假设我有这个文字输入。
tes{}tR{R{abc}aD{mnoR{xyz}}}
我想提取ff输出：
R{abc}
R{xyz}
D{mnoR{xyz}}
R{R{abc}aD{mnoR{xyz}}}
目前，我只能使用msdn中的平衡组方法提取{}组内的内容。这是模式：
^[^{}]*(((?'Open'{)[^{}]*)+((?'Target-Open'})[^{}]*)+)*(?(Open)(?!))$
有人知道如何在输出中包含R {}和D {}吗？
答案 0 :(得分：3)
我认为这里需要采用不同的方法。在匹配第一个较大的群组R{R{abc}aD{mnoR{xyz}}}后（请参阅我对可能的拼写错误的评论），由于正则表达式不允许您捕获个人，因此您无法在内部获取子群组{ {1}}群组。
所以，必须有一些方法来捕获而不是消费，显而易见的方法是使用积极的前瞻。从那里，你可以把你使用的表达，虽然有一些变化，以适应新的焦点变化，我想出了：
R{ ... }[我也改名为＆＃39; Open＆＃39;到了＆＃39; O＆＃39;并删除了近大括号的命名捕获，使其缩短并避免在匹配中产生噪音]
在regexhero.net（我目前唯一知道的免费.NET正则表达式测试程序）中，我得到了以下捕获组：
(?=([A-Z](?:(?:(?'O'{)[^{}]*)+(?:(?'-O'})[^{}]*?)+)+(?(O)(?!))))
正则表达式的细分：
1: R{R{abc}aD{mnoR{xyz}}}
1: R{abc}
1: D{mnoR{xyz}}
1: R{xyz}
我实际上想试试它应该如何在PCRE引擎上运行，因为可以选择使用递归正则表达式，我认为它更容易，因为我对它更熟悉并且产生了更短的正则表达式：）
(?= # Opening positive lookahead
([A-Z] # Opening capture group and any uppercase letter (to match R & D)
(?: # First non-capture group opening
(?: # Second non-capture group opening
(?'O'{) # Get the named opening brace
[^{}]* # Any non-brace
)+ # Close of second non-capture group and repeat over as many times as necessary
(?: # Third non-capture group opening
(?'-O'}) # Removal of named opening brace when encountered
[^{}]*? # Any other non-brace characters in case there are more nested braces
)+ # Close of third non-capture group and repeat over as many times as necessary
)+ # Close of first non-capture group and repeat as many times as necessary for multiple side by side nested braces
(?(O)(?!)) # Condition to prevent unbalanced braces
) # Close capture group
) # Close positive lookahead
(?=([A-Z]{(?:[^{}]|(?1))+}))
答案 1 :(得分：0)
我不确定单个正则表达式是否能够满足您的需求：这些嵌套的子串总是搞乱它。
一个解决方案可能是以下算法（用Java编写，但我想C＃的翻译不会那么难）：
/**
* Finds all matches (i.e. including sub/nested matches) of the regex in the input string.
*
* @param input
* The input string.
* @param regex
* The regex pattern. It has to target the most nested substrings. For example, given the following input string
* <code>A{01B{23}45C{67}89}</code>, if you want to catch every <code>X{*}</code> substrings (where <code>X</code> is a capital letter),
* you have to use <code>[A-Z][{][^{]+?[}]</code> or <code>[A-Z][{][^{}]+[}]</code> instead of <code>[A-Z][{].+?[}]</code>.
* @param format
* The format must follow the <a href= "http://docs.oracle.com/javase/7/docs/api/java/util/Formatter.html#syntax" >format string
* syntax</a>. It will be given one single integer as argument, so it has to contain (and to contain only) a <code>%d</code> flag. The
* format must not be foundable anywhere in the input string. If <code>null</code>, <code>ééé%dèèè</code> will be used.
* @return The list of all the matches of the regex in the input string.
*/
public static List<String> findAllMatches(String input, String regex, String format) {
if (format == null) {
format = "ééé%dèèè";
}
int counter = 0;
Map<String, String> matches = new LinkedHashMap<String, String>();
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
// if a substring has been found
while (matcher.find()) {
// create a unique replacement string using the counter
String replace = String.format(format, counter++);
// store the relation "replacement string --> initial substring" in a queue
matches.put(replace, matcher.group());
String end = input.substring(matcher.end(), input.length());
String start = input.substring(0, matcher.start());
// replace the found substring by the created unique replacement string
input = start + replace + end;
// reiterate on the new input string (faking the original matcher.find() implementation)
matcher = pattern.matcher(input);
}
List<Entry<String, String>> entries = new LinkedList<Entry<String, String>>(matches.entrySet());
// for each relation "replacement string --> initial substring" of the queue
for (int i = 0; i < entries.size(); i++) {
Entry<String, String> current = entries.get(i);
// for each relation that could have been found before the current one (i.e. more nested)
for (int j = 0; j < i; j++) {
Entry<String, String> previous = entries.get(j);
// if the current initial substring contains the previous replacement string
if (current.getValue().contains(previous.getKey())) {
// replace the previous replacement string by the previous initial substring in the current initial substring
current.setValue(current.getValue().replace(previous.getKey(), previous.getValue()));
}
}
}
return new LinkedList<String>(matches.values());
}
因此，在您的情况下：
String input = "tes{}tR{R{abc}aD{mnoR{xyz}}}";
String regex = "[A-Z][{][^{}]+[}]";
findAllMatches(input, regex, null);
返回：
R{abc}
R{xyz}
D{mnoR{xyz}}
R{R{abc}aD{mnoR{xyz}}}
答案 2 :(得分：0)
.Net正则表达式中的平衡组使您可以准确控制要捕获的内容，并且.Net正则表达式引擎保留组的所有捕获的完整历史记录（与仅捕获每个组的最后一次出现的大多数其他类型）不同
MSDN示例有点过于复杂。匹配nestes结构的更简单方法是：
(?>
(?<O>)\p{Lu}\{ # Push to the O stack, and match an upper-case letter and {
| # OR
\}(?<-O>) # Match } and pop from the stack
| # OR
\p{Ll} # Match a lower-case letter
)+
(?(O)(?!)) # Make sure the stack is empty
或单行：
(?>(?<O>)\p{Lu}\{|\}(?<-O>)|\p{Ll})+(?(O)(?!))
<强> Working example on Regex Storm
在您的示例中，它还匹配字符串开头的"tes"，但不要担心，我们还没有完成。
通过小幅修正，我们还可以捕捉 R{ ... }对之间的出现次数：
(?>(?<O>)\p{Lu}\{|\}(?<Target-O>)|\p{Ll})+(?(O)(?!))
每个Match都会有一个名为Group的{{1}}，而且每个"Target"每次出现都会Group - 您只关心这些捕获
Working example on Regex Storm - 点击表标签，查看 Capture