python 获取今天、昨天、明天的时间时间戳

if __name__ == '__main__':
import datetime
import time

# 获取今天日期
print("获取今天日期")
today = datetime.date.today()
print(today) # 2019-09-05
print(type(today)) # <type 'datetime.date'>
print("**" * 10)

# 昨天日期
print("昨天日期")
yesterday = today - datetime.timedelta(days=1)
print(yesterday) # 2019-09-04
print("**" * 10)

# 明天时间
print("明天时间")
tomorrow = today + datetime.timedelta(days=1)
acquire = today + datetime.timedelta(days=2)
print(tomorrow) # 2019-09-06
print(acquire) # 2019-09-07
print("**" * 10)

# 昨天开始时间戳
print("昨天开始时间戳")
yesterday_start_time = int(time.mktime(time.strptime(str(yesterday), '%Y-%m-%d')))
print(yesterday_start_time) # 1567526400 转换后 2019-09-04 00:00:00
print("**" * 10)

# 昨天结束时间戳
print("昨天结束时间戳")
yesterday_end_time = int(time.mktime(time.strptime(str(today), '%Y-%m-%d'))) - 1
print(yesterday_end_time) # 1567612799 转换后 2019-09-04 23:59:59
print("**" * 10)

# 今天开始时间戳
print("今天开始时间戳")
today_start_time = yesterday_end_time + 1
print(today_start_time) # 1567612800 转换后 2019-09-05 00:00:00
print("**" * 10)

# 今天结束时间戳
print("今天结束时间戳")
today_end_time = int(time.mktime(time.strptime(str(tomorrow), '%Y-%m-%d'))) - 1
print(today_end_time) # 1567699199 转换后 2019-09-05 23:59:59
print("**" * 10)

# 明天开始时间戳
print("明天开始时间戳")
tomorrow_start_time = int(time.mktime(time.strptime(str(tomorrow), '%Y-%m-%d')))
print(tomorrow_start_time) # 1567699200 转换后 2019-09-06 00:00:00
print("**" * 10)

# 明天结束时间戳
print("明天结束时间戳")
tomorrow_end_time = int(time.mktime(time.strptime(str(acquire), '%Y-%m-%d'))) - 1
print(tomorrow_end_time) # 1567785599 转换后 2019-09-06 23:59:59
print("**" * 10)

# 最近24小时
print("最近24小时")
time_now = time.time()
before_24 = int(time_now - 24 * 60 * 60)
print(before_24)
print("**" * 10)

# 最近一周
print("最近一周")
time_now = time.time()
before_week = int(time_now - 7 * 24 * 60 * 60)
print(before_week)
print("**" * 10)


# 最近一月
print("最近一月")
time_now = time.time()
before_nonth = int(time_now - 30 * 24 * 60 * 60)
print(before_nonth)
print("**" * 10)

来源：https://blog.csdn.net/a961634066/java/article/details/100554487