如何创建一个名为播放器用户名的文件？ [英] How do I create a file with as name the players username?

问题描述

你好，



我试图创建一个测试登录程序。

（我知道这不安全但是我正在尝试创建一些要学习的东西。）

这一行给出了一个错误：file.open（用户名+。txt）;

我认为+在那条线上不起作用，但我不确定。



这是错误：

c：\program files（x86）\ codeblocks \mingw \bin\..\lib\gcc \mingw32 \4.7.1 \ include \ c ++ \ stream | 702 |注意：void std :: basic_ofstream< _CharT，_Traits> :: open（const char *，std :: ios_base :: openmode）[with _CharT = char; _Traits = std :: char_traits< char> ;; std :: ios_base :: openmode = std :: _ Ios_Openmode] |





这是代码：

  #include   <   iostream  >  
   #include   <   string  >  
   #include   <   fstream  >  
  
 使用 命名空间标准; 
 
 枚举 MainProgramParts {SUCCESS，FAILED，STOP，OPTIONS}; 
 MainProgramParts选项（）; 
 MainProgramParts success（）; 
 MainProgramParts失败（）; 
 
  int  main（）
 {
 MainProgramParts nextPart = OPTIONS; 
 
  while （nextPart！= STOP）{
  switch （ nextPart）{
  case 选项：
 nextPart = options（）; 
  break ; 
  case 成功：
 nextPart = success（）; 
  break ; 
  case  FAILED：
 nextPart = failed（）; 
  break ; 
} 
} 
} 
 
 MainProgramParts选项（）{
  int  choice; 
 
 cout<<   1：注册<< ENDL; 
 cout<<   2：登录<< endl<< ENDL; 
 cin>>选择; 
 
  if （choice ==  1 ）
 {
 string username; 
字符串密码; 
 
 cout<<  用户名：; 
 cin>>用户名; 
 cout<<  密码：; 
 cin>>密码; 
 
 ofstream文件; 
 file.open（用户名+   .txt）; 
 file<<用户名<< endl<<密码; 
 file.close（）; 
 
 main（）; 
 
  if （choice ==  2 ）
 {
 string username; 
字符串密码; 
 string un; 
 string pw; 
 
 cout<<  用户名：; 
 cin>>用户名; 
 cout<<  密码：; 
 cin>>密码; 
 
 ifstream read（用户名+   .txt）; 
 getline（读取，取消）; 
 getline（读，pw）; 
 
  if （un == username&& pw == password）
 {
 返回成功; 
} 
  else  
 {
  return  FAILED; 
} 
} 
} 
} 
 
 MainProgramParts success（）{
 cout<<  已成功登录！; 
} 
 
 MainProgramParts失败（）{
 cout<<  密码/用户名不正确！; 
} 

谢谢，

~ Steff

解决方案

不，这不是连接，这是一个有效的 std :: string 运营商： http://www.cplusplus.com/reference/string/string/operator+ [< a href =http://www.cplusplus.com/reference/string/string/operator+target =_ blanktitle =New Window> ^ ]。



问题是类型，因为预期的参数类型是旧的lame char * 。使用 c_str 操作从 std :: string 实例中获取C字符串：http://www.cplusplus.com/reference/string/string/c_str [ ^ ]。

-SA

SA是正确的。



您可能会发现使用 visual-studio-community [ ^ ]。至少看一下。



你的代码可以正常工作，因为VSC符合C ++ 11标准。



在C ++ 11中允许将std :: string传递给open（）。您可以简单地使用+运算符连接两个std :: strings来创建文件名。

 file.open（用户名+ < span class =code-string>  .txt）; 

在其他编译器中，您必须传递char *（C字符串）。在连接两个字符串后，你可以使用c_str（）从std :: string中获取一个C字符串。

 file.open （（filename +   .txt）。c_str（））; 

Hello,

I tried to create a test login program.
(I know this isn't safe but I'm trying to create some things to learn).
This line gives an error: file.open(username + ".txt");
I think the + in that line doesn't work but I'm not sure.

This is the error:
c:\program files (x86)\codeblocks\mingw\bin\..\lib\gcc\mingw32\4.7.1\include\c++\fstream|702|note: void std::basic_ofstream<_CharT, _Traits>::open(const char*, std::ios_base::openmode) [with _CharT = char; _Traits = std::char_traits<char>; std::ios_base::openmode = std::_Ios_Openmode]|


This is the code:

#include <iostream>
#include <string>
#include <fstream>

using namespace std;

enum MainProgramParts{ SUCCESS, FAILED, STOP, OPTIONS };
MainProgramParts options();
MainProgramParts success();
MainProgramParts failed();

int main()
{
        MainProgramParts nextPart = OPTIONS;

        while(nextPart != STOP){
        switch(nextPart){
        case OPTIONS:
            nextPart = options();
            break;
        case SUCCESS:
            nextPart = success();
            break;
        case FAILED:
            nextPart = failed();
            break;
        }
    }
}

MainProgramParts options(){
    int choice;

    cout << "1: Register" << endl;
    cout << "2: Login" << endl << endl;
    cin >> choice;

    if(choice == 1)
    {
        string username;
        string password;

        cout << "Username: ";
        cin >> username;
        cout << "Password: ";
        cin >> password;

        ofstream file;
        file.open(username + ".txt");
        file << username << endl << password;
        file.close();

        main();

    if(choice == 2)
    {
    string username;
    string password;
    string un;
    string pw;

    cout << "Username: ";
    cin >> username;
    cout << "Password: ";
    cin >> password;

    ifstream read(username + ".txt");
    getline(read, un);
    getline(read, pw);

    if(un == username && pw == password)
    {
        return SUCCESS;
    }
    else
    {
        return FAILED;
    }
}
}
}

MainProgramParts success(){
    cout << "Successfully logged in!";
}

MainProgramParts failed(){
    cout << "Incorrect password / username!";
}

Thanks,
~Steff

解决方案

No, this is not the concatenation, which is a valid std::string operator: http://www.cplusplus.com/reference/string/string/operator+[^].

The problem is the type, as the expected argument type is old lame char*. Use the c_str operation to get a "C string" out of your std::string instance: http://www.cplusplus.com/reference/string/string/c_str[^].

—SA

SA is correct.

You may find it better to use visual-studio-community[^]. At least take a look at it.

Your code would then work fine because VSC is C++11 compliant.

Passing a std::string to open() is allowed in C++11. You may simply concatenate the two std::strings making your file name using the + operator.

file.open(username + ".txt");

In other compilers you will have to pass a char* (C string). You may use c_str() to get a C string from a std::string after concatenating the two strings.

file.open((filename + ".txt").c_str()); 

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