Question

我从数据库中选择基本上连接主实体和子实体，如下面的示例（汽车与零件）作为片段

我想按汽车零件的所有按键进行分组，并有一系列零件，但包括汽车和零件的所有按键。对于groupBy我可以找到的例子，通常它使用groupBy，但它只对一个键组合。我能够使用大量的解决方法，但我确信使用es6或lodash可以实现相同的可管理性（并获得更高的性能）。

有人可以帮我解决这个问题吗？我尝试过多个groupBy并减少组合，但无法正确链接。

＆＃13;

var data =     [{id: 'car1',
name: 'name for car 1',
description: 'description for car1',
partId: 'partId1',
partName: 'partName1'},
  
{id: 'car1',
name: 'name for car 1',
description: 'description for car1',
partId: 'partId2',
partName: 'partName2'},

{id: 'car2',
name: 'name for car 2',
description: 'description for car2',
partId: 'partId3',
partName: 'partName3'},

{id: 'car2',
name: 'name for car 2',
description: 'description for car2',
partId: 'partId4',
partName: 'partName4'}
];
   var dictionary = {};
 data.forEach(function(item, index, array)
  {
    var masterDocument = null;
    if (typeof dictionary[item.id] === 'undefined')
    {
      masterDocument = {
        id: item.id,
        name: item.name,
        description: item.description,
        parts: []
      };
      dictionary[item.id] = masterDocument;
    }
    else {
      var masterDocument = dictionary[item.id];
    }
    masterDocument.parts.push({
      partId: item.partId,
      partName: item.partName
    })
  })


  var asList = [];
  Object.keys(dictionary).forEach((item) => {
    asList.push(dictionary[item])
  });
  console.log(asList);

＆＃13;

.as-console-wrapper {
  min-height: 100%;
  top: 0;
}

＆＃13;

这里是我想要实现的结果的片段。

＆＃13;

[
  {
    "id": "car1",
    "name": "name for car 1",
    "description": "description for car1",
    "parts": [
      {
        "partId": "partId1",
        "partName": "partName1"
      },
      {
        "partId": "partId2",
        "partName": "partName2"
      }
    ]
  },
  {
    "id": "car2",
    "name": "name for car 2",
    "description": "description for car2",
    "parts": [
      {
        "partId": "partId3",
        "partName": "partName3"
      },
      {
        "partId": "partId4",
        "partName": "partName4"
      }
    ]
  }
]

＆＃13;

Answer 1

以下代码可以使用Lodash解决您的问题。基本上你想要做的是：

按id
按照ID分组汽车后，使用地图调用迭代该顶级数组，然后从第一个条目中抓取id，name和description （因为你知道这组中的所有车都是一样的）。保存这些以供稍后返回对象
然后，在仍处于此顶级地图迭代中的同时，还会迭代每个carGrouping（嵌套地图）中的各个车辆，以获取他们的partId和partName，然后将其放入{ {1}}数组
最后，获取所有对象属性，将它们全部放入顶级地图调用中的返回对象中，然后将它们全部返回
请勿忘记在链条末尾拨打parts以启动Lodash序列

＆＃13;

valueOf()

＆＃13;

let data =  [{id: 'car1',
name: 'name for car 1',
description: 'description for car1',
partId: 'partId1',
partName: 'partName1'},

{id: 'car1',
name: 'name for car 1',
description: 'description for car1',
partId: 'partId2',
partName: 'partName2'},

{id: 'car2',
name: 'name for car 2',
description: 'description for car2',
partId: 'partId3',
partName: 'partName3'},

{id: 'car2',
name: 'name for car 2',
description: 'description for car2',
partId: 'partId4',
partName: 'partName4'}
];

const carsInfo = _(data)
    .groupBy('id')
    .map(carGrouping => {
      // all cars in this array have the same id, name, description, so just grab them from the first one
      const firstCarInGroup = _.first(carGrouping); 
      const id = firstCarInGroup.id;
      const name = firstCarInGroup.name;
      const description = firstCarInGroup.description;

      // do a nested map call to iterate over each car in the carGrouping, and grab their partId and partName, and return it in an object
      const parts = _.map(carGrouping, car => {
        return {
          partId: car.partId,
          partName: car.partName
        }
      });

      return {
        id,
        name,
        description,
        parts
      }
    })
    .valueOf();


    console.log(carsInfo);

＆＃13;

Answer 2

这个不使用依赖项。简单的ES6 +。

＆＃13;

const data = [{
    id: 'car1',
    name: 'name for car 1',
    description: 'description for car1',
    partId: 'partId1',
    partName: 'partName1'
  },
  {
    id: 'car1',
    name: 'name for car 1',
    description: 'description for car1',
    partId: 'partId2',
    partName: 'partName2'
  },
  {
    id: 'car2',
    name: 'name for car 2',
    description: 'description for car2',
    partId: 'partId3',
    partName: 'partName3'
  },
  {
    id: 'car2',
    name: 'name for car 2',
    description: 'description for car2',
    partId: 'partId4',
    partName: 'partName4'
  }
];

const nested = data.reduce((acc, part) => {
  let index = acc.findIndex(car => car.id === part.id)
  const { partId, partName, ...car } = part

  if (index === -1) {
    acc.push({
      ...car,
      parts: [],
    })

    index = acc.length - 1
  }

  acc[index].parts.push({
    partId,
    partName,
  })

  return acc
}, [])

console.log(JSON.stringify(nested, null, '  '));

＆＃13;

Lodash - 为父母使用N（重复）键的儿童

2 个答案: