根据第二个数组中的值将数组排序为多个

Question

我正在尝试将对象从一个数组移动到七个不同的数组中。它们从一开始就在一个阵列中。我想根据另一个数组中具有一个相关属性的值对它们进行排序。

我要排序的数组 1

var serialList = [
    {
        "OrderNumber": "1",
        "ArticleNumber": "CLIENT",
        "SerialNumber": "111"

    },
    {
        "OrderNumber": "2",
        "ArticleNumber": "IPAD",
        "SerialNumber": "222"

    },
    {
        "OrderNumber": "3",
        "ArticleNumber": "PHONE",
        "SerialNumber": "333"

    },
    {
        "OrderNumber": "4",
        "ArticleNumber": "SWITCH",
        "SerialNumber": "444"

    },
    {
        "OrderNumber": "5",
        "ArticleNumber": "HARDWARE",
        "SerialNumber": "555"

    },
    {
        "OrderNumber": "6",
        "ArticleNumber": "MAC",
        "SerialNumber": "666"

    }, 
    {
        "OrderNumber": "7",
        "ArticleNumber": "PRINTER",
        "SerialNumber": "777"

    }, 
    {
        "OrderNumber": "8",
        "ArticleNumber": "MAC",
        "SerialNumber": "888"

    }
];

我要比较的数组

var articleNumberList = [
    {
        "Article": "CLIENT",
        "Model": "client"

    },
    {
        "Article": "IPAD",
        "Model": "ipad"

    },
    {
        "Article": "PHONE",
        "Model": "phone"

    },
    {
        "Article": "SWITCH",
        "Model": "switch"

    },
    {
        "Article": "HARDWARE",
        "Model": "hardware"

    },
    {
        "Article": "MAC",
        "Model": "mac",


    }, {
        "Article": "PRINTER",
        "Model": "printer"


    }
];

我想检查第一个数组属性 ArticleNumber 并将其与第二个数组属性 Article 进行比较。一旦找到匹配项，将其排序到另一个数组中，对应于 Article 属性

我尝试过的（目前正在工作）

for (i = 0; i < serialList.length; i++) {

    for (u = 0; u < articleNumberList.length; u++) {
        if (serialList[i].ArticleNumber == articleNumberList[u].Article) {
            if (serialList[i].ArticleNumber == "CLIENT") {
                clientList.push(serialList[i]);

            } else if (serialList[i].ArticleNumber == "IPAD") {
                ipadList.push(serialList[i]);

            } else if (serialList[i].ArticleNumber == "PHONE") {
                phoneList.push(serialList[i]);

            } else if (serialList[i].ArticleNumber == "SWITCH") {
                switchList.push(serialList[i]);

            } else if (serialList[i].ArticleNumber == "HARDWARE") {
                hardwareList.push(serialList[i]);

            } else if (serialList[i].ArticleNumber == "MAC") {
                macList.push(serialList[i]);

            } else if (serialList[i].ArticleNumber == "PRINTER") {
                printerList.push(serialList[i]);

            }
        }
    }
};

为什么我不想使用此解决方案： 它是一个嵌套的for循环，最终将用于各种硬件的数据库，数组serialList今天有几千长并将继续增长，数组articleNumberList目前在40左右并且还将继续增长。我也不太喜欢 IF、ELSE IF 语句，我觉得它可能可以用 array.filter() 处理，但我真的不知道如何处理它。

Answer 1

如果您有一个有效文章类型的列表，则可以初始化一个包含一系列序列项目数组的映射（存储桶）。

const serialList = [
  { "OrderNumber": "1", "ArticleNumber": "CLIENT", "SerialNumber": "111" },
  { "OrderNumber": "2", "ArticleNumber": "IPAD","SerialNumber": "222" },
  { "OrderNumber": "3", "ArticleNumber": "PHONE", "SerialNumber": "333" },
  { "OrderNumber": "4", "ArticleNumber": "SWITCH", "SerialNumber": "444"  },
  { "OrderNumber": "5", "ArticleNumber": "HARDWARE", "SerialNumber": "555" },
  { "OrderNumber": "6", "ArticleNumber": "MAC", "SerialNumber": "666" }, 
  { "OrderNumber": "7", "ArticleNumber": "PRINTER", "SerialNumber": "777" }, 
  { "OrderNumber": "8", "ArticleNumber": "MAC", "SerialNumber": "888"}
];

const articleNumberList = [
  { "Article": "CLIENT", "Model": "client" },
  { "Article": "IPAD", "Model": "ipad" },
  { "Article": "PHONE", "Model": "phone" },
  { "Article": "SWITCH", "Model": "switch" },
  { "Article": "HARDWARE", "Model": "hardware" },
  { "Article": "MAC", "Model": "mac", },
  { "Article": "PRINTER", "Model": "printer" }
];

const buckets = articleNumberList
  .reduce((acc, { Article }) => acc.set(Article, []), new Map);

for (let serialItem of serialList) {
  buckets.get(serialItem.ArticleNumber)?.push(serialItem);
}

console.log(Object.fromEntries([...buckets]));

console.log(buckets.get('MAC')); // Get only the MAC items

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