如何获取多个嵌套对象数组的公共值

Question

我想从每个视图中按名称而不是按 ID 获取公共值。例如：这里 'view 6' 和 'view 2' 很常见。

-这是我的回复 (allDeviceView)：

[
    {
        "ip": "111.11.11.11",
        "views": [
            {
                "id": 3,
                "name": "View 2",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            },
            {
                "id": 4,
                "name": "View 3",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            },
            {
                "id": 5,
                "name": "View 4",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            },
            {
                "id": 6,
                "name": "View 5",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            },
            {
                "id": 7,
                "name": "View 6",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            }
        ]
    },
    {
        "ip": "222.22.22.22",
        "views": [
            {
                "id": 2,
                "name": "View 1",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            },
            {
                "id": 3,
                "name": "View 2",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            },
            {
                "id": 7,
                "name": "View 6",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            }
        ]
    },
    {
        "ip": "144.44.44.44",
        "views": [
            {
                "id": 3,
                "name": "View 2",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            },
            {
                "id": 7,
                "name": "View 6",
                "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
            }
        ]
    }
]

-我想通过名称而不是 id 从视图中获取常见对象，所以这里 'view 6' 和 'view 2' 是常见的。目前，我仅在它们出现不止一次时才获取值，但可以有多个对象，因此我必须在每个视图中都通用。

-预期输出：

{
    "id": 3,
    "name": "View 2",
    "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
},
{
    "id": 7,
    "name": "View 6",
    "thumbnail": "http://media.fanconnect.tv/production/app/635_879656_thumbnail.jpg"
}

-我试过这样做：

 const deviceViewsMaps = allDeviceView?.map((d: any) => d.views).flat();
const flattened = [...deviceViewsMaps];

    /*eslint-disable */
    const counts = flattened.reduce(
      (map, { name }) => map.set(name, (map.get(name) || 0) + 1),
      new Map(),
    );

    /*eslint-enable */
    const names: any = [];

    const found = flattened.filter(({ name }) => {
      if (counts.get(name) > 1 && !names.includes(name)) {
        names.push(name);
        return true;
      }
      return false;
    });

    console.log(found, 'data');

    setFilteredDeviceView(found);

Answer 1

const deviceViewsMaps = allDeviceView?.map((d: any) => d.views || '[]').flat();

const deviceNameMap = allDeviceView?.map((d: any) => d.ip);

const flattened = [...deviceViewsMaps];

/*eslint-disable */
const counts = flattened.reduce(
  (map, { name }) => map.set(name, (map.get(name) || 0) + 1),
  new Map(),
);

/*eslint-enable */
const names: any = [];

const found = flattened.filter(({ name }) => {
//---------------------
      //here i have increased count with number of ip length. i.e : deviceNameMap.length - 1
//---------------------
  if (counts.get(name) > deviceNameMap.length - 1 && !names.includes(name)) {
    names.push(name);
    return true;
  }
  return false;
});

setFilteredDeviceView(found);

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