Question

所以我有如下数据：

[
     {
      "id": 0,
      "title": "happy dayys",
      "owner": {"id": "1", "username": "dillonraphael"},
      "tags": [{"value": "Art", "label": "Art"}],
      "items": []
     },
     {
      "id": 1,
      "title": "happy dayys",
      "owner": {"id": "1", "username": "dillonraphael"},
      "tags": [{"value": "Architecture", "label": "Architecture"}],
      "items": []
     },
]

我正在尝试过滤一个数组，并且仅在标签数组包含另一个字符串的值==时返回。

这是我想出的，但似乎仍然要发回整个数组：

const tagMoodboards = _moodboards.filter(mb => { return mb.tags.filter(t => t.value == name) })

Answer 1

您不希望在filter中使用filter-而是在过滤器中，检查some对象中的tags是否具有.value您想要的属性

const _moodboards = [
     {
      "id": 0,
      "title": "happy dayys",
      "owner": {"id": "1", "username": "dillonraphael"},
      "tags": [{"value": "Art", "label": "Art"}],
      "items": []
     },
     {
      "id": 1,
      "title": "happy dayys",
      "owner": {"id": "1", "username": "dillonraphael"},
      "tags": [{"value": "Architecture", "label": "Architecture"}],
      "items": []
     },
];
const name = 'Architecture';
console.log(_moodboards.filter(({ tags }) => (
  tags.some(({ value }) => value === name)
)));

Answer 2

要使用filter()，您需要返回true或false的值（即布尔值）。那应该是您开始的第一个地方。因此，给定类似对象

{
 "id": 0,
 "title": "happy dayys",
 "owner": {"id": "1", "username": "dillonraphael"},
 "tags": [{"value": "Art", "label": "Art"}],
 "items": []
},

如果您想决定是否应该使用它，可以在tags数组上尝试Array.some()。这将返回一个布尔值。

let tags = [{"value": "Art", "label": "Art"}]

console.log(tags.some(tag => tag.value = "Art")) // true

有了这些，您现在可以一起使用filter()和some()。 some()将为数组中的每个项目返回true或false，并确定是否对其进行过滤：

let arr =  [{"id": 0,"title": "happy dayys","owner": {"id": "1", "username": "dillonraphael"},"tags": [{"value": "Art", "label": "Art"}],"items": []},{"id": 1,"title": "happy dayys","owner": {"id": "1", "username": "dillonraphael"},"tags": [{"value": "Architecture", "label": "Architecture"}],"items": []},]

console.log(arr.filter(obj => obj.tags.some(o => o.value == 'Art') ))

.filter中的ES6 .filter

2 个答案: