如何删除深层对象中的项目？

Question

Suppose I have the following array of objects:

[
   { 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"} ],
]

I want remove all the items which doesn't have as category_id those not included in this array of references: 1, 4, 5 .

So the expected output should be: 1, 3 , because the id 2 doesn't have any category id contained in teh references array.

I wrote this code:

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

but this will return the same items

Expected result:

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

Answer 1

You could use Array#filter , Array#some and the value with Array#includes .

 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是一个数组，您必须以某种方式对其进行迭代：

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

Answer 3

Here you have another approach that uses Array::findIndex()

 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);