有条件地替换字符串中的正则表达式匹配项

问题描述

我正在尝试用不同的替换模式替换字符串中的某些模式.

I am trying to replace certain patterns in a string with different replacement patters.

示例:

string test = "test replacing "these characters"";

我想要做的是将所有的 ' ' 替换为 '_'，并将所有其他非字母或数字字符替换为空字符串.我创建了以下正则表达式，它似乎正确标记化，但我不确定如何(如果可能)使用 regex_replace 执行条件替换.

What I want to do is replace all ' ' with '_' and all other non letter or number characters with an empty string. I have the following regex created and it seems to tokenize correctly, but I am not sure how to (if possible) perform a conditional replace using regex_replace.

string test = "test replacing "these characters"";
regex reg("(\s+)|(\W+)");

替换后的预期结果是:

string result = "test_replacing_these_characters";

我不能使用 boost，这就是为什么我把它排除在标签之外.所以请不要回答包括提升.我必须用标准库来做到这一点.可能是不同的正则表达式可以实现目标，或者我只是坚持做两次.

I cannot use boost, which is why I left it out of the tags. So please no answer that includes boost. I have to do this with the standard library. It may be that a different regex would accomplish the goal or that I am just stuck doing two passes.

我不记得在我原来的正则表达式时 w 中包含了哪些字符，在查找之后我进一步简化了表达式.同样，目标是任何匹配 s+ 的内容都应替换为 '_'，任何匹配的 W+ 均应替换为空字符串.

I did not remember what characters were included in w at the time of my original regex, after looking it up I have further simplified the expression. Again the goal is anything matching s+ should be replaced with '_' and anything matching W+ should be replaced with empty string.

推荐答案

C++ (0x, 11, tr1) 正则表达式 不要在每种情况下都确实有效(stackoverflow)(查找 此页面 上的短语 regex 用于 gcc)，因此最好使用 boost 一段时间.

The c++ (0x, 11, tr1) regular expressions do not really work (stackoverflow) in every case (look up the phrase regex on this page for gcc), so it is better to use boost for a while.

你可以试试你的编译器是否支持所需的正则表达式:

You may try if your compiler supports the regular expressions needed:

#include <string>
#include <iostream>
#include <regex>

using namespace std;

int main(int argc, char * argv[]) {
    string test = "test replacing "these characters"";
    regex reg("[^\w]+");
    test = regex_replace(test, reg, "_");
    cout << test << endl;
}

以上适用于 Visual Studio 2012Rc.

The above works in Visual Studio 2012Rc.

编辑 1:要在一次传递中替换两个不同的字符串(取决于匹配)，我认为这在这里不起作用.在 Perl 中，这可以在计算的替换表达式(/e 开关)中轻松完成.

Edit 1: To replace by two different strings in one pass (depending on the match), I'd think this won't work here. In Perl, this could easily be done within evaluated replacement expressions (/e switch).

因此，正如您已经怀疑的那样，您需要两次通过:

Therefore, you'll need two passes, as you already suspected:

 ...
 string test = "test replacing "these characters"";
 test = regex_replace(test, regex("\s+"), "_");
 test = regex_replace(test, regex("\W+"), "");
 ...

编辑 2:

如果可以在 regex_replace 中使用 回调函数 tr()，那么您可以修改那里的替换，例如:

If it would be possible to use a callback function tr() in regex_replace, then you could modify the substitution there, like:

 string output = regex_replace(test, regex("\s+|\W+"), tr);

用 tr() 做替换工作:

 string tr(const smatch &m) { return m[0].str()[0] == ' ' ? "_" : ""; }

问题就解决了.不幸的是，在某些 C++11 正则表达式实现中没有这样的重载，但是 Boost 有一个.以下将与 boost 一起使用并使用一次传递:

the problem would have been solved. Unfortunately, there's no such overload in some C++11 regex implementations, but Boost has one. The following would work with boost and use one pass:

...
#include <boost/regex.hpp>
using namespace boost;
...
string tr(const smatch &m) { return m[0].str()[0] == ' ' ? "_" : ""; }
...

string test = "test replacing "these characters"";
test = regex_replace(test, regex("\s+|\W+"), tr);   // <= works in Boost
...

也许有一天这将适用于 C++11 或接下来的任何数字.

Maybe some day this will work with C++11 or whatever number comes next.

问候

rbo

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