![]() ![]() Yesterday = datetime. You can solve this problem by making a list of the days that are. Note that instead of calling the datetime.strftime function, you can also directly use strftime method of datetime objects: > (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')Īs a function: from datetime import datetime, timedeltaĭef yesterday(frmt='%Y-%m-%d', string=True): If the day before the day before yesterday is 3 days after Sunday, today is Saturday. > datetime.strftime(yesterday, '%Y-%m-%d') ereyesterday ( not comparable ) ( obsolete) On the day before yesterday. > yesterday = datetime.now() - timedelta(1) After you subtracted the objects you can use datetime.strftime in order to convert the result -which is a date object- to string format based on your format of choice: > from datetime import datetime, timedelta to retrive records from yesterday : SELECT FROM table WHERE DATE CURDATE () - INTERVAL 1 DAY. ![]() When asked about his birthday, a man said: The day before yesterday I was only 25 and next year. You can just use a simpler method, without DATEADD or DATESUB : SELECT FROM table WHERE DATE CURDATE () - INTERVAL 2 DAY. In Python datetime.timedelta object lets you create specific spans of time as a timedelta object.ĭatetime.timedelta(1) gives you the duration of "one day" and is subtractable from a datetime object. A very common and interesting Whatsapp puzzle on Birthday. You just need to subtract one day from today's date.
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