Question

简单的IList更新问题：给定以下嵌套对象，如何在给定主键的情况下更新最深的嵌套属性？

public class Recipe {
    public int RecipeID {get;set;}     // PK

    public string name {get;set;}
    public IList<RecipeStep> RecipeSteps {get;set;}
}

public class RecipeStep {
    public int RecipeID {get;set;}     // PK
    public int RecipeStepID {get;set;} // PK

    public string name {get;set;}
    public IList<Ingredient> {get;set;}
}

public class Ingredient {
    public int RecipeID {get;set;}     // PK
    public int RecipeStepID {get;set;} // PK
    public int IngredientID {get;set;} // PK

    public string name {get;set;}
}

那么如果RecipeID = 2，RecipeStepID = 14和IngredientID = 5（这是int的值，而不是索引），我怎么能设置Recipe.RecipeStep.Ingredient.name。希望有一些直接的方法来引用这些项目没有循环。 Linq表达式很好。（当我尝试使用Linq时，我最终更改了值的副本，而不是值本身.LOL）。

Answer 1

你正在寻找selectmany，它可以采用多个深度的可数，并将它们变成一个联合的结果集，在一个ienumerable，一个selectmany（）中获取所有成分，然后为你的条件做一个where（）

Ingredient ing = theRecipe.RecipeSteps.SelectMany((recipeStep) => recipeStep.Ingredient)
    .FirstOrDefault((ingredient) =>
        ingredient.RecipeId == 2 &&
        ingredient.RecipeStepId == 14 &&
        ingredient.IngredientId == 5);

ing.name = "pretty flowers";

Answer 2

Ingredient theIngredient =
(
  from r in Recipes
  where r.RecipeId == 2
  from rs in r.RecipeSteps
  where rs.RecipeStepID == 14
  from ing in rs.Ingredients
  where ing.IngredientId == 5
  select ing
).Single()

theIngredient.name = theName;

Answer 3

Linq查询：

Ingredient ing = theRecipe.RecipeSteps.Ingredients
    .FirstOrDefault(i =>
        i.RecipeId == 2 &&
        i.RecipeStepId == 14 &&
        i.IngredientId == 5);

if (ing != null) ing.name = "New Name";

简单的C＃List＆lt;＆gt;：给定主键，如何在List＆lt;＆gt;？中查找和更新嵌套对象属性

3 个答案: