我希望我能在这里找到一些帮助。我下周要完成一项任务,包括从txt文件中读取一堆数据到数组中,然后打印出结果。数据采用以下格式:
“麦克白”,“威廉·莎士比亚”,“41.04”,“161”,“23”,“978-88-5985-004-5”
“圣诞颂歌”,“查尔斯狄更斯”,“98.74”,“167”,“547”,“978-26-2885-780-7”。
每行有六条信息需要存储以供日后使用。我应该编写代码来计算我们拥有的文本行数,以便创建一个正确大小的动态数组。我已经满足了。我有39行参赛作品。然后我应该创建一个函数来读取txt文件并将所有数据保存到我创建的数组中的相应对象。
我不知道使用什么方法,我一直在寻找教程和解释几天。我对文件和解析的经验非常有限,所以如果我有点缺乏经验,请原谅。到目前为止,这是我的代码:
#include <fstream>
#include <iomanip>
#include <iostream>
#include <string>
using namespace std;
class Author
{
public:
private:
string fname, lname;
};
class Book
{
friend ofstream& operator<<(ofstream&, Book);
public:
Book();
private:
string bookName;
Author author;
double price;
int qtyOnHand;
int qtySold;
double revenue;
string ISBN;
};
Book :: Book()
{
}
int getLineNumber(ifstream &);
void parseData(ifstream &, Book []);
//void sortBookList(Book[], int, int);
int main()
{
int numberOfBooks;
//open the file from which to read the data
ifstream myFile;
myFile.open("Book List.txt");
//function to find out how many objects to create
numberOfBooks = getLineNumber(myFile);
//create an array with that many objects
Book *bptr;
bptr = new Book[numberOfBooks];
//function to read information from file into array of objects
parseData(myFile, bptr);
//loop to call sorting function and output function based on 4 different criteria
//close the file explicitly
return 0;
}
int getLineNumber(ifstream &myFile)
{
int counter = 0;
string myString;
while(!myFile.eof())
{
getline(myFile, myString);
counter++;
}
myFile.close();
counter --;
return counter;
}
void parseData(ifstream &myFile, Book bookPtr[])
{
}
因此,总结一下我的问题,我不明白如何将文本文件中的数据解析到我的数组中。 非常感谢任何有帮助的人!欢呼声。
编辑:我已经尝试过愚弄代码,我想我朝着正确的方向迈出了一步,但我仍然有点失落。这是我对parseData函数所拥有的。void parseData(ifstream &myFile, Book bookPtr[])
{
string dummyLine;
string word, line;
myFile.open("Book List.txt");
getline(myFile, dummyLine);
string data[6];
while(!myFile.eof())
{
getline(myFile, line, '\n');
for (size_t i = 0; i < line.size(); ++i)
{
char c = line[i];
if(c == ',' || c == '\n')
{
if(!word.empty())
{
data[i] = word;
word.clear();
}
}
else
{
word += c;
}
}
if(!word.empty())
{
//cout << word << endl;
}
}
}
答案 0 :(得分:2)
也许你只需要知道如何对字符串中的每个字符做些什么?
这是一些代码,它遍历字符串中构建单词的每个字符,然后单独打印它们。您会注意到string
与vector
(str[i]
,str.push_back(char)
,str.size()
等具有相同的界面。)
// You'll need to include <iostream> and <string>
std::string example = "This is an example string";
std::string word;
// Notice how you can loop through a string just like a vector<char>
for(size_t i = 0; i < example.size(); ++i) {
char c = example[i];
// When we see whitespace, print the current word and clear it
if(c == ' ' || c == '\t' || c == '\n') {
// Don't print anything if we don't have a word
if(!word.empty()) {
std::cout << word << std::endl;
word.clear();
}
} else {
// Append the current character to the end of the string
word += c; // or word.push_back(c)
}
}
// In case the line doesn't end with whitespace
if(!word.empty()) {
std::cout << word << std::endl;
}
答案 1 :(得分:0)
您可以使用矢量数据结构来保存图书类。 矢量记录;
答案 2 :(得分:0)
(我强烈建议使用矢量(或列表),因为它会避免双重读取文件,因为你根本不需要知道行数。)
要解析具有固定数量字段的行,原则上很容易:
int counter = 0;
string myString;
while(!myFile.eof())
{
getline(myFile, myString);
counter++;
}
counter --;
//Clear the error state flag
myFile.clear()
//Return to the beginning of the file:
myFile.seekg(ios_base::beg);
const int fieldCount = 5;
string field[fieldCount ];
string buffer= "";
char c = '\0';
for( int i = 0; i < counter; ++i ) {
for( int j = 0; j < fieldCount; ++j ) {
myFile.ignore(); //Ignore the first '"'
//Read each character up to the second '"'
while( myFile.good() && (c = myfile.get()) != '"' ) {
buffer += c;
}
field[j] = buffer;
buffer = "";
if( j != fieldCount - 1 ) {
myFile.ignore(); //Ignore the first ','
}
}
//Use the fields here.
}
我没有测试这段代码,我知道缺少错误测试,但它显示了一种方法。