Question

我需要将特定格式的字符串转换为另一个。

Source:
2018-08-10 9:00 PM - 9:40 PM

Target:
2018-08-10 21:00:00

我写了以下内容：

def startofappt(strv):
    return strv.split(' - ')[0]

def convCalDate(dts, tms):
#     2018-08-08 5:50 PM
    from datetime import datetime, date, time
    dt = datetime.strptime(str(dts) + ' ' + str(startofappt(tms)), "%Y-%m-%d %H:%M %p")
    return str(dt)

import datetime
datetoday = datetime.date.today()
from datetime import datetime, date, time
appointments_set = appointment.objects.filter(docid=11)
for appt in appointments_set:
    if appt.date > datetoday:
        print("Upcoming: ", appt.date, appt.time, convCalDate(appt.date, appt.time))
    else:
        print("Other:",appt.date, appt.time, convCalDate(appt.date, appt.time ))

不幸的是我在24小时内遇到问题。

输出：

Upcoming:  2018-08-10 9:00 PM - 9:40 PM 2018-08-10 09:00:00
Other: 2018-08-07 9:40 PM - 10:20 PM 2018-08-07 09:40:00
Other: 2018-08-07 9:00 PM - 9:40 PM 2018-08-07 09:00:00
Upcoming:  2018-08-08 5:50 PM - 6:10 PM 2018-08-08 05:50:00
Upcoming:  2018-08-08 6:10 PM - 6:30 PM 2018-08-08 06:10:00

我在做什么错了？

Answer 1

能否请您尝试将日期时间读取为12小时格式的日期时间，并在返回日期时将其返回为

def convCalDate(dts, tms):
    from datetime import datetime, date, time
    dt = datetime.strptime(str(dts) + ' ' + str(startofappt(tms)), "%Y-%m-%d %I:%M %p")
    return str(dt)

print convCalDate("2018-08-10", "9:00 PM - 9:40 PM") #returns 2018-08-10 21:00:00

print convCalDate("2018-08-10", "9:00 AM - 9:40 PM") #returns 2018-08-10 09:00:00

希望有帮助

Python datetime格式的字符串转换24小时时间问题

1 个答案: