Question

我具有以下数据结构：

const data = {
  "firstName": "A",
  "lastName": "B",
  "address": [{
    "country": "France",
    "city": "Paris"
  },
    {
      "country": "Italy",
      "city": "Rome"
    }
  ],
};

我想使用Ramda将其转换为：

const result = [
  {
    "firstName": "A",
    "lastName": "B",
    "address": {
      "country": "France",
      "city": "Paris"
    },
  },
  {
    "firstName": "A",
    "lastName": "B",
    "address": {
      "country": "Italy",
      "city": "Rome"
    },
  },
];

Answer 1

您可以使用converge函数派生道具address，然后将其与列表中每个地址的主要对象连接起来：

/**
 * R.pick could be replaced with R.omit
 * to let you black list properties:
 * R.omit(['address']); https://ramdajs.com/docs/#omit
**/
const createByAddress = R.converge(R.map, [
  R.pipe(R.pick(['firstName', 'lastName']), R.flip(R.assoc('address'))),
  R.prop('address'),
]);

const data = {
  "firstName": "A",
  "lastName": "B",
  "address": [{
    "country": "France",
    "city": "Paris"
    },
    {
      "country": "Italy",
      "city": "Rome"
    }
  ],
};

console.log(createByAddress(data));

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js" integrity="sha256-xB25ljGZ7K2VXnq087unEnoVhvTosWWtqXB4tAtZmHU=" crossorigin="anonymous"></script>

Answer 2

我的问题是为什么“和Ramda一起”？我是Ramda的创始人和忠实粉丝，但这只是一种工具，除非这是Ramda的学习练习，否则似乎没有必要针对此问题使用它。

我会使用现代JS技术做到这一点：

const transform = ({address, ...rest}) => 
  address .map (a => ({...rest, address: a}))

const data = {firstName: "A", lastName: "B", address: [{country: "France", city: "Paris"}, {country: "Italy", city: "Rome"}]}

console .log (
  transform (data)
)

请注意您的问题所附的评论。 StackOverflow不是代码编写服务。这里的社区希望您能尝试一下并展示出自己陷入困境的地方。

Answer 3

我不确定这是否对您有帮助，但是如果您想基于adress生成多个对象，也许会有所帮助

const obj = {
  firstName: "a",
  lastName: "b",
  adresses: [{
    country: "France",
    city: "Paris"
  }, {
    country: "Italy",
    city: "Rome"
  }]
};

adressAmount = obj.adresses.length;

const adressObjects = [];

for (let i = 0; i < adressAmount; i++) {
  const {
    adresses,
    ...objWithoutAdresses
  } = obj;
  objWithoutAdresses.adress = obj.adresses[i];
  adressObjects.push(objWithoutAdresses);
}

console.log(adressObjects);

Answer 4

您可以迭代地址数组并根据需要创建对象

let obj = {
  "firstName": "A",
  "lastName": "B",
  "address": [{
      "country": "France",
      "city": "Paris"
    },
    {
      "country": "Italy",
      "city": "Rome"
    }
  ]
}


let newData = obj.address.map(function(item) {

  return {
    firstName: obj.firstName,
    lastName: obj.lastName,
    address: {
      country: item.country,
      city: item.city
    }
  }


});
console.log(newData)

Answer 5

我发现这很简单而且简短。

const data = {
  "firstName": "A",
  "lastName": "B",
  "address": [{
    "country": "France",
    "city": "Paris"
  },
    {
      "country": "Italy",
      "city": "Rome"
    }
  ],
};



let requiredData = data.address.map(element=>{
return {...data,address:element}
})

console.log(requiredData);

Answer 6

1）创建一个空字典

2）用于循环数组并将每个数组的索引存储在字典中作为值

将具有数组的对象转换为数组

6 个答案: