【问题标题】:Boost, how to parse following string to date/timeBoost,如何将以下字符串解析为日期/时间
【发布时间】:2015-01-10 03:12:34
【问题描述】:

我有以下毫秒/微秒精度字符串来解析为某种提升日期时间。

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

我看过有关方面的文档。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 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;
}

【问题讨论】:

  • 有什么异常?有什么消息吗?请提供一个最小的例子。

标签: c++ boost-date-time


【解决方案1】:

boost::posix_time::time_from_string 在解析格式方面非常严格。

您正在寻找从std::string 创建boost::posix_time::ptime 的不同方法。您想为stringstream 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;

给了

2009-Dec-01 00:00:04.437000

【讨论】:

  • 如果我现在想将 t 转换回字符串以确保它等于原始的单元格,我该怎么称呼?
  • 如何检查转换是否成功(字符串是否匹配指定格式)?
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