Printing all substrings between/after occurrence of a character in C++ - c++

Question

I am trying to grab all substrings in between or after the occurrences of a certain character.
Specifically with for search query urls (grabbing the options), for example if I have:
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"
I need to output the strings in between and after (for the last occurrence) the "&" character
so the output would be:
rlz=1C1CHBF_enUS851US851
oq=i+need+help
aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7
sourceid=chrome
ie=UTF-8
I understand how to do this with one string, but I am stuck trying to implement it into a loop. This has to be done with several urls of different lengths and number of options.
So far I can only grab one substring, between the first and second occurrences of character, but I need to grab all of them in any given url.
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

Just find the next & from the previous one in a loop and exit the loop if you cannot find any & more and take care of the first element:
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;
}
works for your example: https://ideone.com/SiRZQB

Answer 2

Tbh I genuinely think that you should use a proper URI parser for this job as there could be a lot of fringe cases. But here you go:
#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;
}
}
Playground: http://cpp.sh/8pshy7

Answer 3

You can try the following code which uses the regular expressions to parse the url.
#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 4

You can use just an ordinary for loop as for example
#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';
}
The program output is
rlz=1C1CHBF_enUS851US851
oq=i+need+help
aqs=chrome.0.69i59j0l3j69i60l2.4646j0j7
sourceid=chrome
ie=UTF-8
Or you can write a separate function as for example
#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';
}