如何合并相同selectionId的数组对象并保留全部topicName

问题说明 #

原有代码通过Object.assign直接覆盖同selectionId对应对象的属性，因此只能保留最后一个topicName值。另外你给出的预期输出结构不符合JS对象语法规范：对象不能直接包含["topicName":"xxx",...]这种结构，建议统一将topicName字段设为数组类型，后续循环逻辑无需额外兼容字段类型，使用更便捷。

实现代码 #

方案1：统一topicName为数组类型（推荐） #

const result = Object.values(
  list.reduce((accumulator, current) => {
    const id = current.selectionId
    if (!accumulator[id]) {
      accumulator[id] = {
        selectionId: id,
        topicName: [current.topicName]
      }
    } else {
      accumulator[id].topicName.push(current.topicName)
    }
    return accumulator
  }, {})
)

输出结果 #

[
  {
    "selectionId": 1,
    "topicName": ["topic created or not validate", "hghhhg test test"]
  },
  {
    "selectionId": 2,
    "topicName": ["topic test"]
  },
  {
    "selectionId": 3,
    "topicName": ["new topic for test topic name and description name(test check)"]
  },
  {
    "selectionId": 4,
    "topicName": ["topic check check", "topic check popup"]
  },
  {
    "selectionId": 5,
    "topicName": ["test"]
  }
]

方案2：兼容单个topicName为字符串类型 #

如果需要严格匹配你预期的结构，即单个值为字符串、多个值才为数组，可以用以下实现：

const result = Object.values(
  list.reduce((accumulator, current) => {
    const id = current.selectionId
    if (!accumulator[id]) {
      accumulator[id] = {
        selectionId: id,
        topicName: current.topicName
      }
    } else {
      if (!Array.isArray(accumulator[id].topicName)) {
        accumulator[id].topicName = [accumulator[id].topicName]
      }
      accumulator[id].topicName.push(current.topicName)
    }
    return accumulator
  }, {})
)

输出结果 #

[
  {
    "selectionId": 1,
    "topicName": ["topic created or not validate", "hghhhg test test"]
  },
  {
    "selectionId": 2,
    "topicName": "topic test"
  },
  {
    "selectionId": 3,
    "topicName": "new topic for test topic name and description name(test check)"
  },
  {
    "selectionId": 4,
    "topicName": ["topic check check", "topic check popup"]
  },
  {
    "selectionId": 5,
    "topicName": "test"
  }
]

内容的提问来源于stack exchange，提问作者nihar