C ++正则表达式:哪个组匹配? [英] C++ regex: Which group matched?

问题描述

我有一个正则表达式，包含通过or条件连接的各个子组:

I have a regex containig various sub-groups which are connected through an or condition:

([[:alpha:]]+)|([[:digit:]]+)

当我匹配字符串 1 a 2 时，得到三个匹配项: 1 ， a 和 2

When I match the string 1 a 2, I get three matches: 1, a and 2.

C ++中是否有一种方法可以确定哪些子模式匹配?

Is there a way in C++ to determine which of the sub-patterns matched?

推荐答案

不直接.

使用 std :: regex 库， match_result 类负责子匹配，它具有一个名为 std :: match_results :: size ，然后您可以找到子项的数量-比赛.

with the std::regex library, match_result class takes care of the sub-match and it has a method named std::match_results::size and with that you can find the number of sub-match.

例如:

std::string str( "one two three four five" );
std::regex rx( "(\\w+)(\\w+)(\\w+)(\\w+)(\\w+)" );
std::match_results< std::string::const_iterator > mr;

std::regex_search( str, mr, rx );

std::cout << mr.size() << '\n'; // 6  

此处的输出为 6 而不是 5 ，因为匹配本身也被计算在内.您可以通过 .str(number)方法或 operator []

here the output is 6 not 5 because the match itself is counted as well. You can access them by .str( number ) method or operator[]

因此，由于子匹配是从从左到右进行计数的，因此您应该在看到 size 方法的输出后才能确定匹配组.

So because sub-match are counted form left-to-right you should after seeing the output of size method figure out witch group was matched.

如果将 rx 更改为(\\ w +)(\\ d +)(\\ w +)" ，则大小= 0

If you change the rx to "(\\w+)(\\d+)(\\w+)" then the size = 0

如果将 rx 更改为(\\ w +).+" ，则大小为 2 .这意味着您有一个完全成功匹配和一个总和匹配

If you change the rx to "(\\w+).+" then the size is 2. That means you have a whole successful match and a sum-match

例如:

std::string str( "one two three four five" );
std::regex rx( "(\\w+).+" );
std::match_results< std::string::const_iterator > mr;

std::regex_search( str, mr, rx );

std::cout << mr.str( 1 ) << '\n'; // one
std::cout << mr[ 1 ] << '\n';     // one

两者的输出为:一个

如果您只想打印子匹配项，则可以使用一个具有索引的简单循环，该索引从 1 开始strong>不是 0

And also if you want to print only the sub-match you can use a simple loop that has an index and this index starts from 1 not 0

例如:

std::string str( "one two three four five" );
std::regex rx( "(\\w+) \\w+ (\\w+) \\w+ (\\w+)" );
std::match_results< std::string::const_iterator > mr;

std::regex_search( str, mr, rx );

for( std::size_t index = 1; index < mr.size(); ++index ){
    std::cout << mr[ index ] << '\n';
}

输出为:

one
three
five  

---

通过说确定哪些子模式匹配
如果您的意思是指定应从搜索引擎返回哪个子匹配，则使用 std:答案为是:regex_token_iterator ，您可以确定:

---

By saying determine which of the sub-patterns matched
if you mean specify which sub-match should be return from the search-engine then the answer is yes by using std::regex_token_iterator you can determine that:

Ex :(迭代每个匹配项的 second 个子匹配项)

Ex: (Iterate over second sub-match of each match )

std::string str( "How are you today ? I am fine . How about you ?" );
std::regex rx( "(\\w+) (\\w+) ?" );
std::match_results< std::string::const_iterator > mr;

std::regex_token_iterator< std::string::const_iterator > first( str.begin(), str.end(), rx, 2 ), last;

while( first != last ){
    std::cout << first->str() << '\n';
    ++first;
} 

最后一个参数为2 :(str.begin()，str.end()，rx，2)，这意味着您只需要次子匹配.所以输出是:

the last parameter is 2 : ( str.begin(), str.end(), rx, 2 ) and it means you want only the second sub-match. So the output is:

are
today
am
about

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