blocks|key|941848|text|你只需要从今天的日期减去一天。在Python中，datetime.timedelta对象允许您作为对象创建特定的时间跨度。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|941849|datetime.timedelta(1)给出了“一天”的持续时间，并且可以从datetime对象中减去。在减去对象之后，可以使用datetime.strftime将结果(即日期对象)转换为基于选择的格式的字符串格式。|941850|>>>+from+datetime+import+datetime,+timedelta
>>>+yesterday+=+datetime.now()+-+timedelta(1)

>>>+type(yesterday)++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>>>+datetime.datetime++++

>>>+datetime.strftime(yesterday,+'%25Y-%25m-%25d')
'2015-05-26'|code-block|syntax|javascript|941851|注意，与调用datetime.strftime函数不同，您还可以直接使用datetime对象的strftime方法：|941852|>>>+(datetime.now()+-+timedelta(1)).strftime('%25Y-%25m-%25d')
'2015-05-26'|941853|作为一项职能：|941854|from+datetime+import+datetime,+timedelta


def+yesterday(frmt='%25Y-%25m-%25d',+string=True):
++++yesterday+=+datetime.now()+-+timedelta(1)
++++if+string:
++++++++return+yesterday.strftime(frmt)
++++return+yesterday|941855|示例：|941856|In+[10]:+yesterday()
Out[10]:+'2022-05-13'

In+[11]:+yesterday(string=False)
Out[11]:+datetime.datetime(2022,+5,+13,+12,+34,+31,+701270)|941857|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.5/library/datetime.html#datetime.timedelta|1|https://docs.python.org/3.5/library/datetime.html#datetime.datetime.strptime|2|https://docs.python.org/3.5/library/datetime.html#strftime-and-strptime-behavior^0|O|I|1D|2|0|0|0|L|13|8|1T|H|1T|H|1|2P|5|2|0|0|6|H|10|8|1B|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|F|3|G|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@$9|1N|A|1O|1|1P]|$9|1Q|A|1R|1|1S]]|E|$]]|$1|H|3|I|5|J|7|1T|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1U|8|@$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|21|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|22|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|23|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|24|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|25|8|@]|D|@]|E|$K|L]]|$1|Y|3|-4|5|6|7|26|8|@]|D|@]|E|$]]]|Z|$10|$5|11|12|13|E|$14|15]]|16|$5|11|12|13|E|$14|17]]|18|$5|11|12|13|E|$14|19]]]]

You Just need to subtract one day from today's date. In Python <code>datetime.timedelta</code> object lets you create specific spans of time as a <a href="https://docs.python.org/3.5/library/datetime.html#datetime.timedelta" rel="noreferrer"><code>timedelta</code> object</a>.
<code>datetime.timedelta(1)</code> gives you the duration of &quot;one day&quot; and is subtractable from a <code>datetime</code> object. After you subtracted the objects you can use <a href="https://docs.python.org/3.5/library/datetime.html#datetime.datetime.strptime" rel="noreferrer"><code>datetime.strftime</code></a> in order to convert the result --which is a date object-- to string format based on your <a href="https://docs.python.org/3.5/library/datetime.html#strftime-and-strptime-behavior" rel="noreferrer">format of choice</a>:
<pre><code>&gt;&gt;&gt; from datetime import datetime, timedelta
&gt;&gt;&gt; yesterday = datetime.now() - timedelta(1)

&gt;&gt;&gt; type(yesterday) 
&gt;&gt;&gt; datetime.datetime 

&gt;&gt;&gt; datetime.strftime(yesterday, '%Y-%m-%d')
'2015-05-26'
</code></pre>
Note that instead of calling the <code>datetime.strftime</code> function, you can also directly use <code>strftime</code> method of <code>datetime</code> objects:
<pre><code>&gt;&gt;&gt; (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')
'2015-05-26'
</code></pre>
As a function:
<pre><code>from datetime import datetime, timedelta


def yesterday(frmt='%Y-%m-%d', string=True):
 yesterday = datetime.now() - timedelta(1)
 if string:
 return yesterday.strftime(frmt)
 return yesterday
</code></pre>
example:
<pre><code>In [10]: yesterday()
Out[10]: '2022-05-13'

In [11]: yesterday(string=False)
Out[11]: datetime.datetime(2022, 5, 13, 12, 34, 31, 701270)
</code></pre>

blocks|key|77448|text|>>>+import+datetime
>>>+datetime.date.fromordinal(datetime.date.today().toordinal()-1).strftime("%25F")
'2015-05-26'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|77449|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>&gt;&gt;&gt; import datetime
&gt;&gt;&gt; datetime.date.fromordinal(datetime.date.today().toordinal()-1).strftime("%F")
'2015-05-26'
</code></pre>

blocks|key|77460|text|另一个使用today()方法计算当前日期，然后使用timedelta()减去日期的替代答案。其余步骤保持不变。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|77461|https://docs.python.org/3.7/library/datetime.html#timedelta-objects|77462|from+datetime+import+date,+timedelta
today+=+date.today()
yesterday+=+today+-+timedelta(days+=+1)
print(today)
print(yesterday)

Output:+
2019-06-14
2019-06-13|code-block|syntax|javascript|77463|entityMap|0|LINK|mutability|MUTABLE|url^0|5|7|P|B|0|0|1V|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|G]]]]

An alternative answer that uses <code>today()</code> method to calculate current date and then subtracts one using <code>timedelta()</code>. Rest of the steps remain the same.

<a href="https://docs.python.org/3.7/library/datetime.html#timedelta-objects" rel="noreferrer">https://docs.python.org/3.7/library/datetime.html#timedelta-objects</a>

<pre><code>from datetime import date, timedelta
today = date.today()
yesterday = today - timedelta(days = 1)
print(today)
print(yesterday)

Output: 
2019-06-14
2019-06-13
</code></pre>

blocks|key|1984123|text|对一个.isoformat()对象调用date将为您提供YYYY-MM-DD|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1984124|from+datetime+import+date,+timedelta
(date.today()+-+timedelta(1)).isoformat()|code-block|syntax|javascript|1984125|entityMap^0|3|C|J|4|S|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Calling <code>.isoformat()</code> on a <code>date</code> object will give you <code>YYYY-MM-DD</code>
<pre><code>from datetime import date, timedelta
(date.today() - timedelta(1)).isoformat()
</code></pre>

blocks|key|941936|text|我试图只使用基于import+datetime的这个答案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|941937|import+datetime

oneday+=+datetime.timedelta(days=1)
yesterday+=+datetime.date.today()+-+oneday|code-block|syntax|javascript|941938|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/15707574/15982746^0|8|F|O|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

I'm trying to use only <code>import datetime</code> based on <a href="https://stackoverflow.com/a/15707574/15982746">this answer</a>.
<pre><code>import datetime

oneday = datetime.timedelta(days=1)
yesterday = datetime.date.today() - oneday
</code></pre>

As an input to an API request I need to get yesterday's date as a string in the format <code>YYYY-MM-DD</code>. I have a working version which is:

<pre><code>yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)
report_date = str(yesterday.year) + \
 ('-' if len(str(yesterday.month)) == 2 else '-0') + str(yesterday.month) + \
 ('-' if len(str(yesterday.day)) == 2 else '-0') + str(yesterday.day)
</code></pre>

There must be a more elegant way to do this, interested for educational purposes as much as anything else!

Python - Get Yesterday's date as a string in YYYY-MM-DD format

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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作为API请求的输入，我需要将昨天的日期作为YYYY-MM-DD格式的字符串。我有一个工作版本，就是：yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)report_date = str(yesterday.year) +...

问Python -以YYYY-MM-DD格式将昨天的日期作为字符串获取
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问Python -以YYYY-MM-DD格式将昨天的日期作为字符串获取EN