我目前正在使用C#开发应用程序。想象一下店面正面有结账。我有一个字典结构,其中对象为键, int对象计数器为值。
结构如下:
Dictionary<myObject, int> items.
基本思想是将项目字典传递给方法。我只在字典中添加了唯一的myObjects。 myObject附加了一个计数器规则。一旦计数器规则完全填满,我想用字典中的所有myObect进行计算。
myObject看起来像这样:
public class myObject
{
string ItemId { get; set; }
Discount Discount { get; set; }
}
public class Discount
{
public int Count { get; set; }
public decimal Price { get; set; }
public IDiscountHandler DiscountHandler => new DiscountHandler();
}
示例myObject可能如下所示:
var myObectA = new myObject()
{
ItemId = "A"
};
var discountA = new Discount()
{
Count = 2,
Price = 12 // special price, if 2 myObjects were added to the Dictionary
};
myObjectA.Discount = discountA;
1)我填写项目Dictionary并将其传递给Handler方法:
private decimal _totalDiscountedValue { get; set; } = 0;
if (!_items.ContainsKey(myObject))
{
_items.Add(myObject, 1);
}
else
{
_items[myObject]++;
}
_totalDiscountedValue += _discountHandler.CalculateDiscount(_items);
2)在我的处理程序中,一旦计数器规则完全填满,我会尝试总结所有折扣值。但不幸的是,我在这里挣扎着:
public class DiscountHandler : DiscountHandler
{
private decimal _totalDiscount { get; set; } = 0;
public override decimal CalculateDiscount(IDictionary<myObject, int> items)
{
if (items == null) throw new ArgumentNullException(nameof(items));
// I'm struggeling here:
// check if Dictionary[i].Dicount.Count = Dictionary.Value
// then _totalDiscount += Dictionary[i].Discount.Price
return _totalDiscount;
}
}
您知道如何解决这个问题,或者您对如何解决这个问题有所了解吗?
非常感谢!!
答案 0 :(得分:4)
您可以使用foreach迭代字典,如下所示:
public override decimal CalculateDiscount(IDictionary<myObject, int> items)
{
if (items == null) throw new ArgumentNullException(nameof(items));
foreach (var kvp in items)
{
if (kvp.Key.Discount.Count == kvp.Value)
_totalDiscount += kvp.Key.Discount.Price;
}
return _totalDiscount;
}
答案 1 :(得分:2)
使用Linq
//check if yourDictonary is not null
var sum = yourDictonary.Select(x => x.Key.Discount.Count == x.Value).Sum(x => x.Value)
答案 2 :(得分:0)
如果我理解了问题,也许可以这样做
foreach (var item in items)
{
if (item.Key.Discount.Count == item.Value)
_totalDiscount += item.Key.Discount.Price;
}
return __totalDiscount;