如何在 xpressive 中迭代命名组？

发布于 2024-12-22 09:01:02 字数 267 浏览 1 评论 0原文

假设我有一个像这样的 sregex 对象：

boost::xpressive::sregex::compile("(?P<firstword>\\w+) (?<secondword>\\w+)!");

尽管 xpressive 支持命名组，但我无法在 xpressive 文档中找到任何与此相关的参考。

我知道可以迭代组就可以了，但是我将如何访问组名（如果组有名称）？

那么，我将如何迭代命名组？

原文

Say I have a sregex object like this one:

boost::xpressive::sregex::compile("(?P<firstword>\\w+) (?<secondword>\\w+)!");

I have not been able to find any reference in the xpressive documentation regarding this, despite xpressive supporting named groups just fine.

I know that it is possible to iterate through groups is just fine, but how would I access the groupname (if the group has a name at all)?

So, How would I iterate through the named groups?

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∞梦里开花 2024-12-29 09:01:02

假设我们有您正在处理的整个正则表达式，
在我看来，您似乎正在尝试创建一个与这两个匹配的正则表达式
命名捕获，因此尝试迭代命名捕获是没有用的。

你只需要尝试类似的事情。

std::string str("foo bar");
sregex rx = sregex::compile("(?P<firstword>\\w+) (?<secondword>\\w+)!");
smatch what;
if(regex_search(str, what, rx))
{
    std::cout << "char = " << what["firstword"]  << what["secondword"] std::endl;
}

如果正则表达式是更复杂模式的一部分，为什么不使用静态命名的 capture ：
http://www.boost.org/doc/libs/1_41_0/doc/html/xpressive/user_s_guide.html#boost_xpressive.user_s_guide.named_captures.static_named_captures

Supposing that we have the whole regular expression you are working on,
iIt seems to my point of view that you are trying to create a regular expression that match both of the
the named capture, so it is useless to try to iterate over named capture.

You simply have to try something like that.

std::string str("foo bar");
sregex rx = sregex::compile("(?P<firstword>\\w+) (?<secondword>\\w+)!");
smatch what;
if(regex_search(str, what, rx))
{
    std::cout << "char = " << what["firstword"]  << what["secondword"] std::endl;
}

If the regular expression is part of a more complex pattern why not use the static named capture :
http://www.boost.org/doc/libs/1_41_0/doc/html/xpressive/user_s_guide.html#boost_xpressive.user_s_guide.named_captures.static_named_captures

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