将int转换为字符串

时间:2017-05-11 20:02:41

标签: c# arrays visual-studio object

我很难理解这一点。我做了一个瓶子对象,我有整个瓶子对象的重量。但我试图只指定一种“Cola Cola”瓶子。但是,我知道如何查看我制作的清单并计算总重量。但如果我只想算“可口可乐”的物品,我就迷路了。有什么帮助吗?

class Flaske
{
    public String name = "Coca Cola";
    public int weight = 14;

    public Bottles(String na, int ve)
    {
         name = na;
         weight = ve;
    }
}
     //Total Bottle Weight
    public int findCocaColaWeight(Flaske[] kasse)
    {
            int result = 0;                
            int i = 0;
            while (i < kasse.Length)
            {
                if (kasse[i] != null)
                {
                    result = result + kasse[i].Weight.Equals("Coca Cola");
                }
                i = i + 1;
            }
            return result;            
    }
    //Find Coca Cola Weight
    public int findCocaColaWeight(Flaske[] kasse)
    {
            int result = 0;                
            int i = 0;
            while (i < kasse.Length)
            {
                if (kasse[i] != null)
                {
                    result = result + kasse[i].Weight;
                }
                i = i + 1;
            }
            return result;            
    }

所以基本上,我只需要知道在遇到特定对象时如何计算它们,对吧?

2 个答案:

答案 0 :(得分:7)

如果您使用Linq,只需一行代码。

public int findWeightByName(string name, Flaske[] kasse)
{
    return kasse.Where(x => x.name == name).Sum(w => w.weight);
}

Where使用传递的名称提取Flaske序列,Sum对序列进行处理以总结权重

答案 1 :(得分:0)

根据评论,我假设您的代码应如下所示:

//Total Bottle Weight
public int findTotalWeight(Flaske[] kasse)
{
        int result = 0;                
        int i = 0;
        while (i < kasse.Length)
        {
            if (kasse[i] != null)
                result = result + kasse[i].weight;
            i = i + 1;
        }
        return result;            
}

//Find Coca Cola Weight
public int findCocaColaWeight(Flaske[] kasse)
{
        int result = 0;                
        int i = 0;
        while (i < kasse.Length)
        {
            if (kasse[i] != null && kasse[i].name.Equals("Coca Cola"))
                result = result + kasse[i].weight;
            i = i + 1;
        }
        return result;            
}
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