JQ：将单个对象分组为键

Question

我有以下数据：

[
  {
    "company.u_ats_region": "Region1",
    "hostname": "host1",
    "install_status": "1",
    "os": "Windows",
    "os_domain": "test.com"
  },
  {
    "company.u_ats_region": "Region2",
    "hostname": "host2",
    "install_status": "1",
    "os": "Windows",
    "os_domain": "test.com"
  },
  {
    "company.u_ats_region": "Region3",
    "hostname": "host3",
    "install_status": "7",
    "os": "Windows",
    "os_domain": "test.com"
  }
]

我一直在使用此查询

{count: length,
 regions: [group_by(."company.u_ats_region")[] | 
          {( [0]."company.u_ats_region"): [.[] | 
          {name: (.hostname+"."+.os_domain), 
      os: .os}]}]}

将数据转换为以下内容：

{
  "count": 3,
  "regions": [
    {
      "Region1": [
        {
          "name": "host1.test.com",
          "os": "Windows"
        }
      ]
    },
    {
      "Region2": [
        {
          "name": "host2.test.com",
          "os": "Windows"
        }
      ]
    },
    {
      "Region3": [
        {
          "name": "host3.test.com",
          "os": "Windows"
        }
      ]
    }
  ]
}

这接近我要实现的目标，但我希望“区域”成为一个对象，每个区域是一个对象这样的对象中的关键：

{
  "count": 3,
  "regions": {
      "Region1": [
        {
          "name": "host1.test.com",
          "os": "Windows"
        }
      ],
      "Region2": [
        {
          "name": "host2.test.com",
          "os": "Windows"
        }
      ],
      "Region3": [
        {
          "name": "host3.test.com",
          "os": "Windows"
        }
      ]
    }
}

我尝试使用“添加”来玩耍，但这仍然没有使我更接近我试图实现的结果。任何帮助都将受到赞赏！

Answer 1

使用键和value字段创建对象，然后使用from_entries是一种方法：

{
  count: length,
  regions: group_by(."company.u_ats_region")
  | map({
      key: .[0]."company.u_ats_region",
      value: map({name: "\(.hostname).\(.os_domain)", os})
    })
  | from_entries
}

{
  "count": 3,
  "regions": {
    "Region1": [
      {
        "name": "host1.test.com",
        "os": "Windows"
      }
    ],
    "Region2": [
      {
        "name": "host2.test.com",
        "os": "Windows"
      }
    ],
    "Region3": [
      {
        "name": "host3.test.com",
        "os": "Windows"
      }
    ]
  }
}

demo

Answer 2

您可以定义a 自定义函数执行分组，然后将结果转换为tranfsorm。使用功能避免重复选择器：

def group_to_obj(f):
  group_by(f) | map({key:first|f, value:.}) | from_entries;

{
  count: length,
  regions: group_to_obj(."company.u_ats_region")
    | map_values(map({name: "\(.hostname).\(.os_domain)", os}))
}

输出：

{
  "count": 3,
  "regions": {
    "Region1": [
      {
        "name": "host1.test.com",
        "os": "Windows"
      }
    ],
    "Region2": [
      {
        "name": "host2.test.com",
        "os": "Windows"
      }
    ],
    "Region3": [
      {
        "name": "host3.test.com",
        "os": "Windows"
      }
    ]
  }
}

Answer 3

使用降低迭代构建数组是另一种方法：

{
  count: length,
  regions: (
    reduce .[] as $i ({};
      .[$i."company.u_ats_region"] += [$i | {name: "\(.hostname).\(.os_domain)", os}]
    )
  )
}

{
  "count": 3,
  "regions": {
    "Region1": [
      {
        "name": "host1.test.com",
        "os": "Windows"
      }
    ],
    "Region2": [
      {
        "name": "host2.test.com",
        "os": "Windows"
      }
    ],
    "Region3": [
      {
        "name": "host3.test.com",
        "os": "Windows"
      }
    ]
  }
}

demo