[英]How to convert date/time string to epoch?
Have the following string: 具有以下字符串:
"date Thursday June 03 12:02:56 2017"
What would be the proper way of convert it to epoch time? 将其转换为纪元时间的正确方法是什么?
You can use datetime.strptime()
to parse your date and then just do delta with the epoch: 您可以使用datetime.strptime()
来解析您的日期,然后仅对纪元进行增量操作:
from datetime import datetime as dt
epoch = dt(1970, 1, 1)
date = "date Thursday June 03 12:02:56 2017"
epoch_time = int((dt.strptime(date, "date %A %B %d %H:%M:%S %Y") - epoch).total_seconds())
# 1496491376
Another solution is to create a time.struct_time
structure by parsing with time.strptime()
and then pass it into calendar.timegm()
to convert to epoch time. 另一种解决方案是通过使用time.strptime()
解析来创建一个time.struct_time
结构,然后将其传递到calendar.timegm()
以转换为纪元时间。
import time
import calendar
timestr = "date Thursday June 03 12:02:56 2017"
calendar.timegm(time.strptime(timestr, "date %A %B %d %H:%M:%S %Y"))
# returns 1496491376
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