python - 工资单 odoo 中的休假扣除错误

Question

我尝试为一名员工生成工资单。员工休了半天假(0.5)，但在计算工资单时，其自动填充为 1。

从模块 hr_payroll.py 中已有的代码来看，它看起来如下

 def get_worked_day_lines(self, cr, uid, contract_ids, date_from, date_to, context=None):
        """
        @param contract_ids: list of contract id
        @return: returns a list of dict containing the input that should be applied for the given contract between date_from and date_to
        """
        def was_on_leave(employee_id, datetime_day, context=None):
            res = False
            day = datetime_day.strftime("%Y-%m-%d")
            holiday_ids = self.pool.get('hr.holidays').search(cr, uid, [('state','=','validate'),('employee_id','=',employee_id),('type','=','remove'),('date_from','<=',day),('date_to','>=',day)])
            if holiday_ids:
                res = self.pool.get('hr.holidays').browse(cr, uid, holiday_ids, context=context)[0].holiday_status_id.name
            return res

        res = []
        for contract in self.pool.get('hr.contract').browse(cr, uid, contract_ids, context=context):
            if not contract.working_hours:
                #fill only if the contract as a working schedule linked
                continue
            attendances = {
                 'name': _("Normal Working Days paid at 100%"),
                 'sequence': 1,
                 'code': 'WORK100',
                 'number_of_days': 0.0,
                 'number_of_hours': 0.0,
                 'contract_id': contract.id,
            }
            leaves = {}
            day_from = datetime.strptime(date_from,"%Y-%m-%d")
            day_to = datetime.strptime(date_to,"%Y-%m-%d")
            nb_of_days = (day_to - day_from).days + 1
            for day in range(0, nb_of_days):
                working_hours_on_day = self.pool.get('resource.calendar').working_hours_on_day(cr, uid, contract.working_hours, day_from + timedelta(days=day), context)
                if working_hours_on_day:
                    #the employee had to work
                    leave_type = was_on_leave(contract.employee_id.id, day_from + timedelta(days=day), context=context)
                    if leave_type:
                        #if he was on leave, fill the leaves dict
                        if leave_type in leaves:
                            leaves[leave_type]['number_of_days'] += 1.0
                            leaves[leave_type]['number_of_hours'] += working_hours_on_day
                        else:
                            leaves[leave_type] = {
                                'name': leave_type,
                                'sequence': 5,
                                'code': leave_type,
                                'number_of_days': 1.0,
                                'number_of_hours': working_hours_on_day,
                                'contract_id': contract.id,
                            }
                    else:
                        #add the input vals to tmp (increment if existing)
                        attendances['number_of_days'] += 1.0
                        attendances['number_of_hours'] += working_hours_on_day
            leaves = [value for key,value in leaves.items()]
            res += [attendances] + leaves
        return res

我不确定这是否是问题所在。有人对此有什么建议吗？

Answer 1

让我先说一下:我没有使用过 odoo，也没有你的其余代码可供测试，所以我还没有验证这一点。

你这里肯定有问题:

if leave_type in leaves:
    leaves[leave_type]['number_of_days'] += 1.0
    leaves[leave_type]['number_of_hours'] += working_hours_on_day
else:
    leaves[leave_type] = {
        'name': leave_type,
        'sequence': 5,
        'code': leave_type,
        'number_of_days': 1.0,
        'number_of_hours': working_hours_on_day,
        'contract_id': contract.id,
    }
else:
    #add the input vals to tmp (increment if existing)
    attendances['number_of_days'] += 1.0
    attendances['number_of_hours'] += working_hours_on_day

查看您引用休假和出勤的位置，虽然这些是 float 的，但它们仅在全天中增加，而不是根据您度过的一天的分数来计算。您需要将其更改为:

leaves[leave_type]['number_of_days'] += time_off / hours_per_workday # the 1.0 might make sense here depending on if you count time off as an attendance
leaves[leave_type]['number_of_hours'] += time_off

和

attendances['number_of_days'] += 1.0 - time_off / hours_per_workday
attendances['number_of_hours'] += working_hours_on_day - time_off

显然，我插入的虚拟变量需要在某处定义和计算。

此外，正如安东尼·罗西(Anthony Rossi)在评论中指出的那样，您通常需要几天而不是几个小时才能工作。示例:

day_from = datetime.strptime(date_from,"%Y-%m-%d")
day_to = datetime.strptime(date_to,"%Y-%m-%d")

请注意，您只有 YY/MM/DD 而没有时间。