c++ - 为什么 Visual Studio 既可以工作又会为看起来稳定的 c++ 数据库程序提供错误？

Question

我决定在提交数据库相关任务之前测试一个程序。我应该创建一个管理员工表的系统，接受各种输入并将它们应用于数据库。

在开发过程中，我确保每个输入都工作多次。初始测试，每次运行程序时它都有效。我能够添加一个 id 为 1010 的用户（尚未在系统中。如果该 ID 已经存在，则系统会丢弃该信息并通知用户），然后创建一个无意义的用户，并且它起作用了。

我返回程序添加了一些最终编辑，例如正确显示输出、更好的格式等以及一般的清理和注释，当我注意到在运行应用程序并尝试相同的功能时，程序显示了一些奇怪的行为。

1. 在第一次运行时，该程序检测到我输入的所有员工 ID（甚至是 99999 之类的无意义的 ID）都已经存在。当我退出并返回程序时，这个ID突然被接受了，好像什么也没发生一样
2. 当我成功地将用户输入数据库时​​，它会立即自动退出程序（功能的选择是通过 switch/case 语句完成的）。
3. 2.发生后，我尝试重新运行程序，只是它反复输入“名称”输入，然后跳过所有其他输入并立即退出程序。没有错误信息，什么都没有
4. 当我试图恢复我的所有更改时，它继续发生，即使是在我几天前制作的最稳定的版本上，然后突然再次重复这种行为。

我不完全确定我做错了什么导致它这样做。这在VS中很常见吗？

供参考，我的代码如下

这是案例 3 的代码，即处理入口的代码

        case 3:
            cout << "Please Enter the employee number of the employee to add" << endl;
            cin >> empNum;
            failsafe = findEmployee(conn, empNum, emp);
            if (failsafe == 1) {
                if (emp.empNum == 0) {

                    cout << "Employee does not exist or is not in our records. Entry can proceed" << endl;
                    cout << "proceeding with entry" << endl;
                    cout << "-------------" << endl;
                    cout << "Enter the employee's Number: ";
                    cin >> emp.empNum;
                    cin.ignore();
                    cout << "Enter the employee's first name: ";
                    cin >> emp.firstName;
                    cout << "Last name: ";
                    cin >> emp.lastName;
                    cout << "Enter Employee Email: ";
                    cin >> emp.email;
                    cout << "Enter Employee's Extension: ";
                    cin >> emp.extension;
                    cout << "Enter the employee number of the person this employee reports to: ";
                    cin >> emp.reportsTo;
                    cout << "What is this employees' current position? ";
                    cin >> emp.jobTitle;
                    cout << "What city is this employee working within? ";
                    cin >> emp.city;
                    cout << "Compiling employee data" << endl;
                    insertEmployee(conn, emp);

                }
                else {
                    cout << empNum;
                    cout << "An employee with the same number exists." << endl;
                }
            }
            else if (failsafe == 0) {
            cout << "Judging from the way this program reads data, you shouldn't be able to see this"<<endl; //In case the function for some reason *doesnt* return a 1 like it's been doing every time
            }
            break;

这是处理将员工添加到数据库的代码

    void insertEmployee(MYSQL* conn, struct Employee& emp) {
        int queResult;
        MYSQL_RES* res;
        cout << "This and the below line is just here for testing to see what exactly it's skipping over" << endl;
        pause();
        if (conn) {
            string query = ("INSERT INTO schoolschema.employees(employeeNumber, lastName, firstName, extension, email, officeCode, reportsTo, jobTitle)"
                " VALUES (" + std::string(" \' " ) + to_string(emp.empNum)  + std::string(" \' ") + " , " + " \' " + emp.lastName + " \' " + " , " + " \' " + emp.firstName + " \' " + " , " + std::string(" \' ") + emp.extension + " \' " + " , " + " \' " + emp.email + " \' " + " , " + to_string(1) + " , " + std::string(" \' ") + emp.reportsTo + " \' "+ " , " + " \' " + emp.jobTitle + " \' " + ");");
            const char* q = query.c_str();
            queResult = mysql_query(conn, q);
            if (!queResult) {
                cout << "Employee successfully added, check to verify by checking the number of the employee from the main menu, or using your DBMS application" << endl;
                return;
            }
            else {
                cout << "Error occurred: " << mysql_errno(conn) << " " << mysql_error(conn) << endl;
            }
        }
    }

是的，我知道整个

+" \' "+

事情有点难看，但这是我遇到的另一个问题的解决方法。

如果有人认为错误可能在其他地方，我可以发布更多代码片段。

作为补充说明，我认为问题不在于“查找员工”，因为该函数似乎正确地返回了信息，只是编译器似乎在挑选这些数据是否正确。（您可以在下面看到我可以删除用户 1010 但我无法将其添加回来）

发生的错误

********************* HR Menu *********************
1)  Find Employee
2)  Employees Report
3)  Add Employee
4)  Update Employee
5)  Remove Employee
0)  Exit
5
Enter employee number to delete: 1010
Employee found, deleting
Employee successfully deleted
********************* HR Menu *********************
1)  Find Employee
2)  Employees Report
3)  Add Employee
4)  Update Employee
5)  Remove Employee
0)  Exit
3
Please enter number to add
1010
employee does not exist, proceeding with entry
Enter Employee number: 1010
Last name: Enter Employee Email: Enter Employee's Extension: Enter the employee number of the person this employee reports to: What is this employees' current position? What city is this employee working within? Compiling employee data
Press enter to continue...

编辑：添加注释：出现上述错误后（说按回车继续），它会显示一个sql错误，然后立即假设用户输入“0”并退出

标签： c++mysql