我试图创建可以获取表名和值并将插入返回到查询中的泛型函数,并提出类似下面的内容:
struct any {
enum type {Int, Float, String};
any(int e) { m_data.INT = e; m_type = Int;}
any(float e) { m_data.FLOAT = e; m_type = Float;}
any(char* e) { m_data.STRING = e; m_type = String;}
type get_type() const { return m_type; }
int get_int() const { return m_data.INT; }
float get_float() const { return m_data.FLOAT; }
char* get_string() const { return m_data.STRING; }
private:
type m_type;
union {
int INT;
float FLOAT;
char *STRING;
} m_data;
};
template<typename ...Args>
std::string GetInsertString(const std::string& tableName, Args... args)
{
std::string insertString = "INSERT INTO ";
insertString += tableName;
insertString += " VALUES(";
std::vector<any> vec = {args...};
std::ostringstream ss;
for (unsigned i = 0; i < vec.size(); ++i)
{
switch(vec[i].get_type())
{
case any::Int:
ss.str("");
ss << vec[i].get_int();
insertString += ss.str() + ",";
break;
case any::Float:
ss.str("");
ss << vec[i].get_float();
insertString += ss.str() + ",";
break;
case any::String:
ss.str("");
insertString += "'" + std::string(vec[i].get_string()) + "'," ;
break;
}
}
insertString.pop_back();
insertString += ");";
return insertString;
}
其中any是基于此链接How can I iterate over a packed variadic template argument list?
的类但问题是我无法将std::string
类型作为variadic参数传递给此函数,因为我们在任何类中都有联合,所以需要你们的帮助来传递{{1键入作为插入查询的值的参数
答案 0 :(得分:1)
您可以使用虚拟基类,模板包装器和智能指针写一些内容,如下所示
#include <string>
#include <vector>
#include <memory>
#include <sstream>
#include <iostream>
struct anyBase
{ virtual int unusedVirt () { return 0; }; };
template <typename T>
struct anyW : public anyBase
{
T val;
anyW (T const & v0) : val{v0} { }
};
struct any
{
public:
enum type { Int, Float, String };
any (int e)
: m_type{Int}, m_data{new anyW<int>(e)} { }
any (float e)
: m_type{Float}, m_data{new anyW<float>(e)} { }
any (char const * e)
: m_type{String}, m_data{new anyW<std::string>(e)} { }
any (std::string const & e)
: m_type{String}, m_data{new anyW<std::string>(e)} { }
any (any const & a) : m_type{a.m_type}, m_data{nullptr}
{
switch ( m_type )
{
case Int:
m_data.reset(new anyW<int>(a.get_int()));
break;
case Float:
m_data.reset(new anyW<float>(a.get_float()));
break;
case String:
m_data.reset(new anyW<std::string>(a.get_string()));
break;
}
}
type get_type () const { return m_type; }
int get_int () const
{ return dynamic_cast<anyW<int>*>(m_data.get())->val; }
float get_float () const
{ return dynamic_cast<anyW<float>*>(m_data.get())->val; }
std::string const & get_string () const
{ return dynamic_cast<anyW<std::string>*>(m_data.get())->val; }
private:
type m_type;
std::unique_ptr<anyBase> m_data;
};
template<typename ...Args>
std::string GetInsertString(const std::string& tableName, Args... args)
{
std::string insertString = "INSERT INTO ";
insertString += tableName;
insertString += " VALUES(";
std::vector<any> vec = {args...};
std::ostringstream ss;
for (unsigned i = 0; i < vec.size(); ++i)
{
switch(vec[i].get_type())
{
case any::Int:
ss.str("");
ss << vec[i].get_int();
insertString += ss.str() + ",";
break;
case any::Float:
ss.str("");
ss << vec[i].get_float();
insertString += ss.str() + ",";
break;
case any::String:
ss.str("");
insertString += "'" + std::string(vec[i].get_string()) + "'," ;
break;
}
}
insertString.pop_back();
insertString += ");";
return insertString;
}
int main ()
{
std::cout << GetInsertString("fooTable", 1, 2.2f, "3", std::string("4"))
<< std:: endl;
// print INSERT INTO fooTable VALUES(1,2.2,'3','4');
}
观察:
此解决方案也适用于C ++ 11
我统一了char *
和std::string
个案;我认为注册chat *
使用std::unique_ptr
您需要copy_constructor,因为std::unique_ptr
中的复制构造函数已被删除,因此删除了任何
更好的解决方案可能是等待传入(C ++ 17,如果我没错)std::variant
答案 1 :(得分:-2)
typeof
您可以使用typeof
并添加切换案例。例如,在新的java中,你可以自动化一个半循环,如果我没有弄错的话,同样可以看到它。