基于另一数组对Applications数组进行多条件排序的需求

现有两个数组Applications和ApplicationOrder，需根据ApplicationOrder中的Order值对Applications数组排序；当Order值重复时，再按Applications的Title字段排序，ID为两个数组的共同关联字段。

原始数据 #

const Applications = [
    {
        "ID": 30,
        "Title": "Balance",
        "Acronym": null,
        "Link": {
            "$2_1": "https:abc.com",
            "$1_1": "https:abc.com"
        }
    },
    {
        "ID": 12,
        "Title": "Scorecard",
        "Acronym": null,
        "Link": {
            "$2_1": "https:xyz.com",
            "$1_1": "https:xyz.com"
        }
    },
    {
        "ID": 62,
        "Title": "Best Practices",
        "Acronym": null,
        "Link": {
            "$2_1": "https:xyz.com",
            "$1_1": "https:xyz.com"
        }        
    },
    {
        "ID": 15,
        "Title": "User Actions",
        "Acronym": null,
        "Link": {
            "$2_1": "https:xyz.com",
            "$1_1": "https:xyz.com"
        }        
    }
];

const ApplicationOrder = [
    {"Id":"30","Order":"4"},
    {"Id":"12","Order":"4"},
    {"Id":"62","Order":"2"},
    {"Id":"15","Order":"1"}
];

排序规则 #

1. 优先按Order值从小到大排序
2. 当Order值相同时，按Title字段的字典序从小到大排序

期望输出 #

const sortedApplications = [
    {
        "ID": 15,
        "Title": "User Actions",
        "Acronym": null,
        "Link": {
            "$2_1": "https:xyz.com",
            "$1_1": "https:xyz.com"
        }        
    },
    {
        "ID": 62,
        "Title": "Best Practices",
        "Acronym": null,
        "Link": {
            "$2_1": "https:xyz.com",
            "$1_1": "https:xyz.com"
        }        
    },
    {
        "ID": 30,
        "Title": "Balance",
        "Acronym": null,
        "Link": {
            "$2_1": "https:abc.com",
            "$1_1": "https:abc.com"
        }
    },
    {
        "ID": 12,
        "Title": "Scorecard",
        "Acronym": null,
        "Link": {
            "$2_1": "https:xyz.com",
            "$1_1": "https:xyz.com"
        }
    }
];

通过以下步骤实现需求：

1. 将ApplicationOrder转换为ID与Order的映射表，提升排序时的查找效率
2. 使用数组sort()方法，先按Order数值排序，Order相同时按Title字典序排序

代码实现 #

// 构建ID到Order的映射，字符串ID转数字后存储，避免类型不匹配
const orderMap = new Map();
ApplicationOrder.forEach(item => {
    orderMap.set(Number(item.Id), parseInt(item.Order, 10));
});

// 排序原数组（创建副本避免修改原数据）
const sortedApplications = [...Applications].sort((appA, appB) => {
    const orderA = orderMap.get(appA.ID);
    const orderB = orderMap.get(appB.ID);

    // 优先按Order从小到大排序
    if (orderA !== orderB) {
        return orderA - orderB;
    }

    // Order相同时，按Title字典序排序
    return appA.Title.localeCompare(appB.Title);
});

console.log(sortedApplications);

代码说明 #

• 用Map存储ID与Order的对应关系，单次遍历构建映射，后续查找Order的时间复杂度为O(1)
• 用[...Applications]创建原数组副本，避免排序操作修改原数组
• 使用localeCompare()处理字符串排序，支持多语言场景，比直接用字符串比较运算符更可靠
• 将Order字符串转为整数后比较，避免字符串排序的异常（比如"10"会错误排在"2"前面）

内容的提问来源于stack exchange，提问作者Learner