以天为单位的两个时间戳之间的差异 C++ [英] Difference between two timestamps in days C++

问题描述

我在 XML 文件中有格式 (Year.Month.Day) 的时间戳.

I have timestamps in the format (Year.Month.Day) in an XML file.

我需要找出两个时间戳之间的天数差异.

I need to find out the difference between two timestamps in days.

示例时间戳:

<Time Stamp="20181015">

<Time Stamp="20181012">

如何找到上述时间戳之间的天数?

How can I find the number of days between the above timestamps?

天数 = date2 - date1.我正在考虑所有的日子(不需要跳过周末或任何其他日子).时区也无关紧要.

Number of days = date2 - date1. I am considering all the days (don't need to skip weekends or any other day). Time-zone does not matter as well.

PS: 我知道我必须从 XML 解析时间戳.解析值后，我陷入了逻辑中.

PS: I understand that I have to parse the timestamp from XML. I am stuck in the logic after parsing value.

Update-1: std::chrono::year 和其他类似的东西是 C++20 的一部分.但是我得到一个编译错误:

Update-1: std::chrono::year and other such things are part of C++20. But I get a compilation error:

命名空间std::chrono"没有成员year"

namespace "std::chrono" has no member "year"

推荐答案

您现在可以通过下载 Howard Hinnant 的免费开源日期/时间库.语法如下所示:

You can use C++20's syntax today (with C++11/14/17) by downloading Howard Hinnant's free, open-source date/time library. Here is what the syntax would look like:

#include "date/date.h"
#include <iostream>
#include <sstream>

int
main()
{
    using namespace date;
    using namespace std;
    istringstream in{"<Time Stamp=\"20181015\">\n<Time Stamp=\"20181012\">"};
    const string fmt = " <Time Stamp=\"%Y%m%d\">";
    sys_days date1, date2;
    in >> parse(fmt, date1) >> parse(fmt, date2);
    cout << date2 - date1 << '\n';
    int diff = (date2 - date1).count();
    cout << diff << '\n';
}

输出:

-3d
-3

如果您不需要时区支持(如本示例中所示)，则 date.h 是单个标头、仅限标头的库.这里是完整的文档.

If you don't need time zone support (as in this example), then date.h is a single header, header-only library. Here is the full documentation.

如果您需要时区支持，则需要一个带有标题和源的附加库:tz.h/tz.cpp.这是时区库的文档.

If you need time zone support, that requires an additional library with a header and source: tz.h/tz.cpp. Here is the documentation for the time zone library.

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