json – 如何在Scala中执行嵌套groupBy

2023年4月12日 224次阅读

我有一个像这样的案例对象列表：

AppInfo(client = "client1", project = "project1", version = "version1")
AppInfo(client = "client1", project = "project1", version = "version2")
AppInfo(client = "client2", project = "project3", version = "version1")
AppInfo(client = "client2", project = "project4", version = "version1")

并且需要创建一个这样的嵌套结构：

Map(
  "clients" -> List(
    Map(
      "name" -> "client1",
      "projects" -> List(
        Map(
          "name" -> "project1",
          "versions" -> List(
            "version1",
            "version2"
          )
        )
      )
    ),
    Map(
      "name" -> "client2",
      "projects" -> List(
        Map(
          "name" -> "project3",
          "versions" -> List(
            "version1"
          )
        ),
        Map(
          "name" -> "project4",
          "versions" -> List(
            "version1"
          )
        )
      )
    )
  )
)

这看起来很糟糕,但它将序列化为这个非常简单的JSON：

{
  "clients": [
    {
      "name": "client1",
      "projects": [
        {
          "name": "project1",
          "versions": [
            "version1",
            "version2"
          ]
        }
      ]
    },
    {
      "name": "client2",
      "projects": [
        {
          "name": "project3",
          "versions": [
            "version1"
          ]
        },
        {
          "name": "project4",
          "versions": [
            "version1"
          ]
        }
      ]
    }
  ]
}

有没有合理的方法来做到这一点？现在我在列表中的groupBys中的地图中有地图.

编辑

像Clojure的Specter库可能会有所帮助.

最佳答案 怎么样：

for {
    (clientName, clientInfos) <- infoList.groupBy(_.client)
} yield {
    val clientProjects = clientInfos.groupBy(_.project)
    val projectSection = clientProjects.map { case(name, infos) => Map("name" -> name, "versions" -> infos.map(_.version)) }

    Map("name" -> clientName, "projects" -> projectSection)
}

它不会减少map或groupBy调用的数量,但这是我能够组织代码的最方便的方式.

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