使用boost :: regex获取子match_results [英] Getting sub-match_results with boost::regex

问题描述

嘿，假设我有这个正则表达式：（test [0-9]）+

Hey, let's say I have this regex: (test[0-9])+

我匹配它： test1test2test3test0

const bool ret = boost::regex_search(input, what, r);

for (size_t i = 0; i < what.size(); ++i)
    cout << i << ':' << string(what[i]) << "\n";

现在， what [1] test0 （最后一次出现）。让我们说，我需要得到 test1 ，2和3：我该怎么办？

Now, what[1] will be test0 (the last occurrence). Let's say that I need to get test1, 2 and 3 as well: what should I do?

注意：真正的正则表达式非常复杂，必须保持一个整体匹配，所以将示例regex更改为（test [0-9]）将无法工作。

Note: the real regex is extremely more complex and has to remain one overall match, so changing the example regex to (test[0-9]) won't work.

推荐答案

我认为Dot Net能够创建单个捕获组集合，以便（grp）+将在group1上创建一个集合对象。 boost引擎的regex_search（）将像任何普通的匹配函数一样。你坐在一个while（）循环匹配最后一个匹配留下的模式。您使用的表单不​​使用bid-itterator，因此该函数将不会启动最后一个匹配的下一个匹配。

I think Dot Net has the ability to make single capture group Collections so that (grp)+ will create a collection object on group1. The boost engine's regex_search() is going to be just like any ordinary match function. You sit in a while() loop matching the pattern where the last match left off. The form you used does not use a bid-itterator, so the function won't start the next match where the last match left off.

您可以使用itterator表单：

（编辑 - 您也可以使用令牌迭代器，定义要迭代的组）。

You can use the itterator form:
(Edit - you can also use the token iterator, defining what groups to iterate over. Added in the code below).

#include <boost/regex.hpp> 
#include <string> 
#include <iostream> 

using namespace std;
using namespace boost;

int main() 
{ 
    string input = "test1 ,, test2,, test3,, test0,,";
    boost::regex r("(test[0-9])(?:$|[ ,]+)");
    boost::smatch what;

    std::string::const_iterator start = input.begin();
    std::string::const_iterator end   = input.end();

    while (boost::regex_search(start, end, what, r))
    {
        string stest(what[1].first, what[1].second);
        cout << stest << endl;
        // Update the beginning of the range to the character
        // following the whole match
        start = what[0].second;
    }

    // Alternate method using token iterator 
    const int subs[] = {1};  // we just want to see group 1
    boost::sregex_token_iterator i(input.begin(), input.end(), r, subs);
    boost::sregex_token_iterator j;
    while(i != j)
    {
       cout << *i++ << endl;
    }

    return 0;
}

输出：

test1

test2

test3

test0

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