today=datetime.date.today() oneday=datetime.timedelta(days=1) yesterday=today-oneday return yesterday #2、返回今天日期 def getToday(): return datetime.date.today() #3、获取给定参数的前几天的日期,返回一个list def getDaysByNum(num): today=datetime.date.today() oneday=datetime.timedelta(days=...
然后,我们可以使用一个循环来生成日期范围。在循环中,我们可以使用timedelta函数来逐步增加日期。 fromdatetimeimporttimedelta date_range=[]current_date=start_datewhilecurrent_date<=end_date:date_range.append(current_date)current_date+=timedelta(days=1) 1. 2. 3. 4. 5. 6. 7. 8. 3.3 遍历日期范围,...
fromdatetimeimportdate,timedeltadefget_next_day():current_date=date.today()next_day=current_date+timedelta(days=1)print(next_day)get_next_day() 1. 2. 3. 4. 5. 6. 7. 8. 总结 通过以上步骤,我们使用Python成功获取了日期的后一天。首先,我们导入了datetime模块,然后获取当前日期,接着使用timedelta...
date_day - timedelta(daysdate_day.weekday()) print(f'***历史上的这周的周一是 {_weekdate} ***') if __name__ == '__main__': # 获取当前时间及格式化时间 get_current_time() # datetime.now()联系time_year() # 给定时间日期对应星期几 :历史上的 2022-08- 是星期 3 get_weekday...
#timedeltab1=datetime.timedelta(days=30, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=4, weeks=8)b2=datetime.timedelta(days=3, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=4, weeks=8)b3=b2-b1print(type(b3))print("The resultant duration = ",b3,end='n-...
>>> delta = timedelta(days=+3, hours=-4) >>> now + delta datetime.datetime(2020, 1, 29, 5, 37, 46, 380905) 在此示例中,您添加了三天并减去了四小时,因此新的时间datetime是 1 月 29 日凌晨 5:37。timedelta在这种方式下非常有用,但它有一定的局限性,因为它不能添加或减去大于一天的间隔,...
class datetime.timedelta([days[, seconds[, microseconds[, milliseconds[, minutes[, hours[, weeks]]]) 其没有必填参数,简单控制的话第一个整数就是多少天的间隔的意思: datetime.timedelta(10) 两个时间间隔对象可以彼此之间相加或相减,返回的仍是一个时间间隔对象。而更方便的是一个datetime对象如果减去一个...
# Filename : test.py# author by : www.runoob.com# 引入 datetime 模块importdatetimedefgetYesterday():today=datetime.date.today()oneday=datetime.timedelta(days=1)yesterday=today-onedayreturnyesterday# 输出print(getYesterday()) 执行以上代码输出结果为: ...
dt+=datetime.timedelta(days=1) s="{}{:0>2d}{:0>2d}".format(dt.year,dt.month,dt.day) if s[:]==s[::-1]: if flag: print(s) flag=False if s[0]==s[2]==s[5]==s[7] and s[1]==s[3]==s[4]==s[6]: print(s) ...
defget_year_first_and_last_day(now_time):this_year_start=datetime(now_time.year,1,1)this_year_end=datetime(now_time.year+1,1,1)-timedelta(days=1)returnthis_year_start,this_year_end defget_this_week_start_and_end_day():today=date.today()returntoday-timedelta(days=today.weekday())de...