JS：从嵌套数组中删除对象并返回父数组

Question

如何删除嵌套在另一个数组中的数组 subBrands 中的对象，其中对象的 id 属性为 = 31。我试图在不删除该子品牌的情况下恢复整个父数组。

数组是：

[
  {
    "id": 10,
    "name": "Parent Brand 1",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 09:55:51",
    "updated_at": "2017-02-02 09:55:51",
    "subBrands": [
      {
        "id": 31,
        "name": "Sub Brand 6",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:24:49",
        "updated_at": "2017-02-02 11:42:02"
      },
      {
        "id": 32,
        "name": "Sub Brand 7",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:24:57",
        "updated_at": "2017-02-02 11:42:18"
      },
      {
        "id": 33,
        "name": "Sub Brand 8",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:25:04",
        "updated_at": "2017-02-02 11:42:34"
      },
      {
        "id": 34,
        "name": "Sub Brand 9",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:25:39",
        "updated_at": "2017-02-02 11:42:43"
      },
      {
        "id": 35,
        "name": "Sub Brand 10",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:25:46",
        "updated_at": "2017-02-02 11:42:52"
      },
      {
        "id": 36,
        "name": "Sub Brand 4",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:43:53",
        "updated_at": "2017-02-02 11:43:53"
      }
    ]
  },
  {
    "id": 12,
    "name": "Parent Brand 2",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 09:56:16",
    "updated_at": "2017-02-02 09:56:16",
    "subBrands": []
  },
  {
    "id": 16,
    "name": "Brand no children",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 10:37:40",
    "updated_at": "2017-02-02 10:37:40",
    "subBrands": []
  },
  {
    "id": 37,
    "name": "Whoops brand",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 11:44:10",
    "updated_at": "2017-02-02 11:44:10",
    "subBrands": []
  }
]

我想要得到的是：

[
  {
    "id": 10,
    "name": "Parent Brand 1",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 09:55:51",
    "updated_at": "2017-02-02 09:55:51",
    "subBrands": [
      {
        "id": 32,
        "name": "Sub Brand 7",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:24:57",
        "updated_at": "2017-02-02 11:42:18"
      },
      {
        "id": 33,
        "name": "Sub Brand 8",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:25:04",
        "updated_at": "2017-02-02 11:42:34"
      },
      {
        "id": 34,
        "name": "Sub Brand 9",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:25:39",
        "updated_at": "2017-02-02 11:42:43"
      },
      {
        "id": 35,
        "name": "Sub Brand 10",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:25:46",
        "updated_at": "2017-02-02 11:42:52"
      },
      {
        "id": 36,
        "name": "Sub Brand 4",
        "parent": 10,
        "author": 1,
        "deleted_at": null,
        "created_at": "2017-02-02 11:43:53",
        "updated_at": "2017-02-02 11:43:53"
      }
    ]
  },
  {
    "id": 12,
    "name": "Parent Brand 2",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 09:56:16",
    "updated_at": "2017-02-02 09:56:16",
    "subBrands": []
  },
  {
    "id": 16,
    "name": "Brand no children",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 10:37:40",
    "updated_at": "2017-02-02 10:37:40",
    "subBrands": []
  },
  {
    "id": 37,
    "name": "Whoops brand",
    "parent": null,
    "author": 1,
    "deleted_at": null,
    "created_at": "2017-02-02 11:44:10",
    "updated_at": "2017-02-02 11:44:10",
    "subBrands": []
  }
]

我愿意使用下划线。我最接近的是：

    var brands = _.filter(brands, function(n) { 
        return _.some(n.subBrands, function(subBrand){ 
            return subBrand.id != brand.id;
        });
    });

但这会删除不包含 id 为 31 的子品牌的数组。所以它不是我需要的。

干杯!

Answer 1

var arr = [{"id":10,"name":"Parent Brand 1","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 09:55:51","updated_at":"2017-02-02 09:55:51","subBrands":[{"id":31,"name":"Sub Brand 6","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:24:49","updated_at":"2017-02-02 11:42:02"},{"id":32,"name":"Sub Brand 7","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:24:57","updated_at":"2017-02-02 11:42:18"},{"id":33,"name":"Sub Brand 8","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:25:04","updated_at":"2017-02-02 11:42:34"},{"id":34,"name":"Sub Brand 9","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:25:39","updated_at":"2017-02-02 11:42:43"},{"id":35,"name":"Sub Brand 10","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:25:46","updated_at":"2017-02-02 11:42:52"},{"id":36,"name":"Sub Brand 4","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:43:53","updated_at":"2017-02-02 11:43:53"}]},{"id":12,"name":"Parent Brand 2","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 09:56:16","updated_at":"2017-02-02 09:56:16","subBrands":[]},{"id":16,"name":"Brand no children","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 10:37:40","updated_at":"2017-02-02 10:37:40","subBrands":[]},{"id":37,"name":"Whoops brand","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:44:10","updated_at":"2017-02-02 11:44:10","subBrands":[]}];


var id = prompt("Id of subbrands to remove: ");

arr.forEach(function(o) {
  o.subBrands = o.subBrands.filter(s => s.id != id);
});

console.log(arr);