Error with getline

Error with getline

本文关键字：getline with Error      更新时间：2023-10-16

我在这里得到一个错误，我不知道为什么，我是c++的新手，如果你能看看我的代码的其余部分，以确保它是正确的，那将是伟大的。

这两行出现了错误。

getline(in, e.first, ',');
getline(in, e.last, ',');

它说Employee类没有成员First，而我知道它不在那个函数中，我该如何修复它?

下面是我剩下的代码。

#include <iostream>
#include <string>
#include <fstream>
using namespace std;
struct Person {
    string first;
    string last;
};
struct Address {
    string street;
    string city;
    string state;
    string zipcode;
};
struct Employee {
    Person name;
    Address homeAddress;
    int eid;
};
void readEmployee(istream& in, Employee& e);
void displayEmployee(ostream& out, const Employee& e);
int main(int argc, const char* argv[]) {
    Employee e[50];
    ifstream fin;
    ofstream fout;
    fin.open("employeesIn.txt");
    if (!fin.is_open()) {
        cerr << "Error opening employeesIn.txt for reading." << endl;
        exit(1);
    }
    fout.open("employeesOut.txt");
    if (!fout.is_open()) {
        cerr << "Error opening employeesOut.txt for writing." << endl;
        exit(1);
    }
    int EmployeePopulation = 0;
    readEmployee(fin, e[EmployeePopulation]);
    while (!fin.eof()) {
        EmployeePopulation++;
        readEmployee(fin, e[EmployeePopulation]);
    }
    fin.close();
    for (int i = 0; i <= EmployeePopulation - 1; i++) {
        displayEmployee(fout, e[i]);
    }
    fout.close();
    cout << endl;
    return 0;
}
void readEmployee(istream& in, Employee& e) {
    string cidText;
    if (getline(in, cidText, ',')) {
        e.eid = stoi(cidText);
        getline(in, e.first, ',');
        getline(in, e.last, ',');
        getline(in, e.homeAddress.street, ',');
        getline(in, e.homeAddress.city, ',');
        getline(in, e.homeAddress.state, ',');
        string zipcodeText;
        getline(in, zipcodeText, ',');
        e.homeAddress.zipcode = stoi(zipcodeText);
    }
}

我们将Person结构重命名为Name结构如何?
(毕竟它只包含第一个和最后一个)
这将给我们这个:

struct Name {
    string first;
    string last;
};

雇员现在是什么样子?
它看起来像这样:

+-------------+
|  Employee   |
|             |
| +---------+ |
| |  Name   | |
| +---------+ |
|             |
| +---------+ |
| | Address | |
| +---------+ |
|             |
+-------------+

Name是Employee的一部分，但是第一个和最后一个在哪里呢?它们是Name的一部分。
这是相同的图片，除了它更深入地显示第一和最后:

+---------------+
|   Employee    |
|               |
| +-----------+ |
| |   Name    | |
| |           | |
| | +-------+ | |
| | | first | | |
| | +-------+ | |
| |           | |
| | +-------+ | |
| | | last  | | |
| | +-------+ | |
| |           | |
| +-----------+ |
|               |
|  +---------+  |
|  | Address |  |
|  +---------+  |
|               |
+---------------+

需要使用两个点操作符('.')来访问first和last，因为它们的深度是前者的两倍。

Employee e;
e.name.first = "Joe";

您的代码已被重构以反映这些更改:

#include <iostream>
#include <string>
#include <fstream>
using namespace std;
struct Name {
    string first;
    string last;
};
struct Address {
    string street;
    string city;
    string state;
    string zipcode;
};
struct Employee {
    Name name;
    Address homeAddress;
    int eid;
};
void readEmployee(istream& in, Employee& e);
void displayEmployee(ostream& out, const Employee& e);
int main(int argc, const char* argv[]) {
    Employee e[50];
    ifstream fin;
    ofstream fout;
    fin.open("employeesIn.txt");
    if (!fin.is_open()) {
        cerr << "Error opening employeesIn.txt for reading." << endl;
        exit(1);
    }
    fout.open("employeesOut.txt");
    if (!fout.is_open()) {
        cerr << "Error opening employeesOut.txt for writing." << endl;
        exit(1);
    }
    int EmployeePopulation = 0;
    readEmployee(fin, e[EmployeePopulation]);
    while (!fin.eof()) {
        EmployeePopulation++;
        readEmployee(fin, e[EmployeePopulation]);
    }
    fin.close();
    for (int i = 0; i <= EmployeePopulation - 1; i++) {
        displayEmployee(fout, e[i]);
    }
    fout.close();
    cout << endl;
    return 0;
}
void readEmployee(istream& in, Employee& e) {
    string cidText;
    if (getline(in, cidText, ',')) {
        e.eid = stoi(cidText);
        getline(in, e.name.first, ',');
        getline(in, e.name.last, ',');
        getline(in, e.homeAddress.street, ',');
        getline(in, e.homeAddress.city, ',');
        getline(in, e.homeAddress.state, ',');
        string zipcodeText;
        getline(in, zipcodeText, ',');
        e.homeAddress.zipcode = stoi(zipcodeText);
    }
}
void displayEmployee(ostream& out, const Employee& e){
  //you can access the specific name values this way:
  cout <<  e.name.first << " " << e.name.last << endl;
  return;
}
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