JavaScript如何根据customer_id从对象数组中获取最新唯一数据

实现方案 #

你可以借助Map来实现按customer_id去重，同时保留每个customer_id对应的最大id（即最新）条目，代码如下：

// 原始数组
const originalArr = [
  {id: 408, customer_id: 2, bill_no: 381, bill_period: 'weekly', from_date: '2021-10-10'},
  {id: 409, customer_id: 3, bill_no: 382, bill_period: 'weekly', from_date: '2021-10-10'},
  {id: 410, customer_id: 4, bill_no: 383, bill_period: 'weekly', from_date: '2021-10-10'},
  {id: 411, customer_id: 6, bill_no: 384, bill_period: 'weekly', from_date: '2021-10-10'},
  {id: 412, customer_id: 7, bill_no: 385, bill_period: 'weekly', from_date: '2021-10-10'},
  {id: 413, customer_id: 8, bill_no: 386, bill_period: 'weekly', from_date: '2021-10-10'},
  {id: 414, customer_id: 9, bill_no: 387, bill_period: 'weekly', from_date: '2021-10-10'},
  {id: 387, customer_id: 2, bill_no: 360, bill_period: 'weekly', from_date: '2021-10-03'},
  {id: 388, customer_id: 3, bill_no: 361, bill_period: 'weekly', from_date: '2021-10-03'},
  {id: 389, customer_id: 4, bill_no: 362, bill_period: 'weekly', from_date: '2021-10-03'},
  {id: 390, customer_id: 6, bill_no: 363, bill_period: 'weekly', from_date: '2021-10-03'},
  {id: 391, customer_id: 7, bill_no: 364, bill_period: 'weekly', from_date: '2021-10-03'},
  {id: 392, customer_id: 8, bill_no: 365, bill_period: 'weekly', from_date: '2021-10-03'},
  {id: 393, customer_id: 9, bill_no: 366, bill_period: 'weekly', from_date: '2021-10-03'},
  {id: 380, customer_id: 2, bill_no: 353, bill_period: 'weekly', from_date: '2021-09-26'},
  {id: 381, customer_id: 3, bill_no: 354, bill_period: 'weekly', from_date: '2021-09-26'},
  {id: 382, customer_id: 4, bill_no: 355, bill_period: 'weekly', from_date: '2021-09-26'},
  {id: 383, customer_id: 6, bill_no: 356, bill_period: 'weekly', from_date: '2021-09-26'},
  {id: 384, customer_id: 7, bill_no: 357, bill_period: 'weekly', from_date: '2021-09-26'},
  {id: 385, customer_id: 8, bill_no: 358, bill_period: 'weekly', from_date: '2021-09-26'},
  {id: 386, customer_id: 9, bill_no: 359, bill_period: 'weekly', from_date: '2021-09-26'}
]

// 去重逻辑
const customerMap = new Map()
originalArr.forEach(item => {
  // 仅当Map中无当前customer_id，或当前条目id更大时更新
  if (!customerMap.has(item.customer_id) || item.id > customerMap.get(item.customer_id).id) {
    customerMap.set(item.customer_id, item)
  }
})

// 转换为结果数组
const result = Array.from(customerMap.values())
console.log(result)

逻辑说明 #

• 用Map存储已遍历过的customer_id对应的条目，键为customer_id，值为对应的对象
• 遍历原始数组时，每次判断当前条目是否比已存储的同customer_id条目id更大，是就更新
• 最后把Map的所有值转为数组，就是你需要的去重后结果

如果你的原始数组本身已经按id从大到小排序（你给出的示例数组恰好是这个顺序），可以简化判断逻辑，直接判断Map中是否不存在该customer_id再存入即可，不用对比id大小，遍历完结果是一样的。

内容的提问来源于stack exchange，提问作者Mohammed