匹配兩個陣列并合并它們比較JavaScript中的鍵值-有解無憂

我有兩個不同的陣列。一種是這樣的：

const arrOne = [
    [{sender: '0000', to: '1111'}],
    [{sender: '2222', to: '1111'}]
];

另一個陣列就像：

const arrTwo = [
  {
    firstName: 'John',
    LastName: 'Doe',
    num: '1111'
  }
]

現在我想匹配to來自 arrOne 和num陣列 2 的鍵，如果他們計算，第一個陣列應該用名字和姓氏更新。因此，預期的輸出將是這樣的：

  const arrOne = [
        [{sender: '0000', to: '1111', firstName: 'John', lastName: 'Doe'}],
        [{sender: '2222', to: '1111', firstName: 'John', lastName: 'Doe'}]
    ];

這是我嘗試過的：

arrOne = arrOne.map(item => {
  const item2 = arrTwo.find(i2 => i2.num == item.to);
  return item2 ? { ...item, ...item2 } : item;
});

提前致謝。

uj5u.com熱心網友回復：

你arrOne是一個陣列陣列，所以你需要考慮到這一點。我還假設您需要根據預期輸出num從arrTwo元素中排除。

let arrOne = [
    [{sender: '0000', to: '1111'}],
    [{sender: '2222', to: '1111'}]
];

const arrTwo = [
  {
    firstName: 'John',
    lastName: 'Doe',
    num: '1111'
  }
];

arrOne = arrOne.map(item => {
  const item2 = arrTwo.find(i2 => i2.num == item[0].to);
  return item2 ? [{ ...item[0], firstName: item2.firstName, lastName: item2.lastName }] : item;
});

console.log(arrOne);

uj5u.com熱心網友回復：

您幾乎成功了，只需更改item為item[0]（因為資料中的資料arrOne是嵌套陣列），并添加[]以包裝回傳的結果

let arrOne = [
    [{sender: '0000', to: '1111'}],
    [{sender: '2222', to: '1111'}]
]; // it's nested array here
const arrTwo = [
  {
    firstName: 'John',
    LastName: 'Doe',
    num: '1111'
  }
]

arrOne = arrOne.map(item => {
  const item2 = arrTwo.find(i2 => i2.num == item[0].to)
  return item2 ? [{ ...item[0], ...item2 }] : [item[0]]; // need to return as array
});

console.log(arrOne)

uj5u.com熱心網友回復：

試試這段代碼，

 const arrOne = [
    [{ sender: "0000", to: "1111" }],
    [{ sender: "2222", to: "1111" }],
  ];
  const arrTwo = [
    {
      firstName: "John",
      LastName: "Doe",
      num: "1111",
    },
  ];
  const newarrtt = () => {
    let val;
    const newarry = arrOne.map(item => {
      val = arrTwo.find(x => {
        return x.num, item[0].to;
      });

      if (val) {
        return [
          {
            ...item[0],
            firstName: val.firstName,
            LastName: val.LastName,
          },
        ];
      }

      val = null;
    });
    console.log(newarry);
  };
  newarrtt();

你錯過的是，你的映射函式不是回傳物件，它是陣列。所以你需要訪問諸如 item[0] 之類的東西。

uj5u.com熱心網友回復：

此技術將僅過濾掉不需要的“num”屬性：

const arrOne = [[{sender: '0000', to: '1111'}],[{sender: '2222', to: '1111'}]];
const arrTwo = [{firstName: 'John', LastName: 'Doe',num: '1111'}];
console.log(arrOne.map(i=>[{...i[0], ...Object.fromEntries(Object.entries(
  arrTwo.find(j=>j.num===i[0].to)??{}).filter(([k,v])=>k!=='num')
    .map(([k,v])=>[k==='LastName'?'lastName':k, v]))}]));

如果您需要將“姓氏”屬性重命名為“姓氏”，您也可以這樣做：

const arrOne = [[{sender: '0000', to: '1111'}],[{sender: '2222', to: '1111'}]];
const arrTwo = [{firstName: 'John', LastName: 'Doe',num: '1111'}];
console.log(arrOne.map((i,o)=>[{...i[0], ...(o=(arrTwo.find(j=>j.num===i[0].to)),
    (!o?{}:{firstName: o.firstName, lastName: o.LastName}))}]));

這兩種技術都適用于找不到num匹配項的情況。

轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/527406.html

標籤：javascript

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