实际上,标题让我感到困惑,我什至不确定它是否正确。我在这里遇到了以下问题:我有一个列表,其中每个配方都包含一个 requiredItem[] 数组(大小为 15)这个必需的项目数组可以容纳配方所需的项目。我要做的是将所有食谱放在一个单独的列表中,在他们的食谱中使用木材作为必需的项目之一。例如,一个食谱可能需要 10 根木头来制作一张木桌。此配方在配方列表中。快速的谷歌搜索让我找到了以下内容,但我无法弄清楚如何访问所需的项目索引:
List<Recipe> rec = Main.recipe.ToList();
rec.Select((s, i) => new { i, s })
.Where(t => t.s.requiredItem[indexhere].type == 9)
.Select(t => t.i)
.ToList()
很抱歉我没有看到这个,如果答案真的很简单,我不会感到惊讶。
我也尝试了以下方法。它不会带来任何错误,但实际上它没有选择食谱。
Item refItem = new Item();
refItem.SetDefaults(ItemID.Wood, false);
List<int> selectWood = new List<int>(rec.Select((r, i) => new { i, r }).Where(x => x.r.requiredItem.ToList().Contains(refItem)).Select(x => x.i).ToList());
ErrorLogger.Log(selectWood.ToArray().ToString());
List<int> indices = new List<int>();
foreach (var indice in selectWood)
{
for (int i = 0; i < Main.recipe[indice].requiredItem.Length; i++)
{
var item = Main.recipe[indice].requiredItem[i];
if (item.type == ItemID.Wood && item.stack >= 10) indices.Insert(0, indice);
}
}
foreach (var indice in indices)
{
++numberRecipesRemoved;
rec.RemoveAt(indice);
}
我可能误读了这个问题,但并不是完全清楚为什么你需要和索引(indexHere
)作为一个Any()
子句就足够了。
如果没有Recipe
或requiredItem
的详细信息,很难确定,但以下内容似乎可以实现您需要的。
enum MaterialType
{
None,
Wood,
Glass,
Steel,
Cloth
}
class Ingredient
{
public Ingredient(MaterialType type, int amount)
{
Type = type;
Amount = amount;
}
public MaterialType Type { get; }
public int Amount { get; }
}
class Recipe
{
public Recipe(string name, params Ingredient[] ingredients)
: this(name, (IEnumerable<Ingredient>) ingredients)
{
}
public Recipe(string name, IEnumerable<Ingredient> ingredients)
{
Name = name;
Ingredients = ingredients.ToArray();
}
public string Name { get; }
public Ingredient[] Ingredients { get; }
}
[TestClass]
public class FindAllItemsFixture
{
private readonly static IEnumerable<Recipe> AllItemRecipes = new List<Recipe>
{
new Recipe("Sword", new Ingredient(MaterialType.Steel, 3)),
new Recipe("Spear", new Ingredient(MaterialType.Steel, 1), new Ingredient(MaterialType.Wood, 3)),
new Recipe("Table", new Ingredient(MaterialType.Wood, 6)),
new Recipe("Chair", new Ingredient(MaterialType.Wood, 4)),
new Recipe("Flag", new Ingredient(MaterialType.Cloth, 2)),
};
IEnumerable<Recipe> GetAllRecipesUsingMaterial(MaterialType materialType)
{
return AllItemRecipes.Where(r => r.Ingredients.Any(i => i.Type == materialType));
}
[TestMethod]
public void GetAllWoodenRecipes()
{
var expectedNames = new string[] { "Spear", "Table", "Chair" };
var woodenItems = GetAllRecipesUsingMaterial(MaterialType.Wood);
CollectionAssert.AreEqual(expectedNames, woodenItems.Select(i => i.Name).ToArray());
}
[TestMethod]
public void GetAllClothRecipes()
{
var expectedNames = new string[] { "Flag" };
var clothItems = GetAllRecipesUsingMaterial(MaterialType.Cloth);
CollectionAssert.AreEqual(expectedNames, clothItems.Select(i => i.Name).ToArray());
}
}
重要的部分是单元测试中的GetAllRecipesUsingMaterial()
函数。它选择包含指定类型成分的所有配方。
是的,这很简单,如果我理解你的问题,你需要显示所有包含木材的接收器作为必需的项目。所以这样做:
var woodRecipes = recipes.Where(r => r.requiredItems.Contains(wood)).ToList();