Question

您好我是C ++的新手，特别是C ++中的日期..我怎样才能将这些数字29（天）08（月）86（年）与今天的日期进行比较以获得年龄？

以下是我启动功能的方法：

std::string CodeP, day, month, year, age;

std::cout<<"Enter Permanent Code(example: SALA86082914) :\n "; //birthday first six numbers in code

std::cin>>CodeP;

year  = CodeP.substr (4,2);
month = CodeP.substr (6,2);
day = CodeP.substr (8,2);

std::cout<<"day :"<<day <<'\n';
std::cout<<"month :"<<month <<'\n';
std::cout<<"year :"<<year <<'\n';

//then get today's date to compare with the numbers of birthday to get age

Answer 1

首先减去年份以获得差异。现在比较几个月;如果今天的月份大于出生月份，那么你就完成了。如果没有，比较天数;如果今天的日子大于出生日，你就完成了。否则从差异中减去一个，那就是年龄。

Answer 2

尝试下面的代码，我已经为比较和处理您需要的两个日期做了必要的代码。请注意，我的代码的作用只是比较两个日期，并根据这些日期给出结果年龄。

很抱歉，但我必须让您完成创建代码的工作，该代码将根据六位数的生日输入为您提供预期的输出。我想你已经知道你仍然需要为你的问题提出自己的解决方案，我们只是在这里为你提供一个如何解决它的想法。我们只能帮助和支持你。希望我的帖子很有用！

    class age
    {
    private:
    int day;
    int month;
    int year;
    public:
    age():day(1), month(1), year(1)
    {}
    void get()
    {
    cout<<endl;
    cout<<"enter the day(dd):";
    cin>>day;
    cout<<"enter the month(mm):";
    cin>>month;
    cout<<"enter the year(yyyy):";
    cin>>year;
    cout<<endl;
    }
    void print(age a1, age a2)
    {
    if(a1.day>a2.day && a1.month>a2.month)
    {
    cout<<"your age is DD-MM-YYYY"<<endl;
    cout<<"\t\t"<<a1.day-a2.day<<"-"<<a1.month-a2.month<<"-"<<a1.year-a2.year;
     cout<<endl<<endl;
     }
    else if(a1.day<a2.day && a1.month>a2.month)
    {
    cout<<"your age is DD-MM-YYYY"<<endl;
    cout<<"\t\t"<<(a1.day+30)-a2.day<<"-"<<(a1.month-1)-a2.month<<"-"<<a1.year-a2.year;?
    cout<<endl<<endl;
    }
    else if(a1.day>a2.day && a1.month<a2.month)
    {
   cout<<"your age is DD-MM-YYYY"<<endl;
   cout<<"\t\t"<<a1.day-a2.day<<"-"<<(a1.month+12)-a2.month<<"-"<<(a1.year-1)-a2.year;
   cout<<endl<<endl;
   }
   else if(a1.day<a2.day && a1.month<a2.month)
   {
   cout<<"your age is DD-MM-YYYY"<<endl;
    cout<<"\t\t"<<(a1.day+30)-a2.day<<"-"<<(a1.month+11)-a2.month<<"-"<<(a1.year-1)-a2.year;
   cout<<endl<<endl;
   }     
   }
   };
   int main()
   {
   age a1, a2, a3;
  cout<<"\t Enter the current date.";
  cout<<endl<<endl;
  a1.get();
  cout<<"\t enter Date of Birth.";
  cout<<endl<<endl;
  a2.get();
  a3.print(a1,a2);
  return 0;
  }

Answer 3

这将计算年龄（年）的近似值：

#include <ctime>
#include <iostream>
using namespace std;
int main(){
    struct tm date = {0};
    int day, month, year;
    cout<<"Year: ";
    cin>>year;
    cout<<"Month: ";
    cin>>month;
    cout<<"Day: ";
    cin>>day;
    date.tm_year = year-1900;
    date.tm_mon = month-1;
    date.tm_mday = day;
    time_t normal = mktime(&date);
    time_t current;
    time(&current);
    long d = (difftime(current, normal) + 86400L/2) / 86400L;
    cout<<"You have~: "<<d/365.0<<" years.\n";
    return (0);
}

Answer 4

最好的方法是使用Boost.Gregorian：

假设

#include <boost/date_time/gregorian/gregorian.hpp>

using namespace boost::gregorian;

你会这样做：

date bday { from_undelimited_string(std::string("19")+a.substr(4,6)) };
date now = day_clock::local_day();

unsigned years = 0;
year_iterator y { bday };
while( *++y <= *year_iterator(now) )
  ++years;
--y;
std::cout << *y << "\n";

unsigned months = 0;
month_iterator m { *y };
while( *++m <= *month_iterator(now) )
  ++months;
--m;
std::cout << *m << "\n";

unsigned days = 0;
day_iterator d { *m };
while( *++d <= *day_iterator(now) )
  ++days;
--d;
std::cout << *d << "\n";

std::cout << years << " years, " << months << " months, " << days << " days\n";

Answer 5

我找到了另一个（更简单的？）答案，使用Boost.Gregorian

#include <boost/date_time/gregorian/gregorian.hpp>

using namespace boost::gregorian;

template<typename Iterator, typename Date>
struct date_count : public std::pair<unsigned, Date> {
  typedef std::pair<unsigned, Date> p;
  operator unsigned() { return p::first; }
  operator Date() { return p::second; }
  date_count(Date begin, Date end) : p(0, begin) {
    Iterator b { begin };
    while( *++b <= end )
      ++p::first;
    p::second = *--b;
  }
};

你必须这样使用它：

  date bday { from_undelimited_string(std::string("19")+a.substr(4,6)) };
  date now = day_clock::local_day();

  date_count<year_iterator, date> years { bday, now };
  date_count<month_iterator, date> months { years, now };
  date_count<day_iterator, date> days { months, now };

  std::cout << "From " << bday << " to " << now << " there are " <<
    std::setw(5) << years << " years, " <<
    std::setw(3) << months << " months, " <<
    std::setw(3) << days << " days\n";

如果您只想要年数，可以这样做：

date_count<year_iterator> years(bday, now);

并按照＆＃34;年＆＃34;的年数运行变量。：d

C ++比较日期以获得年龄

5 个答案: