TypeError: 必须是字符串，而不是datetime.timed

Question

这是我的代码，第一次打印很好，给我这个错误 类型错误：必须是str，而不是datetime.timedelta

import datetime
hour = datetime.datetime.now().hour
tomorrow = datetime.datetime.now()
for i in range(50):
    hour += 1
    print(hour)
    if hour >= 23:
        tomorrow = tomorrow + datetime.timedelta(days=1)
        tomorrow = tomorrow.strftime('%d/%m/%Y')
        hour = 9
        print (tomorrow)

输出如下：

13
14
15
16
17
18
19
20
21
22
23
07/12/2017
10
11
12
13
14
15
16
17
18
19
20
21
22
23
Traceback (most recent call last):
  File "C:\Users\hamza.salhi\Desktop\datetime test.py", line 8, in <module>
    tomorrow = tomorrow + datetime.timedelta(days=1)
TypeError: must be str, not datetime.timedelta

请有人告诉我为什么一开始它工作得很好，然后我得到那个错误

Answer 1

它第一次工作是因为tomorrow是一个datetime.datetime实例，然后：

tomorrow = datetime.datetime.now()

但是，在第一次之后，用字符串替换tomorrow：

tomorrow = tomorrow.strftime('%d/%m/%Y')

不能将timedelta()对象添加到字符串。不要重新绑定tomorrow。将字符串指定给其他名称，或仅格式化要打印的对象（而不将其指定给名称）：

if hour >= 23:
    tomorrow = tomorrow + datetime.timedelta(days=1)
    tomorrow_formatted = tomorrow.strftime('%d/%m/%Y')
    hour = 9
    print(tomorrow_formatted)  # different name
    print(tomorrow.strftime('%d/%m/%Y'))  # print the `strftime()` result

Answer 2

tomorrow = tomorrow.strftime('%d/%m/%Y') 将明天的类型更改为string，因此无法将其添加到datetime.timedelta。

而不是：

tomorrow = tomorrow + datetime.timedelta(days=1)
tomorrow = tomorrow.strftime('%d/%m/%Y')
...
print (tomorrow)

尝试：

tomorrow = tomorrow + datetime.timedelta(days=1)
...
print('{:%d/%m/%Y}'.format(tomorrow))