这是我的代码：
import datetime
today = datetime.date.today()
print(today)
这张照片是：2008-11-22，这正是我想要的。
但是，我有一个列表，我要将其附加到列表中，然后突然一切都变得“不稳定”。代码如下：
import datetime
mylist = []
today = datetime.date.today()
mylist.append(today)
print(mylist)
这将打印以下内容：
[datetime.date(2008, 11, 22)]
我怎样才能得到像2008-11-22这样的简单约会？
当前回答
您需要将datetime对象转换为str。
以下代码适用于我：
import datetime
collection = []
dateTimeString = str(datetime.date.today())
collection.append(dateTimeString)
print(collection)
如果你需要更多帮助，请告诉我。
其他回答
date、datetime和time对象都支持strftime（格式）方法，在显式格式的控制下创建表示时间的字符串一串
以下是格式代码及其指令和含义的列表。
%a Locale’s abbreviated weekday name.
%A Locale’s full weekday name.
%b Locale’s abbreviated month name.
%B Locale’s full month name.
%c Locale’s appropriate date and time representation.
%d Day of the month as a decimal number [01,31].
%f Microsecond as a decimal number [0,999999], zero-padded on the left
%H Hour (24-hour clock) as a decimal number [00,23].
%I Hour (12-hour clock) as a decimal number [01,12].
%j Day of the year as a decimal number [001,366].
%m Month as a decimal number [01,12].
%M Minute as a decimal number [00,59].
%p Locale’s equivalent of either AM or PM.
%S Second as a decimal number [00,61].
%U Week number of the year (Sunday as the first day of the week)
%w Weekday as a decimal number [0(Sunday),6].
%W Week number of the year (Monday as the first day of the week)
%x Locale’s appropriate date representation.
%X Locale’s appropriate time representation.
%y Year without century as a decimal number [00,99].
%Y Year with century as a decimal number.
%z UTC offset in the form +HHMM or -HHMM.
%Z Time zone name (empty string if the object is naive).
%% A literal '%' character.
这是我们可以使用Python中的datetime和time模块所做的
import time
import datetime
print "Time in seconds since the epoch: %s" %time.time()
print "Current date and time: ", datetime.datetime.now()
print "Or like this: ", datetime.datetime.now().strftime("%y-%m-%d-%H-%M")
print "Current year: ", datetime.date.today().strftime("%Y")
print "Month of year: ", datetime.date.today().strftime("%B")
print "Week number of the year: ", datetime.date.today().strftime("%W")
print "Weekday of the week: ", datetime.date.today().strftime("%w")
print "Day of year: ", datetime.date.today().strftime("%j")
print "Day of the month : ", datetime.date.today().strftime("%d")
print "Day of week: ", datetime.date.today().strftime("%A")
这将打印出如下内容：
Time in seconds since the epoch: 1349271346.46
Current date and time: 2012-10-03 15:35:46.461491
Or like this: 12-10-03-15-35
Current year: 2012
Month of year: October
Week number of the year: 40
Weekday of the week: 3
Day of year: 277
Day of the month : 03
Day of week: Wednesday
您需要将datetime对象转换为str。
以下代码适用于我：
import datetime
collection = []
dateTimeString = str(datetime.date.today())
collection.append(dateTimeString)
print(collection)
如果你需要更多帮助，请告诉我。
# convert date time to regular format.
d_date = datetime.datetime.now()
reg_format_date = d_date.strftime("%Y-%m-%d %I:%M:%S %p")
print(reg_format_date)
# some other date formats.
reg_format_date = d_date.strftime("%d %B %Y %I:%M:%S %p")
print(reg_format_date)
reg_format_date = d_date.strftime("%Y-%m-%d %H:%M:%S")
print(reg_format_date)
输出，输出
2016-10-06 01:21:34 PM
06 October 2016 01:21:34 PM
2016-10-06 13:21:34
import datetime
print datetime.datetime.now().strftime("%Y-%m-%d %H:%M")
编辑：
根据Cees的建议，我也开始利用时间：
import time
print time.strftime("%Y-%m-%d %H:%M")
由于print today返回您想要的内容，这意味着today对象的__str__函数返回您要查找的字符串。
因此，您也可以执行mylist.append（today.__str__（））。
