Question

我总是在Question.objects.filter()使用过滤器但这一次，我想使用过滤器从类中获取值，如：

class Expense():
    amount = ''
    currency = ''
    date = ''

这是我的方法：

def get_expense_list(self, taskId):
    #get expense list
    self.expenseList = [Expense()]
    # get all activity from task 
    self.activiytList = Activity.objects.filter(task_id=taskId)

    for activity in self.activiytList:
        try:    
            self.expenseList.filter(currency=activity.expense_currency, self.expenseList)
            self.expenseList.amount = self.expenseList.expense.amount + activity.expense
        except:
            newExpense = Expense()
            newExpense.currency = activity.currency
            newExpense.amount = activity.amount
            newExpense.date = activity.date
            self.expenseList.append(newExpense)

我正在从活动列表中创建费用清单。在activity.model中，有些对象是：

amount =  models.DecimalField(max_digits=16, decimal_places=2)
currency =  models.CharField(max_length=5)
date = models.DateField()

我想在费用清单中显示，如果货币和日期相同，金额将会添加，如果没有现有的日期和货币，则会append()。

请帮帮我。

Answer 1

列表中没有filter()方法，因此您编写的代码无效。

所以，忽略你的代码，如果你想要的是：

我想在费用清单中显示，如果他们的货币和日期相同，金额将会增加，如果没有现有的日期和货币，它会追加（）。

那么你想要做的事情是：

def get_expense_list(self, taskId):
    self.expenseList = []
    self.activityList = Activity.objects.filter(task_id=taskId)

    for activity in self.activityList:
        # all existing expenses for this currency and date
        comparisonFunc = lambda x: x.currency == activity.currency and x.date == activity.date
        expenses = filter(comparisonFunc, self.expenseList)

        # if there are no expenses, create a new
        if len(expenses) == 0:
            expense = Expense()
            expense.amount = activity.amount
            expense.date = activity.date
            expense.currency = activity.currency
            self.expenseList.append(expense)
        # otherwise, increase the amount for expenses that were matching
        else:
            for expense in expenses:
                expense.amount += activity.amount

您也可以使用聚合在数据库中执行所有这些操作。

我可以使用不使用object.filter python的过滤器

1 个答案: