如何使用 sqlite3 c++ 将整数转换为字符串

How to convert integer to string using sqlite3 c++

本文关键字：转换 字符串 整数 何使用 sqlite3 c++      更新时间：2023-10-16

我正在尝试将整数数据转换为字符串，但出现错误cannot convert 'int*' to 'const char*' for argument '2' to 'char* strcat(char*, const char*)'

嘿伙计们请解决这些问题，我哪里做错了？

    #include <iostream>
    using namespace std;
    #include "sqlite3.h"
    #include<string.h>
    #include <sstream>
    #define STRING_MAX 32
    typedef struct metadata_t
    {
      char session_username[256];   
      int user_id[256];
      int session_id[256];
      char buffer[256];
    }metadata_t;
    void  insertdata1(metadata_t *data);
  int main (int argc, const char * argv[])
      {
      struct metadata_t *data;
        cin >> *data->user_id;
        cin >> data->session_username;
        cin >> *data->session_id;
        cin >> data->buffer;

         insertdata1(data);
         }
    void insertdata1(metadata_t *data)
        {
         // char *buff1,*buff2;
             sqlite3 *db;
             sqlite3_open("test1.db", & db);
             string createQuery = "CREATE TABLE IF NOT EXISTS items (user_id INTEGER PRIMARY KEY,   session_username TEXT,Session_id INTEGER,buffer TEXT,));";
             sqlite3_stmt *createStmt;
             cout << "Creating Table Statement" << endl;
             sqlite3_prepare(db, createQuery.c_str(), createQuery.size(), &createStmt, NULL);
             cout << "Stepping Table Statement" << endl;
             if (sqlite3_step(createStmt) != SQLITE_DONE) cout << "Didn't Create Table!" << endl;

            //itoa(*data->userid,buff1,10);  // not work
            //itoa(*data->userphone,buff2,10);// Integer to String Conversion
        /*   std::string  ss;
           ss << *data->user_id;
           ss << *data->session_id;
           std::string user_Id = ss.str();
           std::string session_Id = ss.str();  */ // not working
                   char *a="(";
                   char *d=")";
                   char *b="'";
                   char *c=",";
                   char str1[1000];
                   char * str2 = "";
                   char * g = ";"; 
               strcpy(str1, "INSERT INTO items (user_id,session_username,session_id,buffer)VALUES");
               strcat(str1,a);
               strcat(str1,b);
               strcat(str1,data->user_id);
               strcat(str1,b);
               strcat(str1,c);
               strcat(str1,b);
               strcat(str1,data->session_username);
               strcat(str1,b);
               strcat(str1,c);
               strcat(str1,b);
               strcat(str1,data->session_id);
               strcat(str1,b);
               strcat(str1,c);
               strcat(str1,b);
               strcat(str1,data->buffer);
               strcat(str1,b);
               strcat(str1,d);
               strcat(str1,g);
            std::string insertQuery = str1; // WORKS!
            sqlite3_stmt *insertStmt;
            cout << "Creating Insert Statement" << endl;
            sqlite3_prepare(db, insertQuery.c_str(), insertQuery.size(), &insertStmt, NULL);
            cout << "Stepping Insert Statement" << endl;
            if (sqlite3_step(insertStmt) != SQLITE_DONE) cout << "Didn't Insert Item!" << endl;


      }**

建议使用std::ostringstream（您需要#include <sstream>）来构造查询：

std::ostringstream str1;
str1 << "INSERT INTO items (user_id,session_username,session_id,buffer)VALUES"
     << a
     << b // and so on
     << g;
const std::string insertQuery = str1.str();

以下内容不分配任何内存，而只是声明一个指针（C++中不需要struct）：

struct metadata_t *data;

任何取消引用它的尝试都会导致未定义的行为，很可能是分段错误：

cin >> *data->user_id;

只需在堆栈上声明它：

metadata_t data;

并将其地址（或通过引用传递）传递给insertdata1()：

insertdata1(&data);

更喜欢std::string而不是char*或char[]：

struct metadata_t
{
    std::string session_username;
    int user_id[256];  // Did you really mean an array of int here ?
    int session_id[256];
    std::string buffer;
};
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