通过引用的文档字段进行Mongoid搜索(Mongoid search by referenced documents fields)

 你使用固定尺寸的工作，如果它们在你的情况下大于7个字符，它将产生垃圾。 
 
 猜猜你想用C ++编写代码，你应该远离C“字符串函数” 
 
#include <iostream>
#include <string>

using namespace std;
void main( ... )
{
   int my_int = 666;
   cout << "string: ";
   string my_string;
   cin >> my_string;

   // concat - separated by semicolon in this example
   stringstream ss;
   ss << my_string << ";" << my_int;

   cout << "string and int in one string: "
         << ss.str().c_str() << endl;

   // split
   string tmpStr = ss.str();
   size_t found = tmpStr.find(";");
   while( found != string::npos  )
   {
      cout << tmpStr.substr(0, found).c_str() << endl;
      tmpStr = tmpStr.substr( found+1 );
      found = tmpStr.find(";");

      // this case is true for the last part
      if( found == string::npos )
         cout << tmpStr.c_str() << "\n";
   }

   return 0;
}
 
 根据你想要用它做什么，你可能需要尝试将每个字符串部分转换为整数，如果失败，你就知道它是一个字符串，否则你可以将它分配给一个整数。 
 
 我认为，即使这种方法不是我在生产代码中真正做的，它应该给你一个提示，告诉你如何从那里做你想做的事情。 

Your working with fixed sizes and if they are larger then 7 chars in your case, it will produce garbage.
 
Guess since you want to code in C++, you should stay away from C "string-functions"
 
#include <iostream>
#include <string>

using namespace std;
void main( ... )
{
   int my_int = 666;
   cout << "string: ";
   string my_string;
   cin >> my_string;

   // concat - separated by semicolon in this example
   stringstream ss;
   ss << my_string << ";" << my_int;

   cout << "string and int in one string: "
         << ss.str().c_str() << endl;

   // split
   string tmpStr = ss.str();
   size_t found = tmpStr.find(";");
   while( found != string::npos  )
   {
      cout << tmpStr.substr(0, found).c_str() << endl;
      tmpStr = tmpStr.substr( found+1 );
      found = tmpStr.find(";");

      // this case is true for the last part
      if( found == string::npos )
         cout << tmpStr.c_str() << "\n";
   }

   return 0;
}
 
Depending on what you want to do with it you may need to try to convert each string-parts to an integer and if it fails you know its a string otherwise you can assign it to an integer.
 
I think, even this approach is not what i would really do in productive code, it should give you a hint how to do what you want to do from there.