odoo 11 / Python 3:如果排除某些工作日,如何查找订阅的到期日
odoo 11 / Python 3: How to find expiry date of subscription if some weekdays are excluded
我正在研究订阅管理系统。我在计算订阅到期日期时遇到问题。场景如下,
如果一个人订阅 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呢?
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 returns 09/12/2018 而不是 12/12/2018。这有什么问题吗??
@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)
return 结束
在您的问题中,没有 day_selection
和 weekday_count
的定义,因此很难看出发生了什么。问题可能出在递归上,这不是必需的。
如果 day_selection
定义为选择排除的天数([=15= 的列表],如 ['Mon', 'Tue']
),则:
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
而且使用 from odoo.tools import DEFAULT_SERVER_DATE_FORMAT
比 dayformat = '%Y-%m-%d'
更安全。而对于 start
和 end
这样的参数,如果它们是必填的,可以在 Odoo 中将这些字段设置为必填。
你想做的逻辑很难
据我了解你是运行一个订阅系统
基于天数而不是日期所以为了计算
结束日期你只需要这个:
- 开始日期(例如:01/11/2018)
- 天数(例如:30 天)
- 排除天数(例如:周五、周六)
# day are like this
# Fri
# Sat
# Sun
# Mon
# Tue
# Wed
# Thu
# start day
def get_expiry_date(start_date, number_of_days, excluded_days = None):
""" compute end date of subcription period """
if excluded_days is None:
excluded_days = []
# check params
start_date = str(start_date)
if number_of_days < 1:
raise exception.UserError(_('Number of days should be > 0!!')) # import the translate "_" method
if len(excluded_days) > 5:
raise exception.UserError(_('To much excluded days!!')) # import the translate "_" method
date_format = '%Y-%m-%d'
end_date = datetime.strptime(start_date, date_format)
# compute end date
# keeping adding one day until you finish your days
add_one_day = timedelta(days=1)
while number_of_days > 1:
end_date += add_one_day
if end_date.strftime('%a') not in excluded_days:
# day is not excluded compute it
number_of_days += -1
return end_date.strftime(date_format)
这是一个您可以检查的测试值:
print get_expiry_date('2018-11-01', 30, ['Fri', 'Sat']) # 2018-12-12
print get_expiry_date('2018-11-01', 30, ['Fri']) # 2018-12-05
print get_expiry_date('2018-11-01', 30, []) # 2018-11-30
print get_expiry_date('2018-11-01', 90, []) # 2019-01-29
print get_expiry_date('2018-11-01', 30, ['Mon', 'Thu']) # 2018-12-11
如果您已经准备好一个 运行 系统,您可以使用这个:
def get_expiry_date(start_date, end_date, excluded_days = None):
if excluded_days is None:
excluded_days = []
start_date = str(start_date)
end_date = str(end_date)
date_format = '%Y-%m-%d'
# compute number of days
number_of_days = (datetime.strptime(end_date, date_format) - datetime.strptime(start_date, date_format)).days
expiry_date = datetime.strptime(start_date, date_format)
if len(excluded_days) > 5:
raise Exception('To much excluded days!!')
# keeping adding one day until you finish your days
one_day = timedelta(days=1)
while number_of_days > 0:
expiry_date += one_day
# if day is not excluded reduce the number of left days
if expiry_date.strftime('%a') not in excluded_days:
number_of_days += -1
return expiry_date.strftime(date_format)
print get_expiry_date('2018-11-01', '2018-11-30', ['Fri', 'Sat']) # 2018-12-12
我正在研究订阅管理系统。我在计算订阅到期日期时遇到问题。场景如下, 如果一个人订阅 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呢?
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 returns 09/12/2018 而不是 12/12/2018。这有什么问题吗??
@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)
return 结束
在您的问题中,没有 day_selection
和 weekday_count
的定义,因此很难看出发生了什么。问题可能出在递归上,这不是必需的。
如果 day_selection
定义为选择排除的天数([=15= 的列表],如 ['Mon', 'Tue']
),则:
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
而且使用 from odoo.tools import DEFAULT_SERVER_DATE_FORMAT
比 dayformat = '%Y-%m-%d'
更安全。而对于 start
和 end
这样的参数,如果它们是必填的,可以在 Odoo 中将这些字段设置为必填。
你想做的逻辑很难
据我了解你是运行一个订阅系统 基于天数而不是日期所以为了计算 结束日期你只需要这个:
- 开始日期(例如:01/11/2018)
- 天数(例如:30 天)
- 排除天数(例如:周五、周六)
# day are like this
# Fri
# Sat
# Sun
# Mon
# Tue
# Wed
# Thu
# start day
def get_expiry_date(start_date, number_of_days, excluded_days = None):
""" compute end date of subcription period """
if excluded_days is None:
excluded_days = []
# check params
start_date = str(start_date)
if number_of_days < 1:
raise exception.UserError(_('Number of days should be > 0!!')) # import the translate "_" method
if len(excluded_days) > 5:
raise exception.UserError(_('To much excluded days!!')) # import the translate "_" method
date_format = '%Y-%m-%d'
end_date = datetime.strptime(start_date, date_format)
# compute end date
# keeping adding one day until you finish your days
add_one_day = timedelta(days=1)
while number_of_days > 1:
end_date += add_one_day
if end_date.strftime('%a') not in excluded_days:
# day is not excluded compute it
number_of_days += -1
return end_date.strftime(date_format)
这是一个您可以检查的测试值:
print get_expiry_date('2018-11-01', 30, ['Fri', 'Sat']) # 2018-12-12
print get_expiry_date('2018-11-01', 30, ['Fri']) # 2018-12-05
print get_expiry_date('2018-11-01', 30, []) # 2018-11-30
print get_expiry_date('2018-11-01', 90, []) # 2019-01-29
print get_expiry_date('2018-11-01', 30, ['Mon', 'Thu']) # 2018-12-11
如果您已经准备好一个 运行 系统,您可以使用这个:
def get_expiry_date(start_date, end_date, excluded_days = None):
if excluded_days is None:
excluded_days = []
start_date = str(start_date)
end_date = str(end_date)
date_format = '%Y-%m-%d'
# compute number of days
number_of_days = (datetime.strptime(end_date, date_format) - datetime.strptime(start_date, date_format)).days
expiry_date = datetime.strptime(start_date, date_format)
if len(excluded_days) > 5:
raise Exception('To much excluded days!!')
# keeping adding one day until you finish your days
one_day = timedelta(days=1)
while number_of_days > 0:
expiry_date += one_day
# if day is not excluded reduce the number of left days
if expiry_date.strftime('%a') not in excluded_days:
number_of_days += -1
return expiry_date.strftime(date_format)
print get_expiry_date('2018-11-01', '2018-11-30', ['Fri', 'Sat']) # 2018-12-12