如何在 C++ 11 中使用小时分钟和秒来查找两个日期之间的持续时间？答案

【问题标题】：How to find Time Duration between two dates with Hours minutes and seconds in C++ 11?如何在 C++ 11 中使用小时分钟和秒来查找两个日期之间的持续时间？
【发布时间】：2021-08-29 00:48:31
【问题描述】：

我需要找出两个给定日期之间的时间差。

我尝试了以下代码，但无法将其转换为小时分钟秒

         const char *time_details = "06/10/2021 16:35:12";
struct tm tm;
strptime(time_details, "%m/%d/%Y %H:%M:%S", &tm); // Prev date

time_t t = mktime(&tm);

const char *time_details1 = "06/11/2021 14:35:12";
struct tm tm1;
strptime(time_details1, "%m/%d/%Y %H:%M:%S", &tm1); // future date

time_t t33 = mktime(&tm1);

int timediff1 = difftime(t33,t);

    int seconds1 = timediff1%60;
int hour1 =seconds1/ 3600;
int minutes1 = seconds1 / 60 ;

上面总是以小时分秒显示“0”。但获得一些价值

【问题讨论】：

如果你使用的是 c++11，那么看看使用 - stackoverflow.com/questions/31657511/…
数学上面的代码应该是这样的：int hour1 = timediff1 / 3600; int minutes1 = timediff1 % 3600 / 60; 因为int seconds1 = timediff1%60 给出的秒数是从 0 到 59，而你显然不能从这个值计算小时和分钟。
@dewaffled 谢谢你的帮助。我得到了输出。谢谢
@Dean-Jason。谢谢。

标签： c++ visual-studio c++11 mktime time-t

【解决方案1】：

有了 C++2a 的早期支持，您可以做这样的事情

std::tm tm1 = {}, tm2 = {};

std::stringstream ss("06/10/2021 16:35:12 06/11/2021 14:35:12");

ss >> std::get_time(&tm1, "%m/%d/%Y %H:%M:%S");
ss >> std::get_time(&tm2, "%m/%d/%Y %H:%M:%S");
auto tp1 = std::chrono::system_clock::from_time_t(std::mktime(&tm1));
auto tp2 = std::chrono::system_clock::from_time_t(std::mktime(&tm2));
auto diff = tp2 - tp1;

// convert to human-readable form
auto d = std::chrono::duration_cast<std::chrono::days>(diff);
diff -= d;
auto h = std::chrono::duration_cast<std::chrono::hours>(diff);
diff -= h;
auto m = std::chrono::duration_cast<std::chrono::minutes>(diff);
diff -= m;
auto s = std::chrono::duration_cast<std::chrono::seconds>(diff);

std::cout << std::setw(2) << d.count() << "d:"
   << std::setw(2) << h.count() << "h:"
   << std::setw(2) << m.count() << "m:"
   << std::setw(2) << s.count() << 's';

并且为duration 类提供operator<< 的全面支持。对于 C++11，std::chrono::days 和 hours 必须替换为等价物，例如duration<int, std::ratio<86400>> 为days。

【讨论】：

【解决方案2】：

试试这个

void computeTimeDiff(struct TIME t1, struct TIME t2, struct TIME *difference){
    
    if(t2.seconds > t1.seconds)
    {
        --t1.minutes;
        t1.seconds += 60;
    }

    difference->seconds = t1.seconds - t2.seconds;
    if(t2.minutes > t1.minutes)
    {
        --t1.hours;
        t1.minutes += 60;
    }
    difference->minutes = t1.minutes-t2.minutes;
    difference->hours = t1.hours-t2.hours;
}

【讨论】：