只能從文本文件中讀取一個名稱

Question

我有一個包含名稱（char名稱[25]），age（int age）和gpa（float gpa）的struct（RECORD）。我想從這個文本文件中讀取數據：

Jacqueline Kennedy  33 3.5 
 
Claudia Johnson   25 2.5 
 
Pat Nixon    33 2.7 
 
Rosalyn Carter   26 2.6 
 
Nancy Reagan    19 3.5 
 
Barbara Bush    33 3.4 
 
Hillary Clinton   25 2.5

文件中的每名是25個字符長（即數字是25個字符的權利）。我試圖將這些數據複製到RECORD a [7]的數組中。這是我的代碼：

fstream f; 
 
f.open("data.txt", ios::in); 
 
for (int i = 0; i < 7; i++) 
 
{ 
 
\t f.get(a[i].name, 25); //reads the first 25 characters 
 
\t f >> a[i].age >> a[i].gpa; 
 
} 
 
f.close();

只讀取數據的第一線，但沒有了。我如何讓它繼續其餘的線？

Answer 1

我認爲這會有所幫助。首先，您需要讀取數組中的整個文件，並將其轉換爲字符串數組，該數組的長度爲25個字符。然後你只需遍歷該數組來顯示它。

if(f.is_open()) 
{ 
//file opened successfully so we are here 
cout << "File Opened successfully!!!. Reading data from file into array" << endl; 
//this loop run until end of file (eof) does not occ 
    while(!f.eof() && position < array_size) 
    { 
     f.get(array[position]); //reading one character from file to array 
     position++; 
    } 
    array[position-1] = '\0'; //placing character array terminating character 

cout << "Displaying Array..." << endl << endl; 
//this loop display all the charaters in array till \0 
    for(int i = 0; array[i] != '\0'; i++) 
    { 
     cout << array[i]; 
     /*then you can divide the array in your desired length strings, which  here is: Jacqueline Kennedy  33 3.5 
     Claudia Johnson   25 2.5 
     Pat Nixon    33 2.7 
     Rosalyn Carter   26 2.6 
     Nancy Reagan    19 3.5 
     Barbara Bush    33 3.4 
     Hillary Clinton   25 2.5*/ 

    } 
} 
else //file could not be opened 
{ 
    cout << "File could not be opened." << endl; 
}