Question

在State块中执行映射后，我要在delete entry.State;数组中删除Entries属性（reduce）。我觉得这段代码是有气味的。

完成映射后，我不再需要State属性，删除它的正确方法是什么？

您可以看到我正在返回不带状态的状态和数据对象列表（已删除）。

const data = {
  FirstName: "Paul",
  LastName: "Allen",
  Email: "name@domain.com",
  Entries: [
    {
      EntryId: "FF-DD",
      Title: "Item 1",
      State: [
        { Name: "Lock 1", Time: "1563479611" },
        { Name: "Lock 2", Time: "1563479663" }
      ]
    },
    {
      EntryId: "GG-RR",
      Title: "Item 2",
      State: [
        { Name: "Run 1", Time: "116479611" },
        { Name: "Run 2", Time: "126347963" }
      ]
    }
  ]
};

const newData = createArray(data);
console.log(JSON.stringify(newData, null, 2));

function createArray(data) {
  let cloneData = { ...data };
  
  cloneData.EmailAddress = cloneData.Email;
  delete cloneData.Email;

  const states = cloneData.Entries.reduce((stateBlock, entry) => {
    const states = entry.State.map(state => {
      const CreatedTime = state.Time;
      const newState = {
        ...state,
        CreatedTime: CreatedTime,
        EntryId: entry.EntryId
      }

      delete newState.Time;
      return newState;
    });

    stateBlock.push(...states);
    delete entry.State;

    return stateBlock;
  }, []);

  return [cloneData, ...states];
}

输出按预期运行：

[
  {
    "FirstName": "Paul",
    "LastName": "Allen",
    "Entries": [
      {
        "EntryId": "FF-DD",
        "Title": "Item 1"
      },
      {
        "EntryId": "GG-RR",
        "Title": "Item 2"
      }
    ],
    "EmailAddress": "name@domain.com"
  },
  {
    "Name": "Lock 1",
    "CreatedTime": "1563479611",
    "EntryId": "FF-DD"
  },
  {
    "Name": "Lock 2",
    "CreatedTime": "1563479663",
    "EntryId": "FF-DD"
  },
  {
    "Name": "Run 1",
    "CreatedTime": "116479611",
    "EntryId": "GG-RR"
  },
  {
    "Name": "Run 2",
    "CreatedTime": "126347963",
    "EntryId": "GG-RR"
  }
]

时间已重命名为CreatedTime，并且还在每个州都包含EntryId。删除Entries

中的所有状态

Answer 1

您可以使用destructuring assignments with spread operators并嵌套.map()来获得一些创意，以完成此任务。

代码中的详细信息。

const data = { FirstName: "Paul", LastName: "Allen", Email: "name@domain.com", Entries: [ { EntryId: "FF-DD", Title: "Item 1", State: [ { Name: "Lock 1", Time: "1563479611" }, { Name: "Lock 2", Time: "1563479663" } ] }, { EntryId: "GG-RR", Title: "Item 2", State: [ { Name: "Run 1", Time: "116479611" }, { Name: "Run 2", Time: "126347963" } ] } ] };

const newData = (() => {
  let states = [];
  //Loop through Entries
  let Entries = data.Entries.map(({State, ...others}) => {

    if (State) {
      //Map each state to new format
      let state = State.map(({Time, ...obj}) => ({...obj, CreatedTime: Time, EntryId: others.EntryId}));
      states.push(...state);
    }

    //Return Entitites without State
    return others;
  });

  //Remove LastName and Email from root object
  let {LastName, Email, ...dataWithoutName} = data;

  //Change Email to EmailAddress, combine all the pieces
  return [ {EmailAddress: Email, ...dataWithoutName, Entries}, ...states ]; 
})();

console.log(newData);

Answer 2

如果要就地修改列表中每个项目的内容，请使用map。如果要将列表“简化”为一个值，请使用reduce。

function createArray(data) {
  let clone = { ...data };
  clone.Entries = clone.Entries.map(entry => {
      return Object.keys(entry).reduce((a, c) => {
        if(c === 'State') return a;
        a[c] = entry[c]
        return a;
      }, {})
   });
  return clone;
}

在给定数据上运行此操作将返回

{  
  "FirstName":"Paul",
  "LastName":"Allen",
  "Email":"name@domain.com",
  "Entries":[  
    {  
      "EntryId":"FF-DD",
      "Title":"Item 1"
    },
    {  
      "Id":"GG-RR",
      "Title":"Item 2"
    }
  ]
}

编辑：没有意识到您的条目中有两个不同的键，Id和EntryId。

要在reduce和map中删除属性（数组）吗？

2 个答案: