Question

如果用户没有在用户名和密码中添加任何内容，我将如何添加其他条件？（“空用户名和/或密码 - 需要用户名和密码！再试一次？（是/否）”）

这是我的。代码。

#include "stdafx.h"
#include <iostream>;
#include <string>;
using namespace std;
string username;
string password;
string choice;

int main(int argc, const char * argv[])
{
do {

Username:
std::cout << "Enter Username: ";
std::cin >> username;

if (username != "Joven")
{
std::cout << "Invalid Username. Please try again.\n\n\n";
goto choice;
goto Username;

}
Password:
std::cout << "Enter Password: ";
std::cin >> password;

if (password != "Fabricante7188")
{
std::cout << "Invalid Password. Please try again.\n\n\n";
std::cout << "Do you want to try again? y/n \n\n";
cin >> choice;
goto Password;
}
else
{
std::cout << "Correct Username and Password - Log In Successfull.\n";
break;

choice:
std::cout << "Do you want to try again? y/n \n\n";
std::cin >> choice;

}
}while (choice != "y" && choice != "n");

if (choice != "y" && choice != "n")
{
cout << "Invalid choice.\n";
goto choice;
}
system("pause");
return 0;
}`

非常感谢！

Answer 1

使用输入操作>>，您无法做到。它会阻塞，直到有一些实际的非空格输入后跟 Enter 键（或者出现错误或＆＃34;文件结尾＆＃34;）。

C ++标准解决方案（不依赖于OS特定功能）是读取整个内容，例如使用例如std::getline，删除前导（可能是尾随）空格的输入，然后查看结果字符串是否为空。

Answer 2

我的解决方案是这样的:( scorch编辑2017/01/14 10:24 AEST）

//#include "stdafx.h"       I made this a comment because I don't appear to have this file. It wasn't necessary for me. Don't know whether you need it or not.
// include statements must not end with semicolons.
#include <iostream>
#include <string>
#include <cstdlib>  // Has system() function
using namespace std;

int main(int argc, const char * argv[])
{
    // Always declare variables inside the function.
    string username;
    string password;
    string choice;
    bool finished = false; // Declare this.
    bool askingToTryAgain = false; // For when the user is asked whether they want to try again.

    do {

        std::cout << "Enter Username: ";
        std::getline(std::cin, username);

        std::cout << "Enter Password: ";
        std::getline(std::cin, password);

        // Validate username and password.
        if (username != "Joven" && password != "Fabricante7188") {
            // If both the username and password are invalid, report it.
            std::cout << "Invalid Username and Password. Try again? (y/n): ";
        } else if (username == "Joven" && password == "Fabricante7188") {
            // If both fields are valid, login is successful.
            std::cout << "Correct Username and Password - Log In Successful.\n";
            finished = true; // Login is now complete, the loop will end.
        } else {
            // If just one of the fields is invalid, report which one it is.
            if (username != "Joven") {
                std::cout << "Invalid Username. Try again? (y/n): ";
            }
            if (password != "Fabricante7188") {
                std::cout << "Invalid Password. Try again? (y/n): ";
            }
        }

        if (finished == false) {
            // If the login was unsuccessful, await user input for whether they wish to try again or not.
            askingToTryAgain = true;

            do {
                // Fetch user input (y/n)
                std::getline(std::cin, choice);
                // Validate it.
                if (choice != "y" && choice != "n") {
                    std::cout << "Enter 'y' or 'n'\n";
                } else {
                    askingToTryAgain = false;

                    if (choice == "y") {
                        // Nothing to do here. The parent loop will continue after this one stops.
                    } else if (choice == "n") {
                        finished = true; // The user wishes to quit.
                    }
                }
            } while (askingToTryAgain);

        }

    } while (finished == false);

    system("pause"); // During testing I used 'sleep 4' (sleep for 4 seconds) because I'm running Linux.
    return 0;
}

另外，请帮个忙，避免使用'goto'作为控制程序执行流程的方法。在结构化编程语言（如C ++）中，循环和条件是比'goto'语句更好的解决方案。

我建议您查看http://cplusplus.com/作为一个很好的参考工具。

希望这有帮助。

如何检查用户是否输入了任何字符串？

2 个答案: