C++ 使用 boost::date_time 获取当前时区当前时间的最简单方法？
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Simplest way to get current time in current timezone using boost::date_time?
提问by timday
If I do date +%H-%M-%Son the commandline (Debian/Lenny), I get a user-friendly (not UTC, not DST-less, the time a normal person has on their wristwatch) time printed.
如果我date +%H-%M-%S在命令行（Debian/Lenny）上这样做，我会打印一个用户友好的（不是 UTC，不是无 DST，正常人在手表上的时间）时间。
What's the simplest way to obtain the same thing with boost::date_time?
获得相同东西的最简单方法是boost::date_time什么？
If I do this:
如果我这样做：
std::ostringstream msg;
boost::local_time::local_date_time t =
boost::local_time::local_sec_clock::local_time(
boost::local_time::time_zone_ptr()
);
boost::local_time::local_time_facet* lf(
new boost::local_time::local_time_facet("%H-%M-%S")
);
msg.imbue(std::locale(msg.getloc(),lf));
msg << t;
Then msg.str()is an hour earlier than the time I want to see. I'm not sure whether this is because it's showing UTC or local timezone time without a DST correction (I'm in the UK).
然后msg.str()比我想看的时间早了一个小时。我不确定这是否是因为它显示的是 UTC 或本地时区时间而没有 DST 更正（我在英国）。
What's the simplest way to modify the above to yield the DST corrected local timezone time ? I have an idea it involves boost::date_time:: c_local_adjustorbut can't figure it out from the examples.
修改上述内容以产生 DST 校正的本地时区时间的最简单方法是什么？我有一个想法，boost::date_time:: c_local_adjustor但无法从示例中弄清楚。
回答by timday
This does what I want:
这做我想要的：
namespace pt = boost::posix_time;
std::ostringstream msg;
const pt::ptime now = pt::second_clock::local_time();
pt::time_facet*const f = new pt::time_facet("%H-%M-%S");
msg.imbue(std::locale(msg.getloc(),f));
msg << now;
回答by daminetreg
While this is not using boost::date_time it's relatively easy with boost::locale, which is quite more adapted for this task. As your need is simply getting a formatted time from the current locale.
虽然这不是使用 boost::date_time，但使用 boost::locale 相对容易，它更适合此任务。因为您的需要只是从当前语言环境中获取格式化的时间。
IMHO boost::date_time should be used when you deal with softwares like gantt/planning computations, were you have alot of date_time arithmetic. But simply for using time and doing some arithmetic on it, you will faster success with boost::locale.
恕我直言，当您处理甘特图/规划计算等软件时，应该使用 boost::date_time，如果您有很多 date_time 算术。但是仅仅为了使用时间并对其进行一些算术运算，使用 boost::locale 会更快地取得成功。
#include <iostream>
#include <boost/locale.hpp>
using namespace boost;
int main(int argc, char **argv) {
locale::generator gen;
std::locale::global(gen(""));
locale::date_time now;
std::cout.imbue(std::locale());
std::cout << locale::as::ftime("%H-%M-%S") << now << std::endl;
return 0;
}
Right now it should output : 15-45-48. :)
现在它应该输出：15-45-48。:)
回答by Xeverous
I haven't found other answers to be convenient enough, so here is an example that showcases how to get a local or universal time with full control of units:
我还没有找到其他足够方便的答案，所以这里有一个例子，展示了如何在完全控制单位的情况下获得当地时间或世界时间：
#include <boost/date_time/local_time/local_time.hpp>
#include <boost/format.hpp>
#include <iostream>
int main()
{
auto const now = boost::posix_time::microsec_clock::local_time(); // or universal_time() for GMT+0
if (now.is_special()) {
// some error happened
return 1;
}
// example timestamp (eg for logging)
auto const t = now.time_of_day();
boost::format formater("[%02d:%02d:%02d.%06d]");
formater % t.hours() % t.minutes() % t.seconds() % (t.total_microseconds() % 1000000);
std::cout << formater.str();
}
Note: the time_of_daystruct has no .microseconds()or .nanoseconds()functions, there is only .fractional_seconds()which returns an integer that is a multiple of configuration-dependent unit. .num_fractional_digits()can be used to obtain precision information where 10^ frac_digitsis the number of fractional_secondsthat is equal to 1 second.
注意：time_of_day结构体没有.microseconds()or.nanoseconds()函数，只有.fractional_seconds()which 返回一个整数，它是配置相关单元的倍数。.num_fractional_digits()可以用来获得精确的信息，其中10^frac_digits是的数目fractional_seconds等于1秒。
To obtain configuration-independent sub-second units one can perform modulo with the total_ milli/micro/nano _seconds()functions as a workaround.
要获得与配置无关的亚秒单位，可以使用total_ milli/micro/nano _seconds()函数进行取模作为一种解决方法。
