Question

如何在未知大小的重复嵌套数组中查找对象和父对象？使用lodash或本地javascript。

数组可能是这样的：

var modules = [{
    name: 'Module1',
    submodules: [{
        name: 'Submodule1',
        id: 1,
        submodules: [{
          name: 'Submodule11',
          id: 1,
          submodules: []
        }, {
          name: 'Submodule12',
          id: 2,
          submodules: [{
            name: 'Submodule121',
            id: 1,
            submodules: []
          }, {
            name: 'Submodule122',
            id: 2,
            submodules: []
          }]
        }]
      },
      {
        name: 'Submodule2',
        id: 2,
        submodules: []
      }
    ]
  },
  {
    name: 'Module2',
    submodules: [{
      name: 'Submodule1',
      id: 3,
      submodules: []
    }, {
      name: 'Submodule2',
      id: 4,
      submodules: []
    }]
  }
];

并且说数组中的所有“名称”属性都是唯一的。

我想找到：

name: 'Submodule122'

我正在使用此函数查找父对象，但它仅在数组的第一级起作用：

_.find(this.modules , function(item) {
            return _.some(item.submodules, { name: 'Submodule122'});

这可以找到实际的对象，但它也仅适用于数组的第一级：

_(this.modules)
            .thru(function (coll) {
                return _.union(coll, _.map(coll, 'submodules'));
            })
            .flatten()
            .find({ name: 'Submodule122'})

Answer 1

您可以通过将实际对象保留为父对象来采取迭代和递归的方法。

function find(name, array, parent) {
    var result;
    array.some(object =>
        object.name === name && (result = { object, parent }) || 
        (result = find(name, object.submodules, object))
    );
    return result;
}

var modules = [{ name: 'Module1', submodules: [{ name: 'Submodule1', id: 1, submodules: [{ name: 'Submodule11', id: 1, submodules: [] }, { name: 'Submodule12', id: 2, submodules: [{ name: 'Submodule121', id: 1, submodules: [] }, { name: 'Submodule122', id: 2, submodules: [] }] }, { name: 'Submodule2', id: 2, submodules: [] }] }, { name: 'Module2', submodules: [{ name: 'Submodule1', id: 3, submodules: [] }, { name: 'Submodule2', id: 4, submodules: [] }] }] }];

console.log(find('Submodule122', modules));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

var modules = [{
    name: 'Module1',
    submodules: [{
        name: 'Submodule1',
        id: 1,
        submodules: [{
          name: 'Submodule11',
          id: 1,
          submodules: []
        }, {
          name: 'Submodule12',
          id: 2,
          submodules: [{
            name: 'Submodule121',
            id: 1,
            submodules: []
          }, {
            name: 'Submodule122',
            id: 2,
            submodules: []
          }]
        }]
      },
      {
        name: 'Submodule2',
        id: 2,
        submodules: []
      }
    ]
  },
  {
    name: 'Module2',
    submodules: [{
      name: 'Submodule1',
      id: 3,
      submodules: []
    }, {
      name: 'Submodule2',
      id: 4,
      submodules: []
    }]
  }
];

var findByName = (name, module, parent, item) => {
  parent.push(...module.filter(x => Array.isArray(x.submodules) && x.submodules.some(y => y.name == name)));
  item.push(...module.filter(y => y.name == name));

  module.forEach(x => {
    if (Array.isArray(x.submodules) && x.submodules.length > 0) {
      findByName(name, x.submodules, parent, item);
    }
  });

}

var parents = [], items = [];
findByName('Submodule12', modules, parents, items);
console.log(parents);
console.log(items);

Answer 3

要求不完全相同，但我认为仍然值得分享。现在，我们将object-scan用于我们所有的数据处理任务，一旦您将其束之高阁，它就会非常强大。这是您回答问题的方式

请注意，这将返回所有父母，但是很容易选择想要的父母。

const objectScan = require('object-scan');

const search = (name, data) => objectScan(['**.name'], {
  rtn: 'parents',
  abort: true,
  filterFn: ({ value }) => value === name
})(data);

const modules = [{"name":"Module1","submodules":[{"name":"Submodule1","id":1,"submodules":[{"name":"Submodule11","id":1,"submodules":[]},{"name":"Submodule12","id":2,"submodules":[{"name":"Submodule121","id":1,"submodules":[]},{"name":"Submodule122","id":2,"submodules":[]}]}]},{"name":"Submodule2","id":2,"submodules":[]}]},{"name":"Module2","submodules":[{"name":"Submodule1","id":3,"submodules":[]},{"name":"Submodule2","id":4,"submodules":[]}]}];

console.log(search('Submodule122', modules));
// => [ { name: 'Submodule122', id: 2, submodules: [] },
//   [ { name: 'Submodule121', id: 1, submodules: [] },
//     { name: 'Submodule122', id: 2, submodules: [] } ],
//   { name: 'Submodule12',
//     id: 2,
//     submodules: [ [Object], [Object] ] },
//   [ { name: 'Submodule11', id: 1, submodules: [] },
//     { name: 'Submodule12', id: 2, submodules: [Array] } ],
//   { name: 'Submodule1', id: 1, submodules: [ [Object], [Object] ] },
//   [ { name: 'Submodule1', id: 1, submodules: [Array] },
//     { name: 'Submodule2', id: 2, submodules: [] } ],
//   { name: 'Module1', submodules: [ [Object], [Object] ] },
//   [ { name: 'Module1', submodules: [Array] },
//     { name: 'Module2', submodules: [Array] } ] ]

console.log(search('unknown', modules));
// => undefined

在重复嵌套的数组中查找对象和父对象

3 个答案: