從文本文件中讀取數據到結構數組中

Question

我以前見過這個問題，但找不到幫助我解決問題的答案。這裏發生了一些事情。這是顯示菜單，這將允許用戶：退出返回上一級菜單，

1. 查看費用，
2. 更新費用，或
3. 。

該信息被保存到應該輸入和輸出到的數據文件中。嘗試查看文件中的數據時，第一個菜單選項有問題。假設已經存在包含該信息的數據文件C:\danceexpenses.txt：

DJ 
100 
Site 
100 
Food 
100 
Favors 
100 
Security 
100 
Ticket printing 
100 
Etc. 
100 
etc. 
100 

該程序將不顯示該信息。它只給每個空行。有什麼我需要添加配置輸出？

void charges() 
{ 
    //Structures to hold data 
    struct Expenses 
    { 
     string name; 
     double cost; 
    }; 

    int chargesChoice; 
    const int CHARGES_MENU_VIEW = 1, CHARGES_MENU_UPDATE = 2, CHARGES_MENU_QUIT = 3; 
    const int NUM_EXPENSES = 8; //number of expenses 
    Expenses danceExpenses[NUM_EXPENSES]; //array of structures 
    int index; //loop counter 

    cout << "\nWhat would you like to do?\n\n" 
     << "1. View charges\n" 
     << "2. Update charges\n" 
     << "3. Return to main menu\n" 
     << "Enter your choice: "; 
    cin >> chargesChoice; 

    while (chargesChoice < CHARGES_MENU_VIEW || chargesChoice > CHARGES_MENU_QUIT) 
    { 
     cout << "\nPlease enter a valid menu choice: "; 
     cin >> chargesChoice;  
    } 

    switch (chargesChoice) 
    { 
     case CHARGES_MENU_VIEW: //Display the expenses data 
      myFile.open("c:\\danceexpenses.txt", ios::in); //open file 
      if (!myFile) 
      { 
       cout << "Error opening file. Program aborting.\n"; 
       return; 
      } 
      char output[100]; 
      cout << "\nHere are the expected expenses for the dance:\n"; 
      for (index = 0; index < NUM_EXPENSES; index++) 
       cout << danceExpenses[index].name << endl << endl; 
      myFile.close(); //close file 
      break; 
     case CHARGES_MENU_UPDATE: 
      //get expense data 
      myFile.open("c:\\danceexpenses.txt", ios::out); //open file 
      if (!myFile) 
      { 
       cout << "Error opening file. Program aborting.\n"; 
       return; 
      } 
      for (index = 0; index < NUM_EXPENSES; index++) 
      { 
       cout << "Enter the name of the expense: "; 
       cin.ignore(); 
       getline(cin, danceExpenses[index].name); 
       cout << "Enter the expected cost for this expense: "; 
       cin >> danceExpenses[index].cost; 
      } 

      //Write data to file 
      for (index = 0; index < NUM_EXPENSES; index++) 
      { 
       myFile << danceExpenses[index].name << endl; 
       myFile << danceExpenses[index].cost << endl; 
      } 
      myFile.close(); //close file 
      break; 
     case CHARGES_MENU_QUIT: 
      showMenu(); 
      break; 
    } 
    charges(); 
} 

Answer 1

如果你的第一個動作是輸入1，則代碼應該進入：

case CHARGES_MENU_VIEW: //Display the expenses data 
     myFile.open("c:\\danceexpenses.txt", ios::in); //open file 
     if (!myFile) 
     { 
      cout << "Error opening file. Program aborting.\n"; 
      return; 
     } 
     char output[100]; 
     cout << "\nHere are the expected expenses for the dance:\n"; 
     for (index = 0; index < NUM_EXPENSES; index++) 
      cout << danceExpenses[index].name << endl << endl; 
     myFile.close(); //close file 
     break; 

這將打開舞蹈開支文件，但並沒有讀取。然後它遍歷尚未初始化的費用列表，並關閉輸入文件。還有一個未使用的變量output。

打印出來之前，您需要將數據讀入內存。

int num = 0; 
while (myFile >> danceexpenses[num].name >> danceexpenses[num].cost) 
    num++; 

for (index = 0; index < num; index++) 
    cout << danceExpenses[index].name << endl << endl;