how to split a string value that contains characters and numbers - c++

Question

I have a std::string s=n8Name4Surname. How can I obtain in 2 strings the Name and the Surname? THX

Answer 1

One way to do this is using Boost.Tokenizer. See this example:
#include <string>
#include <boost/tokenizer.hpp>
#include <boost/foreach.hpp>
int main()
{
using namespace std;
using namespace boost;
string text="n8Name4Surname.";
char_separator<char> sep("0123456789");
tokenizer<char_separator<char> > tokens(text, sep);
string name, surname;
int count = 0;
BOOST_FOREACH(const string& s, tokens)
{
if(count == 1)
{
name = s;
}
if(count == 2)
{
surname = s;
}
++count;
}
}
EDIT
If you put the results in a vector, its even less code:
#include <string>
#include <boost/tokenizer.hpp>
#include <boost/foreach.hpp>
#include <algorithm>
#include <iterator>
#include <vector>
int main()
{
using namespace std;
using namespace boost;
string text="n8Name4Surname.";
char_separator<char> sep("0123456789");
tokenizer<char_separator<char> > tokens(text, sep);
vector<string> names;
tokenizer<char_separator<char> >::iterator iter = tokens.begin();
++iter;
if(iter != tokens.end())
{
copy(iter, tokens.end(), back_inserter(names));
}
}

Answer 2

You can detect numerical characters in the string using function isdigit(mystring.at(position), then extract substring between those positions.
See:
http://www.cplusplus.com/reference/clibrary/cctype/isdigit/

Answer 3

Use Boost tokenizer with the digits 0-9 as delimiters. Then, throw away the string containing "n". It's overkill, I realize...

Answer 4

Simple STL approach:
#include <string>
#include <vector>
#include <iostream>
int main()
{
std::string s= "n8Name4Surname";
std::vector<std::string> parts;
const char digits[] = "0123456789";
std::string::size_type from=0, to=std::string::npos;
do
{
from = s.find_first_of(digits, from);
if (std::string::npos != from)
from = s.find_first_not_of(digits, from);
if (std::string::npos != from)
{
to = s.find_first_of(digits, from);
if (std::string::npos == to)
parts.push_back(s.substr(from));
else
parts.push_back(s.substr(from, to-from));
from = to; // could be npos
}
} while (std::string::npos != from);
for (int i=0; i<parts.size(); i++)
std::cout << i << ":\t" << parts[i] << std::endl;
return 0;
}

Answer 5

Mandatory Boost Spirit sample:
#include <string>
#include <boost/spirit/include/qi.hpp>
#include <iostream>
int main()
{
std::string s= "n8Name4Surname";
std::string::const_iterator b(s.begin()), e(s.end());
std::string ignore, name, surname;
using namespace boost::spirit::qi;
rule<std::string::const_iterator, space_type, char()>
digit = char_("0123456789"),
other = (char_ - digit);
if (phrase_parse(b, e, *other >> +digit >> +other >> +digit >> +other, space, ignore, ignore, name, ignore, surname))
{
std::cout << "name = " << name << std::endl;
std::cout << "surname = " << surname << std::endl;
}
return 0;
}