Capturing group
* [^\s)]+: match any number of character except ^, ) and spaces.
* ) : Closing of capturing group.
* \s*: matches any white character(0 to any number of character)
* \)*: Exactly matches character ")" at least once.
private static Pattern REGULAR_EXPRESSION = Pattern.compile("\(+\s*([^\s)]+)\s*\)+");
