Question

每当找到*模式时，我都会尝试将替换为\w*\w。这就是我所拥有的

#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，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.

请参阅C++ online demo

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

模式([a-zA-Z])\*(?=[a-zA-Z])匹配并捕获ASCII字母char（带([a-zA-Z])），然后匹配*（带\*）然后需要（不消耗） ASCII字母字符（带(?=[a-zA-Z])）。

$1是对([a-zA-Z])群体捕获的值的反向引用。

从正则表达式匹配中替换另一个特定字符？

1 个答案: