通过给定的键将两个对象数组合并为一个并复制其他值

Question

我有两个包含相似值的对象数组data = [array1, array2]array1 = [{hour: "565", age: "21", id: "1", naban: "sdfsd"}, {hour: "515", age: "25", id: "2", naban: "sf"}]array2 = [{hour: "56454", age: "21", id: "1", too: "898"}, {hour: "8979", age: "25", id: "2", too: "234234"},  {hour: "65465", age: "27", id: "6", too: "123"}]我有一个数组，其中这两个对象集的值将用于合并keys= ['id', 'id']我想合并这些对象并创建一个对象数组，如下所示：result = [{hour: "565", hour2: "56454", age: "21", age2: "21", id: "1", too: "898", naban: "sdfsd"}, {hour: "515", hour2: "8979", age: "25", age2: "25", id: "2", too: "234234", naban: "sf"}, {hour: "65465", age: "27", id: "6", too: "123"} ]标准：我想保留所有不在键数组中的信息，例如。如果您查看结果数组，您将看到 hour 和 hour2 值。如果键不匹配，它将按原样推送对象（例如结果数组中的第三项）这是我到目前为止所做的：mergeOjects = (object, keys) => {const sameKeys = Object.values(keys);const data = Object.values(object);if (data[0].length > data[1].length) {  const yarrak = data[0].map((item, i) => {    if (item[sameKeys[0]] === data[1][i][sameKeys[1]]) {      return Object.assign({}, item, data[1][i]);    }    return Object.assign({}, item);  });  console.log({ sameKeys, data, yarrak });} else {  const yarrak = data[1].map((item, i) => {    if (data[0][i]) {      if (item[sameKeys[1]] === data[0][i][sameKeys[0]]) {        return Object.assign({}, item, data[0][i]);      }    }    return Object.assign({}, item);  });  console.log({ sameKeys, data, yarrak });}};它可能需要一些清洁，但我现在正试图让逻辑工作，所以提前抱歉。我能够完成第二个条件，但它会覆盖小时值，而不是像示例中那样单独存储

Answer 1

您可以减少第二个数组并将所需的键作为查找的通用部分。

var array1 = [{ hour: "565", age: "21", id: "1", naban: "sdfsd" }, { hour: "515", age: "25", id: "2", naban: "sf" }],

    array2 = [{ hour: "56454", age: "21", id: "1", too: "898" }, { hour: "8979", age: "25", id: "2", too: "234234" }, { hour: "65465", age: "27", id: "6", too: "123" }],

    data = [array1, array2],

    keys = ['id', 'id'],

    merged = data[1].reduce((r, o) => {

        var temp = r.find(q => o[keys[0]] === q[keys[1]]);

        if (temp) {

            Object

                .entries(o)

                .forEach(([k, v]) => {

                    if (keys.includes(k)) return;

                    temp[k in temp ? k + 2 : k] = v;

                });

        } else {

            r.push({ ...o });

        }

        return r;

    }, data[0]);

console.log(merged);

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