我在Python中看到了很多关于将日期字符串转换为datetime对象的内容,但我想采用另一种方式。 我有
datetime.datetime(2012, 2, 23, 0, 0)
我想把它转换成'2/23/2012'这样的字符串。
我在Python中看到了很多关于将日期字符串转换为datetime对象的内容,但我想采用另一种方式。 我有
datetime.datetime(2012, 2, 23, 0, 0)
我想把它转换成'2/23/2012'这样的字符串。
你可以使用strftime来帮助你格式化你的日期。
例如,
import datetime
t = datetime.datetime(2012, 2, 23, 0, 0)
t.strftime('%m/%d/%Y')
将收益率:
'02/23/2012'
关于格式的更多信息请参见这里
你可以使用简单的字符串格式化方法:
>>> dt = datetime.datetime(2012, 2, 23, 0, 0)
>>> '{0.month}/{0.day}/{0.year}'.format(dt)
'2/23/2012'
>>> '%s/%s/%s' % (dt.month, dt.day, dt.year)
'2/23/2012'
类型特定的格式也可以使用:
t = datetime.datetime(2012, 2, 23, 0, 0)
"{:%m/%d/%Y}".format(t)
输出:
'02/23/2012'
Date和datetime对象(以及time)支持一种迷你语言来指定输出,并且有两种方式来访问它:
直接方法调用:dt。strftime(这里的格式) 格式化方法(python 2.6+): '{: Format here}'.format(dt) f-strings (python 3.6+): f'{dt:format here}'
所以你的例子可以是这样的:
dt。strftime('日期是%b %d, %Y') '日期为{:%b %d, %Y}'.format(dt) f'日期为{dt:%b %d, %Y}'
在这三种情况下,输出结果是:
日期是2012年2月23日
为了完整起见:你也可以直接访问对象的属性,但这样你只能得到数字:
'The date is %s/%s/%s' % (dt.month, dt.day, dt.year)
# The date is 02/23/2012
学习这种迷你语言所花的时间是值得的。
作为参考,下面是迷你语言中使用的代码:
%a Weekday as locale’s abbreviated name. %A Weekday as locale’s full name. %w Weekday as a decimal number, where 0 is Sunday and 6 is Saturday. %d Day of the month as a zero-padded decimal number. %b Month as locale’s abbreviated name. %B Month as locale’s full name. %m Month as a zero-padded decimal number. 01, ..., 12 %y Year without century as a zero-padded decimal number. 00, ..., 99 %Y Year with century as a decimal number. 1970, 1988, 2001, 2013 %H Hour (24-hour clock) as a zero-padded decimal number. 00, ..., 23 %I Hour (12-hour clock) as a zero-padded decimal number. 01, ..., 12 %p Locale’s equivalent of either AM or PM. %M Minute as a zero-padded decimal number. 00, ..., 59 %S Second as a zero-padded decimal number. 00, ..., 59 %f Microsecond as a decimal number, zero-padded on the left. 000000, ..., 999999 %z UTC offset in the form +HHMM or -HHMM (empty if naive), +0000, -0400, +1030 %Z Time zone name (empty if naive), UTC, EST, CST %j Day of the year as a zero-padded decimal number. 001, ..., 366 %U Week number of the year (Sunday is the first) as a zero padded decimal number. %W Week number of the year (Monday is first) as a decimal number. %c Locale’s appropriate date and time representation. %x Locale’s appropriate date representation. %X Locale’s appropriate time representation. %% A literal '%' character.
字符串连接str.join可用于构建字符串。
d = datetime.now()
'/'.join(str(x) for x in (d.month, d.day, d.year))
'3/7/2016'
另一个选择:
import datetime
now=datetime.datetime.now()
now.isoformat()
# ouptut --> '2016-03-09T08:18:20.860968'
通过直接使用datetime对象的组件,可以将datetime对象转换为字符串。
from datetime import date
myDate = date.today()
#print(myDate) would output 2017-05-23 because that is today
#reassign the myDate variable to myDate = myDate.month
#then you could print(myDate.month) and you would get 5 as an integer
dateStr = str(myDate.month)+ "/" + str(myDate.day) + "/" + str(myDate.year)
# myDate.month is equal to 5 as an integer, i use str() to change it to a
# string I add(+)the "/" so now I have "5/" then myDate.day is 23 as
# an integer i change it to a string with str() and it is added to the "5/"
# to get "5/23" and then I add another "/" now we have "5/23/" next is the
# year which is 2017 as an integer, I use the function str() to change it to
# a string and add it to the rest of the string. Now we have "5/23/2017" as
# a string. The final line prints the string.
print(dateStr)
产出——> 5/23/2017
如果您正在寻找datetime到字符串转换的简单方法,可以省略该格式。您可以将datetime对象转换为str对象,然后使用数组切片。
In [1]: from datetime import datetime
In [2]: now = datetime.now()
In [3]: str(now)
Out[3]: '2019-04-26 18:03:50.941332'
In [5]: str(now)[:10]
Out[5]: '2019-04-26'
In [6]: str(now)[:19]
Out[6]: '2019-04-26 18:03:50'
但是注意下面的事情。如果其他解决方案会在变量为None时引发AttributeError,在这种情况下,您将收到一个'None'字符串。
In [9]: str(None)[:19]
Out[9]: 'None'
如果你也想要时间,那就去吧
datetime.datetime.now().__str__()
打印2019-07-11 19:36:31.118766在控制台为我
到目前为止最性感的版本是格式字符串。
from datetime import datetime
print(f'{datetime.today():%Y-%m-%d}')
end_date = "2021-04-18 16:00:00"
end_date_string = end_date.strftime("%Y-%m-%d")
print(end_date_string)
你可以用下面的方法将datetime转换为字符串:
from datetime import datetime
date_time = datetime(2012, 2, 23, 0, 0)
date = date_time.strftime('%m/%d/%Y')
print("date: %s" % date)
以下是一些可以用来将datetime转换为字符串的模式:
为了更好地理解,您可以参考这篇关于如何在Python或官方strftime文档中将字符串转换为datetime和datetime转换为字符串的文章
一种距离现在有多远的方法
通过传入参数li来支持不同的语言,这是一个对应的时间戳列表。
from datetime import datetime
from dateutil import parser
t1 = parser.parse("Tue May 26 15:14:45 2021")
t2 = parser.parse("Tue May 26 15:9:45 2021")
# 5min
t3 = parser.parse("Tue May 26 11:14:45 2021")
# 4h
t4 = parser.parse("Tue May 26 11:9:45 2021")
# 1day
t6 = parser.parse("Tue May 25 11:14:45 2021")
# 1day4h
t7 = parser.parse("Tue May 25 11:9:45 2021")
# 1day4h5min
t8 = parser.parse("Tue May 19 11:9:45 2021")
# 1w
t9 = parser.parse("Tue Apr 26 11:14:45 2021")
# 1m
t10 = parser.parse("Tue Oct 08 06:00:20 2019")
# 1y7m, 19m
t11 = parser.parse("Tue Jan 08 00:00:00 2019")
# 2y4m, 28m
# create: date of object creation
# now: time now
# li: a list of string indicate time (in any language)
# lst: suffix (in any language)
# long: display length
def howLongAgo(create, now, li, lst, long=2):
dif = create - now
print(dif.days)
sec = dif.days * 24 * 60 * 60 + dif.seconds
minute = sec // 60
sec %= 60
hour = minute // 60
minute %= 60
day = hour // 24
hour %= 24
week = day // 7
day %= 7
month = (week * 7) // 30
week %= 30
year = month // 12
month %= 12
s = []
for ii, tt in enumerate([sec, minute, hour, day, week, month, year]):
ss = li[ii]
if tt != 0:
if tt == 1:
s.append(str(tt) + ss)
else:
s.append(str(tt) + ss + 's')
return ' '.join(list(reversed(s))[:long]) + ' ' + lst
t = howLongAgo(t1, t11, [
'second',
'minute',
'hour',
'day',
'week',
'month',
'year',
], 'ago')
print(t)
# 2years 4months ago
我已经使用此方法将日期插入JSON对象
my_json_string = json.dumps({'date_of_birth': '''{}'''.format(date_of_birth)})