代码
# -*- coding: utf-8 -*-
# Author : markadc
from datetime import datetime, timedelta
# 获取今天任意时刻的时间戳
def today_anytime_tsp(hour, minute, second=0):
now = datetime.now()
today_0 = now - timedelta(hours=now.hour, minutes=now.minute, seconds=now.second)
today_anytime = today_0 + timedelta(hours=hour, minutes=minute, seconds=second)
tsp = today_anytime.timestamp()
print('{} 的时间戳是 {}'.format(today_anytime, tsp))
return tsp
if __name__ == '__main__':
tsp = today_anytime_tsp(18, 15, 23) # 获取今天18点15分23秒的时间戳
print(tsp)
输出
这个问题也是常见问题，可能碰到这种问题我们就会这么写yesterday = today - 1，tomorrow = today + 1；这样写实际会报错，会提示不支持的操作类型。碰到时间差的问题的，优先考虑datetime模块中的timedelta。具体获取很简单
>>> import datetime
>>> today = datetime.date.today() # 今天
>>> yesterday = today - datetime.timedelta(days=1) # 昨天
>>> tomorrow = today + datetime.timedelta(days=1) # 明天
>>> print(yesterday,today,tomorrow)
(datetime.date(2018, 7, 31), datetime.date(2018, 8, 1), datetime.date(2018, 8, 2))
>>>
其他的也可以获取一秒后，一分钟，一小时，甚至一年的具体日期
>>> import datetime
>>> t1 = datetime.datetime.today() # 获取现在时间
>>> t1
datetime.datetime(2018, 8, 1, 21, 34, 19, 924000)
>>> t2 = t1 + datetime.timedelta(seconds=1) # 获取一秒后的时间
>>> t2
datetime.datetime(2018, 8, 1, 21, 34, 20, 924000)
>>> t3 = t1 + datetime.timedelta(seconds=60) # 获取一分后的时间
>>> t3
datetime.datetime(2018, 8, 1, 21, 35, 19, 924000)
>>> t4 = t1 + datetime.timedelta(seconds=3600) # 获取一小时后的时间
>>> t4
datetime.datetime(2018, 8, 1, 22, 34, 19, 924000)
>>> today = datetime.date.today() # 获取当前日期
>>> anday = today + datetime.timedelta(days=365) # 获取一年后日期
>>> print(anday)
2019-08-01
>>>
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