缩短长开关盒 [英] Shorten long switch case

问题描述

所以让我先说我是 C# 的新手.我有一个 switch 语句，目前有 10 个不同的 case，但是，我需要使用它 3 次不同的时间(相同的 10 个 case，每个 case 的结果不同)，并且每个 case 只有一点点变化.

So let me start by saying I am new to C#. I have a switch statement that currently has 10 different cases, however, I need to use it 3 different times (same 10 cases, different results per case), and each case there is only a slight change.

我觉得我只是在重复代码，有没有办法缩短它?

I feel like I'm just repeating code, is there a way to shorten it?

//Set the growth time of the crop based on what cropType.
        switch (cropType) {
            case 1:
                //Potatoes
                growth = 60;
                break;
            case 2:
                //Strawberries
                growth = 80;
                break;
            case 3:
                //Cabbages
                growth = 90;
                break;
            case 4:
                //Carrots
                growth = 40;
                break;
            case 5:
                //Melon
                growth = 120;
                break;
            case 6:
                //Pumpkin
                growth = 130;
                break;
            case 7:
                //Eggplant
                growth = 50;
                break;
            case 8:
                //Mushroom
                growth = 70;
                break;
            case 9:
                //Wheat
                growth = 40;
                break;
            case 10:
                //Truffle
                growth = 150;
                break;
        }

这是我的 1 个部分的代码.在第二部分中，我根据情况分配图像，这必须单独完成，因为它依赖于增长和变化，而增长则不然.不过，我实际上并没有在其他开关上使用它.这是我进一步了解的另一个:

That is my code for 1 section. In the 2nd section, I assign an image depending on the case, this has to be done separately as it relies on the growth and changes, whereas growth does not. I don't actually use it on the other swithces though. This is another one I have further down:

switch (cropType) {
            case 1:
                //Potatoes
                Debug.Log("Potatoes Harvested!");
                Global.potato += 2;
                break;
            case 2:
                //Strawberries
                Debug.Log("Strawberries Harvested!");
                Global.strawberry += 4;
                break;
            case 3:
                //Cabbages
                Debug.Log("Cabbages Harvested!");
                Global.cabbage += 1;
                break;
            case 4:
                //Carrots
                Debug.Log("Carrots Harvested!");
                Global.carrot += 3;
                break;
            case 5:
                //Melon
                Debug.Log("Melons Harvested!");
                Global.melon += 1;
                break;
            case 6:
                //Pumpkin
                Debug.Log("Pumpkins Harvested!");
                Global.pumpkin += 1;
                break;
            case 7:
                //Eggplant
                Debug.Log("Eggplant Harvested!");
                Global.eggplant += 2;
                break;
            case 8:
                //Mushroom
                Debug.Log("Mushrooms Harvested!");
                Global.mushroom += 4;
                break;
            case 9:
                //Wheat
                Debug.Log("Wheat Harvested!");
                Global.wheat += 6;
                break;
            case 10:
                //Truffle
                Debug.Log("Truffles Harvested!");
                Global.truffle += 1;
                break;
        }

基本上它是一个脚本，需要根据其中的cropType 来做不同的事情.

Basically it is a script that needs to do different things based on what cropType is in it.

推荐答案

也许这太多了，但您可以使用枚举和类/结构加上字典(如 ggorlen 建议)

Maybe this would be too much, but you can use enums and classes/structs plus a dictionary (as ggorlen suggested)

为什么要枚举?避免使用硬编码数字；不易出错并提高可读性；

Why enums? to avoid using hardcoded numbers; less error-prone and will improve readability;

private enum CropType
{
    Undefined = 0,
    Cabbages,
    Carrots,
    Eggplant,
    Melon,
    Mushroom,
    Potatoes,
    Pumpkin,
    Strawberries,
    Truffle,
    Wheat
}

private struct Crop
{
    public CropType Type { get; private set; }
    public float GrowthFactor { get; private set; }
    public float HarvestFactor { get; private set; }

    public Crop(CropType type, float growthFactor, float harvestFactor) 
    {
        this.Type = type;
        this.GrowthFactor = growthFactor;
        this.HarvestFactor = harvestFactor;
    }
}

private Dictionary<CropType, Crop> crops;
private Dictionary<CropType, Crop> Crops 
{
    get 
    {
        if (crops == null) 
        {
            crops = new Dictionary<CropType, Crop>() 
            {
                { CropType.Cabbages, new Crop(CropType.Cabbages, 90, 1) },
                { CropType.Carrots, new Crop(CropType.Carrots, 80, 5) }
                // here you can add the rest of your products...
            };
        }
        return crops;
    }
}

public Crop GetCrop(CropType crop) 
{
    if (!Crops.ContainsKey(type)) 
    {
        Debug.LogWarningFormat("GetCrop; CropType [{0}] not present in dictionary ", type);
        return null;
    }

    return Crops[type];
}

在这里(最终)您将检索您想要的值.

Here is where (finally) you will retrieve the values that you want.

public float GetGrowthFactor(CropType type) 
{
    var crop = GetCrop(type);
    return crop == null ? default(float) : crop.GrowthFactor;
}

public float GetHarvestFactor(CropType type) 
{
    var crop = GetCrop(type);
    return crop == null ? default(float) : crop.HarvestFactor;
}

所以你会以这种方式求值；

So you will ask for values in this way;

private void Example()
{
    var carrotsGrowth = GetGrowthFactor(CropType.Carrots);
}

这篇关于缩短长开关盒的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！