如何刪除深層對象中的項目？

Question

假設我有以下對象數組：

[
   { id: "1", categories: [ { category_id: "1"}, { category_id: "2"} ],
   { id: "2", categories: [ { category_id: "2"}, { category_id: "3"} ],
   { id: "3", categories: [ { category_id: "1"}, { category_id: "5"} ],
]

我要刪除所有沒有作為category_id的項目，這些項目不包括在此引用數組中： 1, 4, 5 。

因此，預期的輸出應為： 1, 3 ，因為id 2在reference數組中沒有任何類別id。

我寫了這段代碼：

items.filter(obj => !references.includes(obj.categories.category_id));

但這將返回相同的項目

預期結果：

[
   { id: "1", categories: [ { category_id: "1"}, { category_id: "2"} ],
   { id: "3", categories: [ { category_id: "1"}, { category_id: "5"} ],
]

Answer 1

您可以使用Array#filter ， Array#some和帶有Array#includes的值。

 var array = [{ id: "1", categories: [{ category_id: "1" }, { category_id: "2" }] }, { id: "2", categories: [{ category_id: "2" }, { category_id: "3" }] }, { id: "3", categories: [{ category_id: "1" }, { category_id: "5" }] }], keep = ["1", "4", "5"], result = array.filter(({ categories }) => categories.some(({ category_id }) => keep.includes(category_id))); console.log(result); 

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Answer 2

obj.categories是一個數組，您必須以某種方式對其進行迭代：

items.filter(obj => obj.categories.every(category => !references.includes(category.category_id)));

Answer 3

在這里，您還有另一種使用Array :: findIndex（）的方法

 const input = [ {id: "1", categories: [{category_id: "1"}, {category_id: "2"}]}, {id: "2", categories: [{category_id: "2"}, {category_id: "3"}]}, {id: "3", categories: [{category_id: "1"}, {category_id: "5"}]}, ]; const references = [1, 4, 5]; let res = input.filter( y => y.categories.findIndex(x => references.includes(Number(x.category_id))) >= 0 ); console.log(res);