只能从文本文件中读取一个名称

Question

我有一个包含名称（char名称[25]），age（int age）和gpa（float gpa）的struct（RECORD）。我想从这个文本文件中读取数据：

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个字符长（即数字是25个字符的权利）。我试图将这些数据复制到RECORD a [7]的数组中。这是我的代码：

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();

只读取数据的第一线，但没有了。我如何让它继续其余的线？

Answer 1

我认为这会有所帮助。首先，您需要读取数组中的整个文件，并将其转换为字符串数组，该数组的长度为25个字符。然后你只需遍历该数组来显示它。

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; 
}