ValueError：day日期时间超出范围 [英] ValueError: day is out of range for month datetime

问题描述

我遇到了一些我一直在写的代码的问题。我以四个输入（日，月和年）作为日期，以及他们想重复任务的次数（例如每个星期三3周）。代码很好，但是如果几个月之间的时间不同，我会收到以下错误：

 文件C：\Users\dansi \AppData\Local\Programs\Python\Python36-32\gui test 3.py，第72行，addtimeslot 
 fulldateadd = datetime.date（年，月，日）
 ValueError：日期超出月份的范围
  

相关代码的一部分：

  for i in range（0，times）：
 fulldateadd = datetime.date（年，月，日）
 cursor.execute（'''INSERT INTO dates（Date，Name，Start，End）VALUES（？，？，？，？）;'''（（fulldateadd，name1，starttimehour，endtimehour））
 day = day + 7 
 if day> 31：
 month = month + 1 
  

我试图增加月份天数超过31，但似乎不起作用。

解决方案

>组件，然后创建一个新的不是一个好主意。主要是因为自己处理公历是不是那个愉快的IMHO，而datetime对象可以为你做。

在这一点上，一个更简单的方法是在循环中添加一个timedelta到你的datetime。例如，

 >>>来自datetime import timedelta 
>>>> times = 4 
>>>> cur_date = datetime.date（2017，2，24）
 
>>> for _ in range（times）：
 print（'today is {0}，do something'.format（cur_date））
 cur_date + = timedelta（days = 7）
 
今天是2017-02-24，做某事
今天是2017-03-03，做某事
今天是2017-03-10，做某事
今天是2017-03-17 ，做某事
  

这可以放在一个生成器中，这取决于你的用例。

 >>>对于范围（次）中的x的（cur_date + timedelta（days = x * 7）），dt（）为
 print（'today is {0}，do something'.format（dt））
 
今天是2017-02-24，做某事
今天是2017-03-03，做某事
今天是2017-03-10，做某事
今天是2017-03 -17，做某事
  

或与Pandas pd.date_range

 >>>将大熊猫导入为pd 
>>> list（pd.date_range（start ='2017-02-24'，periods = 4，freq ='7D'））
 
 [Timestamp（'2017-02-24 00:00:00' ，freq ='7D'），
时间戳（'2017-03-03 00:00:00'，频率='7D'），
时间戳（'2017-03-10 00:00： 00'，频率='7D'），
时间戳（'2017-03-17 00:00:00'，频率='7D'）] 
  

---

现在如果您尝试使用此方法，会发生什么情况？

 >>>年，月，日= 2017，2，24 
 
>>>对于我在范围（0，次）：
天=天
 fulldateadd = datetime.date（年，月，日）
打印（'今天是{0}，做某事'。 format（fulldateadd））
 day = day + 7 
 if day> 31：
 day = day  -  31 
 month = month + 1 
 
今天是2017-02-24，做某事
 
 ValueErrorTraceback（最近的call last）
< ipython-input-255-7df608ebbf8e>在< module>（）
 1 for i in range（0，times）：
 2 day = day 
 ----> 3 fulldateadd = datetime.date（年，月，日）
 4 print（'today is {0}，do something'.format（fulldateadd））
 5天=天+ 7 
 
 ValueError：day超出范围为
  

二月没有31天。 ..所以你必须包括一张支票与映射到每个月的天数。包括闰年的逻辑。

I have run into a problem with some code I have been writing. I take in four inputs ( day, month and year ) as a date, and times for how many times they want to repeat the task ( eg. every Monday for 3 weeks ). The code is great however if the weeks differ between months I get this error:

  File "C:\Users\dansi\AppData\Local\Programs\Python\Python36-32\gui test 3.py", line 72, in addtimeslot
fulldateadd = datetime.date(year, month, day)
ValueError: day is out of range for month

Part of code that is relevant:

for i in range(0 , times):
    fulldateadd = datetime.date(year, month, day)
    cursor.execute( '''INSERT INTO dates (Date, Name, Start, End) VALUES( ?,?,?,? );''', (fulldateadd , name1, starttimehour, endtimehour))
    day = day + 7
    if day > 31:
        month = month + 1

I've tried to increment the month when the number of days are more than 31 however it doesn't seem to work.

解决方案

There are several reasons why incrementing the components of a datetime and then creating a new one is not a good idea. Primarily because dealing with the Gregorian calendar yourself isn't that enjoyable IMHO, and datetime objects can do it for you.

On that note, a much more straightforward approach would be to add a timedelta to your datetime within the loop. For instance,

>>> from datetime import timedelta
>>> times = 4
>>> cur_date = datetime.date(2017, 2, 24)

>>> for _ in range(times):
        print('today is {0}, do something'.format(cur_date))
        cur_date += timedelta(days=7)

today is 2017-02-24, do something
today is 2017-03-03, do something
today is 2017-03-10, do something
today is 2017-03-17, do something

This could be placed in a generator as well, depending on your use case.

>>> for dt in (cur_date + timedelta(days=x*7) for x in range(times)):
        print('today is {0}, do something'.format(dt))

today is 2017-02-24, do something
today is 2017-03-03, do something
today is 2017-03-10, do something
today is 2017-03-17, do something

or with Pandas pd.date_range

>>> import pandas as pd
>>> list(pd.date_range(start='2017-02-24', periods=4, freq='7D'))

[Timestamp('2017-02-24 00:00:00', freq='7D'),
 Timestamp('2017-03-03 00:00:00', freq='7D'),
 Timestamp('2017-03-10 00:00:00', freq='7D'),
 Timestamp('2017-03-17 00:00:00', freq='7D')]

---

Now what would happen if you attempted this example with your approach?

>>> year, month, day = 2017, 2, 24

>>> for i in range(0 , times):
        day = day
        fulldateadd = datetime.date(year, month, day)
        print('today is {0}, do something'.format(fulldateadd))
        day = day + 7
        if day > 31:
            day = day - 31
            month = month + 1

today is 2017-02-24, do something

ValueErrorTraceback (most recent call last)
<ipython-input-255-7df608ebbf8e> in <module>()
      1 for i in range(0 , times):
      2     day = day
----> 3     fulldateadd = datetime.date(year, month, day)
      4     print('today is {0}, do something'.format(fulldateadd))
      5     day = day + 7

ValueError: day is out of range for month

February doesn't have 31 days... so you would have to include a check with a mapping to the number of days in each month. Including logic for leap years.

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