如何使用正则表达式递归匹配模式？

Question

The string can be like one of the following:

a(b,c)
a(a(b,c),d)
a(a(a(a(a(b,c),d),a(e,f)),g),h)
etc

I want to match an unlimited number of "a(x,y)". How can I do that using Regex? Here's what I have:

\\w\\(((?:\\([a-zA-Z0-9]+\\))|(?:[a-zA-Z0-9]+)),((?:\\([a-zA-Z0-9]+\\))|(?:[a-zA-Z0-9]+))\\)

It only matches up two recursions of "a(x,y)".

Answer 1

Java's standard regex lib does not support recursion, so you can't match such general nested constructs with it.

But in flavors that do support recursion (Perl, PCRE, .NET, etc) you can use expressions like:

\w+(?:\((?R)(?:,(?R))*\))?

Answer 2

You can also use my Regular Expression library https://github.com/florianingerl/com.florianingerl.util.regex , that supports Recursive Regular Expressions! The API is basically the same as the one of java.util.regex, just the required import statements are different, eg

Pattern p = Pattern.compile("(?<first>a\\((?<second>(?'first')|[a-zA-Z]),(?'second')\\))");
assert p.matcher("a(a(a(a(a(b,c),d),a(e,f)),g),h)").find();

Answer 3

2 个选项 - 1) 使用词法分析自己进行模式匹配和替换 [OR] 2) 如果你想坚持使用正则表达式，那么使用一些 shell 编程（或任何支持语言）并从 Java 中调用它。

Answer 4

The language you're describing isn't a regular language so it can't be matched by a regular expression. Look into lexical analysis (ie use a parser)

Answer 5

I think you are looking for something like:

a(x,y) = [az] ( [az] , [az] )

regex = a(x,y) | a(regex|y) | a(x, regex)

Not sure how you can do in a language.