用正则表达式匹配中的另一个替换特定字符？

Question

每当找到\w*\w 模式时，我都会尝试将* 替换为。这是我所拥有的

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

using namespace std;

int main() {
    string text = "Dear Customer, You have made a Debit Card purchase of INR962.00 on 26 Oct. Info.VPS*Brown House. Your Net Available Balance is INR 5,584.58.";

    regex reg("[\w*\w]");
    text = regex_replace(text, reg, " ");
    cout << text << "\n";
}

但它也将* 替换为，并将w 替换为。

上述程序的输出是

Dear Customer, You have made a Debit Card purchase of INR962.00 on 26 Oct. Info.VPS Bro n House. Your Net Available Balance is INR 5,584.58.

Answer 1

使用

regex reg(R"(([a-zA-Z])\*(?=[a-zA-Z]))");
text = regex_replace(text, reg, "$1 ");
// => Dear Customer, You have made a Debit Card purchase of INR962.00 on 26 Oct. Info.VPS Brown House. Your Net Available Balance is INR 5,584.58.

见C++ online demo

R"(([a-zA-Z])\*(?=[a-zA-Z]))" 是一个原始字符串文字，其中 \ 被视为文字 \ 符号，而不是 \n 或 \r 等实体的转义符号。

([a-zA-Z])\*(?=[a-zA-Z]) 模式匹配并捕获一个 ASCII 字母字符（带有 ([a-zA-Z])），然后匹配一个 *（带有 \*），然后需要（不消耗）一个 ASCII 字母字符（带有 @987654332 @)。

$1 是对使用([a-zA-Z]) 组捕获的值的反向引用。