python-3.x - odoo 11/Python 3 : How to find expiry date of subscription if some weekdays are excluded

Question

我正在研究订阅管理系统。我在计算订阅到期日时遇到问题。场景如下，
如果一个人订阅了 1 个月，比如 1/11/2018 到 30/11/2018 [DD/MM/YYYY] 总共 30 天，但他想从每周 30 天中排除周五和周六。那么我应该如何计算到期日？

Logic is : Say End Date = Expiry Date then find Fri/Sat from 1/11/2018 to 30/11/2018 which comes out 5 Fri & 4 Sat = 9 days. Add to Expiry Date which will be 09/12/2018. Now search Fur & Sat between End Date & Expiry Date which comes out 1 Fri & 2 Sat = 3 days. Now End Date = Expiry Date and Expiry Date + 3 Days = 12/12/2018. Search between End Date & Expiry Date for Fri & Sat which is 0 so the Expiry Date is 12/12/2018 return value <<

以下代码执行此操作，但方法返回 09/12/2018 而不是 12/12/2018。这有什么问题？？

@api.one
def get_expiry_date(self, start, end, day_selection):
    print("i am in 2nd", start, end, day_selection)
    dayformat = '%Y-%m-%d'
    current_date = datetime.now().date()
    if start:
        start = datetime.strptime(str(start), dayformat).date()
    if end:    
        end = datetime.strptime(str(end), dayformat).date()        
    if day_selection:
        selected_days = day_selection        
    if start < current_date:
        start = datetime.strptime(str(start), dayformat).date()
    weekdays = self.weekday_count(start,end)
    print("days for start and end date",start,end, day_selection) 
    adddays = 0
    if weekdays:
        for i in range(len(day_selection)):                          
            for item in weekdays[0]:
                weekdays_dict = item
                print("dict", type(weekdays), type(weekdays[0]), weekdays_dict)
                print("compare", selected_days[i], weekdays_dict, selected_days[i] == weekdays_dict)
                if selected_days[i] == item:
                    adddays = adddays + weekdays[0].get(item)
        new_start = end
        end = datetime.strptime(str(end), dayformat).date() + timedelta(days=adddays)
        start = new_start
        print("New Expiry Date", start, end, adddays)
        if adddays > 0:               
            self.get_expiry_date(start, end, day_selection)
            print("type of end is ", type(end))
            print("selected days are", selected_days[i],weekdays[0], weekdays[0].get(item), adddays)  
            print("last returned values is",end)

返回结束

Answer 1

在你的问题中，没有 day_selection 的定义也不是 weekday_count因此很难看出发生了什么。可能问题在于递归，这不是必需的。

如 day_selection定义为选择排除的天数( strftime('%a') 的列表，如 ['Mon', 'Tue'] )，然后:

from datetime import datetime, timedelta

def get_expiry_date(start, end, day_selection):
    dayformat = '%Y-%m-%d'
    weekdays= {}
    start = datetime.strptime(str(start), dayformat).date()
    end = datetime.strptime(str(end), dayformat).date()
    # Count weekdays
    for i in range((end - start).days+1):
        weekday = (start + timedelta(i)).strftime('%a')
        weekdays[weekday] = 1 + weekdays.setdefault(weekday, 0)
    # Count subscription days to add
    sub_days_to_add = 0
    for selected in day_selection:
        sub_days_to_add += weekdays.setdefault(selected, 0)
    # Count calender days to add
    cal_days_extension = 0
    while sub_days_to_add > 0:
        if (end + timedelta(days=cal_days_extension + 1)).strftime('%a') not in day_selection:
            sub_days_to_add -= 1
        cal_days_extension += 1
    # Add to end day
    return end + timedelta(days=cal_days_extension)

测试:

print (get_expiry_date('2018-11-01', '2018-11-30', ['Fri', 'Sat']))
# --->  2018-12-12
print (get_expiry_date('2018-11-01', '2018-11-30', []))
# --->  2018-11-30

此外，使用 from odoo.tools import DEFAULT_SERVER_DATE_FORMAT 会更安全比 dayformat = '%Y-%m-%d' .对于像 start 这样的参数& end ，如果它们是强制性的，可以在 Odoo 中使这些字段成为必需的。