如何将文件中的行存储为变量 [英] How to store lines from a file as a variable

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本文介绍了如何将文件中的行存储为变量的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想从一个名为organisation.txt的文本文件中显示员工编号名称、职业和部门,并将它们保存在OrganisationRecord类中声明的变量中.

I'd like to show the employee number name, occupation, and department of employees from a text file called organisation.txt, and save them in the variables declared in the class OrganisationRecord.

我该怎么做?

#include <iostream>
#include <string>
#include <vector>
#include <fstream>

#define ORGANISATIONALRECORDSFILE "organisation.txt"
#define HRRECORDSFILE "HR_records.txt"
#define PAYROLLRECORDSFILE "payroll_records.txt"

using namespace std;


class OrganisationRecord
{
private:
public:
    string name;
    string occupation;
    string department;
};

class HRRecord
{
private:
public:
    string address;
    string phonenumber;
    string ninumber;
};

class PayrollRecord
{
private:
public:
    string ninumber;
    double salary;
};

class PayrollProcessing
{
private:
    ifstream inputfile;
    ofstream outputfile;
    vector<OrganisationRecord> OrganisationRecords;
    vector<HRRecord> HRRecords;
    vector<PayrollRecord> PayrollRecords;
public:
    void loadOrganisationRecords(string filename);
    void loadHRRecords(string filename);
    void loadPayrollRecords(string filename);
    void displayEmployeeOfSalaryGTE(double salary);
    //GTE = greater than or equal to
};

void PayrollProcessing::loadOrganisationRecords(string filename)
{
    inputfile.open(ORGANISATIONALRECORDSFILE);

    if (!inputfile)
    {
        cout << "the organisation records file does not exist" << endl;
        return;
    }

        OrganisationRecord _organisationrecord;
        int employeenumber;


        while (inputfile >> employeenumber)
        {   
            while (inputfile >> _organisationrecord.name)
            {
                cout << _organisationrecord.name;
                cout << _organisationrecord.occupation;
                cout << _organisationrecord.department <<endl;
            }

            OrganisationRecords.push_back(_organisationrecord);
        }

}



int main(void)
{
    PayrollProcessing database1;
    database1.loadOrganisationRecords(ORGANISATIONALRECORDSFILE);

    return 0;
}

organisation.txt

0001 
Stephen Jones 
Sales Clerk 
Sales
0002 
John Smith 
Programmer 
OS Development
0003 
Fred Blogs 
Project Manager 
Outsourcing

推荐答案

当你使用 inputfile >>_organisationrecord.name 它在第一个空白字符处停止.因此,只有 Stephen 将被读取并存储在 organisationrecord.name 中.

When you use inputfile >> _organisationrecord.name it stops at the first whitespace character. Hence, only Stephen will be read and stored in organisationrecord.name.

您需要稍微改变一下策略.

You need to change your strategy a bit.

  1. 逐行读取文件内容.当没有更多行时停止.
  2. 按照您认为合适的方式处理每一行.

这是处理输入的一种方法.

Here's one way to deal with the input.

std::string line;
while ( std::getline(inputfile, line) )
{   
   // Extract the employeenumber from the line
   std::istringstream str(line);
   if ( !(str >> employeenumber) )
   {
      // Problem reading the employeenumber.
      // Stop reading.
      break;
   }

   if (!std::getline(inputfile, line) )
   {
      // Problem reading the next line.
      // Stop reading.
      break;
   }

   _organisationrecord.name = line;

   if (!std::getline(inputfile, line) )
   {
      // Problem reading the next line.
      // Stop reading.
      break;
   }
   _organisationrecord.occupation = line;

   if (!std::getline(inputfile, line) )
   {
      // Problem reading the next line.
      // Stop reading.
      break;
   }
   _organisationrecord.department = line;

   std::cout << _organisationrecord.employeenumber << std::endl;
   std::cout << _organisationrecord.name << std::endl;
   std::cout << _organisationrecord.occupation << std::endl;
   std::cout << _organisationrecord.department << endl;

   OrganisationRecords.push_back(_organisationrecord);
}

这篇关于如何将文件中的行存储为变量的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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