Boost，如何将以下字符串解析为日期/时间

Boost, how to parse following string to date/time

本文关键字：日期 时间 字符串 Boost      更新时间：2023-10-16

我有以下毫秒精度的字符串要解析为某种提升日期时间。

std::string cell ="20091201 00:00:04.437";

我看过关于facets的文档。像这样的

date_input_facet* f = new date_input_facet();
f->format("%Y%m%d %F *");

但我不知道如何使用它们。

我用从StackOverflow中清除的代码尝试了这个程序，但我无法显示毫秒：

#include <string>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <boost/algorithm/string.hpp>
#include <boost/date_time/posix_time/posix_time.hpp>
#include <boost/date_time.hpp>
namespace bt = boost::posix_time;
const std::locale formats[] =
{
    std::locale(std::locale::classic(),new bt::time_input_facet("%Y%m%d %H:%M:%S.f")),
    std::locale(std::locale::classic(),new bt::time_input_facet("%Y-%m-%d %H:%M:%S")),
    std::locale(std::locale::classic(),new bt::time_input_facet("%Y/%m/%d %H:%M:%S")),
    std::locale(std::locale::classic(),new bt::time_input_facet("%d.%m.%Y %H:%M:%S")),
    std::locale(std::locale::classic(),new bt::time_input_facet("%Y-%m-%d"))
};
const size_t formats_n = sizeof(formats) / sizeof(formats[0]);
std::time_t pt_to_time_t(const bt::ptime& pt)
{
    bt::ptime timet_start(boost::gregorian::date(1970,1,1));
    bt::time_duration diff = pt - timet_start;
    return diff.ticks()/bt::time_duration::rep_type::ticks_per_second;
}
void seconds_from_epoch(const std::string& s)
{
    bt::ptime pt;
    for(size_t i = 0; i < formats_n; ++i)
    {
        std::istringstream is(s);
        is.imbue(formats[i]);
        is >> pt;
        if(pt != bt::ptime()) break;
    }
    bt::time_duration td = pt.time_of_day();
    long fs = td.fractional_seconds();
    std::cout << " ptime is " << pt << 'n';
    std::cout << " seconds from epoch are " << pt_to_time_t(pt) << " " << fs << 'n';
}
int main(int, char *argv[])
{
    std::string cell ="20091201 00:00:04.437";
    seconds_from_epoch(cell);
    int enterAnumber;
    std::
    cin >> enterAnumber;
}

boost::posix_time::time_from_string在解析格式时非常严格。

您正在寻找一种不同于std::string创建boost::posix_time::ptime的方法。您希望在stringstream中注入以下格式：

const std::string cell = "20091201 00:00:04.437";
const std::locale loc = std::locale(std::locale::classic(), new boost::posix_time::time_input_facet("%Y%m%d %H:%M:%S%f"));
std::istringstream is(cell);
is.imbue(loc);
boost::posix_time::ptime t;
is >> t;

然后

std::cout << t << std::endl;

给出

2009-Dec-01 00:00:04.437000
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