解析ISO 8601持续时间 [英] Parse ISO 8601 durations

问题描述

在ISO 8601中，持续时间以格式 P [n] Y [n] M [n] DT [n] H [n] M [n] S 。

示例：

20秒：

  PT20.0S 
  

一年，2个月，3天，4小时，5分6秒：

  P1Y2M3DT4H5M6S 
  
 
 
 
 问题： 
 
 
 
包含以iso 8601格式的持续时间。我想获得该持续时间的总秒数。 
 
 
 $ b  
 
 b 
例如，在boost DateTime中有ptime from_iso_string（std :: string），不适合这里。 
解决方案
使用标准的regex库，你想要的正则表达式是这样的：
 P\（\（[0-9] + \）Y \）？\ \（[0-9] + \）M \）？\（\（[0-9] + \）D\）？T \（\（[0-9] + \）H \）？\（\（[0-9] + \）M \）？\（\（[0-9] + \（\。 ] + \）？S \）？
  
年，月等并计算总秒数。
 
In ISO 8601, durations are given in the format P[n]Y[n]M[n]DT[n]H[n]M[n]S.

Examples:

20 seconds:
PT20.0S
One year, 2 month, 3 days, 4 hours, 5 minutes, 6 seconds:
P1Y2M3DT4H5M6S
Question:

Given a string that contains a duration in iso 8601 format. I want to obtain the overall number of seconds of that duration. What is the recommended way in standard C++11 to achieve this?

Remarks:

E.g., there is ptime from_iso_string(std::string) in boost DateTime which does not fit here. Is there a similar way without doing a regex by hand?
解决方案
Use the standard regex library, the regex you want is something like:
"P\(\([0-9]+\)Y\)?\(\([0-9]+\)M\)?\(\([0-9]+\)D\)?T\(\([0-9]+\)H\)?\(\([0-9]+\)M\)?\(\([0-9]+\(\.[0-9]+\)?S\)?"
from that you can pull out the number of years, months etc and calculate total seconds.

                        
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