Question

这是我的代码，它适用于第一次打印，并给我这个错误 TypeError：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)

这是输出：

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它起初工作正常然后我得到了那个错误

Answer 1

它首次运行，因为tomorrow是datetime.datetime实例，然后是：

tomorrow = datetime.datetime.now()

然而，在第一次之后，您用字符串替换tomorrow ：

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

您无法将timedelta()对象添加到字符串中。不要重新绑定tomorrow。将字符串分配给不同的名称，或者仅格式化对象以进行打印（而不是将其分配给名称）：

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字符串，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))

TypeError：必须是str，而不是datetime.timedelta

2 个答案: