在C++中提取字符串

幸运的是,字符串包含实现这一点的成员函数,尽管有人可能会想出一个“更酷”的解决方案,但这是可行的(并且打算这样做):

#include <iostream>
#include <string>
#include <sstream>

int main(){
    std::ostringstream oss;
    oss << "GET /analysis HTTP/1.1\n\n"
        << "Host: localhost:4433\n\n"
        << "User-Agent: curl/7.47.0\n\n"
        << "Accept: */*\n\n"
        << "Authorization: Basic MTIzYWxpY2U6bWVyY3VyeQ==\n\n"
        << "Content-Length: 40\n\n"
        << "Content-Type: application/x-www-form-urlencoded\n\n"
        << "{\"u_id\": 62, \"g_id\": 14, \"a_type\": \"LR\"}";

    std::string content = oss.str();

    std::string delimiterStart = "Basic ";
    std::string delimiterEnd = " ";

    int start = content.find(delimiterStart) + delimiterStart.length();
    std::string partial = content.substr(start, content.length());
    partial = partial.substr(0, partial.find(delimiterEnd));

    std::cout<<"STR: "<< partial;
    return 0;
}

这是假设您知道两个定界符,不管是什么情况,您都需要它,否则,如果您不知道从何处到何处“获取”任何内容,您将如何提取任何内容?

我知道正则表达式被过度使用了,但为了子孙后代,这里有一个使用正则表达式的解决方案。

#include <iostream>
#include <regex>
using namespace std;

std::string source = R"(GET /analysis HTTP/1.1

Host: localhost:4433

User-Agent: curl/7.47.0

Accept: */*

Authorization: Basic MTIzYWxpY2U6bWVyY3VyeQ==

Content-Length: 40

Content-Type: application/x-www-form-urlencoded



{"u_id": 62, "g_id": 14, "a_type": "LR"})";

int main() {
    std::smatch match;
    std::regex reg(R"(Authorization:[\s]*Basic[\s]*([^\s]+))");

    std::regex_search(source,match,reg);

    if(match.size() >=2) {
        std::cout << match[1] << std::endl;
    }
    return 0;
}