C++ 将时间字符串从纪元转换为秒
我有一个格式如下的字符串:
I have a string with the following format:
2010-11-04T23:23:01Z
2010-11-04T23:23:01Z
Z 表示时间是 UTC.
我宁愿将其存储为纪元时间，以便于比较.
The Z indicates that the time is UTC.
I would rather store this as a epoch time to make comparison easy.
推荐的方法是什么?
目前(经过快速搜索)最简单的算法是:
Currently (after a quck search) the simplist algorithm is:
1: <Convert string to struct_tm: by manually parsing string>
2: Use mktime() to convert struct_tm to epoch time.
// Problem here is that mktime uses local time not UTC time.
推荐答案
使用 C++11 功能，我们现在可以使用流来解析时间:
Using C++11 functionality we can now use streams to parse times:
iomanip std::get_time将根据一组格式参数转换一个字符串，并将它们转换为 struct tz 对象.
The iomanip std::get_time will convert a string based on a set of format parameters and convert them into a struct tz object.
然后您可以使用 std::mktime() 将其转换为纪元值.
You can then use std::mktime() to convert this into an epoch value.
#include <iostream>
#include <sstream>
#include <locale>
#include <iomanip>
int main()
{
std::tm t = {};
std::istringstream ss("2010-11-04T23:23:01Z");
if (ss >> std::get_time(&t, "%Y-%m-%dT%H:%M:%S"))
{
std::cout << std::put_time(&t, "%c") << "
"
<< std::mktime(&t) << "
";
}
else
{
std::cout << "Parse failed
";
}
return 0;
}
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