如何对含嵌套多日期数组的对象数组按日期拆分重组并排序？

我从API获取了如下对象数组：

const schedules = [
    {
        "_id": "6436b48b875967d0bea245b4",
        "service": "64246dc9a2d61593d103c749",
        "scheduledDates": [
            {
                "date": "2023-04-17T18:00:00.000Z",
                "user": "643701f7e5f61e6760f4d1f3"
            },
            {
                "date": "2023-04-12T18:00:00.000Z",
                "user": "643701f7e5f61e6760f4d1f3"
            }
        ]
    },
    {
        "_id": "6436b48b875967d0bea245b5",
        "service": "64246dc9a2d61593d103c749",
        "scheduledDates": [
            {
                "date": "2023-04-19T10:30:00.000Z",
                "user": "64217a8dcc69c5fa48a5b484"
            },
            {
                "date": "2023-04-12T18:00:00.000Z",
                "user": "6414be936b0bbf2bd8fa964f"
            }
        ]
    },
]

我希望将该数组按scheduledDates中的date字段全局排序，最终拆分为每个scheduledDates元素单独对应一个原结构对象的数组，结果如下：

const newSchedules = [
  {
      "_id": "6436b48b875967d0bea245b4",
      "service": "64246dc9a2d61593d103c749",
      "scheduledDates": [
          {
              "date": "2023-04-12T18:00:00.000Z",
              "user": "643701f7e5f61e6760f4d1f3"
          }
      ]
  },
  {
    "_id": "6436b48b875967d0bea245b5",
    "service": "64246dc9a2d61593d103c749",
    "scheduledDates": [
        {
            "date": "2023-04-12T18:00:00.000Z",
            "user": "6414be936b0bbf2bd8fa964f"
        }
    ]
},
  {
      "_id": "6436b48b875967d0bea245b4",
      "service": "64246dc9a2d61593d103c749",
      "scheduledDates": [
          {
              "date": "2023-04-17T18:00:00.000Z",
              "user": "643701f7e5f61e6760f4d1f3"
          }
      ]
  },
  {
      "_id": "6436b48b875967d0bea245b5",
      "service": "64246dc9a2d61593d103c749",
      "scheduledDates": [
          {
              "date": "2023-04-19T10:30:00.000Z",
              "user": "64217a8dcc69c5fa48a5b484"
          }
      ]
  },
]

我编写的排序函数只能对每个对象内部的scheduledDates数组排序，无法实现全局排序和结构转换：

const handleSort = () => {
    if (sortOrder === "asc") {
      setSortOrder("desc");
      return schedules.map((schedule) =>
      schedule.scheduledDates.sort((a, b) => new Date(a.date) - new Date(b.date))
      );
    } else {
      setSortOrder("asc");
      return schedules.map((schedule) =>
      schedule.scheduledDates.sort((a, b) => new Date(b.date) - new Date(a.date))
    }
  };

请问该如何解决？

要实现需求，需要分三步处理：

1. 扁平化结构：把每个schedule对象的scheduledDates数组拆分成单个元素，每个元素对应一个独立的schedule对象副本
2. 全局排序：基于拆分后每个对象的scheduledDates[0].date进行排序
3. 切换排序方向：根据当前sortOrder控制升序/降序

修改后的handleSort函数如下：

const handleSort = () => {
  // 第一步：扁平化数组，将每个scheduledDates元素转为独立对象
  const flattened = schedules.flatMap(schedule => 
    schedule.scheduledDates.map(dateItem => ({
      ...schedule,
      scheduledDates: [dateItem] // 保留原结构，仅包含单个日期项
    }))
  );

  // 第二步：根据日期排序，切换升降序
  const sorted = flattened.sort((a, b) => {
    const dateA = new Date(a.scheduledDates[0].date);
    const dateB = new Date(b.scheduledDates[0].date);
    return sortOrder === "asc" ? dateB - dateA : dateA - dateB;
  });

  // 第三步：切换排序状态并返回结果
  setSortOrder(prev => prev === "asc" ? "desc" : "asc");
  return sorted;
};

代码说明 #

• flatMap+map：flatMap遍历原数组并将每个对象的scheduledDates拆分为多个对象，避免嵌套数组；map为每个日期项创建原schedule的副本，确保不修改原数据
• 排序逻辑：直接基于拆分后每个对象的唯一日期项进行比较，实现全局排序；通过sortOrder判断是升序（dateA - dateB）还是降序（dateB - dateA）
• 状态切换：使用函数式更新setSortOrder(prev => ...)确保状态更新的准确性

内容的提问来源于stack exchange，提问作者BraulioPortela