在 C++ 中出现字符之间/之后打印所有子字符串

Question

我试图在某个字符出现之间或之后获取所有子字符串。 特别是搜索查询网址（获取选项），例如，如果我有：

std::string url = "https://www.google.com/search?q=i+need+help&rlz=1C1CHBF_enUS851US851&oq=i+need+help&aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7&sourceid=chrome&ie=UTF-8"

我需要在“&”字符之间和之后（最后一次出现）输出字符串 所以输出将是：

rlz=1C1CHBF_enUS851US851 
oq=i+need+help
aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7
sourceid=chrome 
ie=UTF-8

我了解如何使用一个字符串来执行此操作，但我一直在尝试将其实现为循环。这必须通过几个不同长度和选项数量的 url 来完成。

到目前为止，我只能在第一次和第二次出现字符之间抓取一个子字符串，但我需要在任何给定的 url 中抓取所有子字符串。

int a = url.find("&") + 1;
int b = url.find("&", url.find("&") + 1);
int c = (b - a);
std::string option = url.substr(a, c);

Answer 1

例如，您可以只使用普通的 for 循环

#include <iostream>
#include <string>
#include <vector>
#include <iterator>
#include <algorithm>

int main() 
{
    std::string url = "https://www.google.com/search?q=i+need+help"
                      "&rlz=1C1CHBF_enUS851US851"
                      "&oq=i+need+help"
                      "&aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7"
                      "&sourceid=chrome&ie=UTF-8";

    char c = '&';

    size_t n = std::count_if( std::begin( url ), std::end( url ),
                              [=]( const auto &item )
                              {
                                 return item == c;
                              } );

    std::vector<std::string> v;
    v.reserve( n );

    for ( auto pos = url.find( c, 0 );  pos != std::string::npos; )
    {
        auto next = url.find( c, ++pos );

        auto n = ( next == std::string::npos ? url.size() : next ) - pos;

        v.push_back( url.substr( pos, n ) ); 

        pos = next;                    
    }

    for ( const auto &s : v ) std::cout << s << '\n';

}

程序输出是

rlz=1C1CHBF_enUS851US851
oq=i+need+help
aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7
sourceid=chrome
ie=UTF-8

或者您可以编写一个单独的函数，例如

#include <iostream>
#include <string>
#include <vector>
#include <iterator>
#include <algorithm>

std::vector<std::string> split_url( const std::string &url, char c = '&' )
{
    size_t n = std::count_if( std::begin( url ), std::end( url ),
                              [=]( const auto &item )
                              {
                                 return item == c;
                              } );

    std::vector<std::string> v;
    v.reserve( n );

    for ( auto pos = url.find( c, 0 );  pos != std::string::npos; )
    {
        auto next = url.find( c, ++pos );

        auto n = ( next == std::string::npos ? url.size() : next ) - pos;

        v.push_back( url.substr( pos, n ) ); 

        pos = next;                    
    }

    return v;
}

int main() 
{
    std::string url = "https://www.google.com/search?q=i+need+help"
                      "&rlz=1C1CHBF_enUS851US851"
                      "&oq=i+need+help"
                      "&aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7"
                      "&sourceid=chrome&ie=UTF-8";


    auto v = split_url(url );

    for ( const auto &s : v ) std::cout << s << '\n';

}

Answer 2

您可以尝试以下使用正则表达式解析url的代码。

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

int main(){
    string url = "https://www.google.com/search?q=i+need+help&rlz=1C1CHBF_enUS851US851&oq=i+need+help&aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7&sourceid=chrome&ie=UTF-8";
    regex rg("[\?&](([^&]+)=([^&]+))");
    for(smatch sm; regex_search(url, sm, rg); url=sm.suffix())
        cout << sm[1] <<endl;
    return 0;
}

Answer 3

Tbh 我真的认为您应该为这项工作使用适当的 URI 解析器，因为可能会有很多边缘情况。但是给你：

#include <iostream>
#include <string>

int main()
{
  std::string url = "https://www.google.com/search?q=i+need+help&rlz=1C1CHBF_enUS851US851&oq=i+need+help&aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7&sourceid=chrome&ie=UTF-8";
  char delimiter = '&';
  size_t start = url.find(delimiter);
  size_t end;
  while (start != std::string::npos) {
      end = url.find(delimiter, start + 1);
      std::cout << url.substr(start + 1, end - start - 1) << std::endl;
      start = end;
  }
}

游乐场：http://cpp.sh/8pshy7

Answer 4

只需在循环中从上一个& 中找到下一个&，如果找不到更多&，则退出循环并处理第一个元素：

vector<string> foo(const string& url)
{
    vector<string> result;
    auto a = url.find("?");
    if (a == string::npos) return result;

    auto b = url.find("&");
    if (b == string::npos)
    {
        result.push_back(url.substr(a + 1, string::npos));
        return result;
    }
    result.push_back(url.substr(a + 1, b - a - 1));
    do
    {
        a = b;
        b = url.find("&", a + 1);
        result.push_back(url.substr(a + 1, b - a - 1));
    } while (b != string::npos);

    return result;
}

适用于您的示例：https://ideone.com/SiRZQB