使用jq展平JSON文档

Question

我正在考虑以下JSON对象数组:

[
  {
    "index": "index1",
    "type": "type1",
    "id": "id1",
    "fields": {
      "deviceOs": [
        "Android"
      ],
      "deviceID": [
        "deviceID1"
      ],
      "type": [
        "type"
      ],
      "country": [
        "DE"
      ]
    }
  },
  {
    "index": "index2",
    "type": "type2",
    "id": "id2",
    "fields": {
      "deviceOs": [
        "Android"
      ],
      "deviceID": [
        "deviceID2"
      ],
      "type": [
        "type"
      ],
      "country": [
        "US"
      ]
    }
  }
]

我想弄平它得到:

[
  {
    "index": "index1",
    "type": "type",
    "id": "id1",
    "deviceOs": "Android",
    "deviceID": "deviceID1",
    "country": "DE"
  },
  {
    "index": "index2",
    "type": "type",
    "id": "id2",
    "deviceOs": "Android",
    "deviceID": "deviceID2",
    "country": "US"
  }
]

我正在尝试与之合作,jq但我没有将其弄平"fields".我该怎么办？目前我对命令行工具感兴趣,但我也对其他建议持开放态度.

Answer 1

这个是一个棘手的工艺.

map
(
    with_entries(select(.key != "fields"))
    +
    (.fields | with_entries(.value = .value[0]))
)

让我们分解并解释它的各个部分

1. 对于阵列中的每个项目......

   map(...)
   

2. 创建一个包含除fields属性之外的所有值的新对象.

   with_entries(select(.key != "fields"))
   

3. 结合......

   +
   

4. 每个fields值将每个值投影到每个阵列的第一项

   (.fields | with_entries(.value = .value[0]))
   

Answer 2

有一个名为的工具gron，您可以通过管道将该 json 导入，以获得类似的内容。

$ gron document.json
json = [];
json[0] = {};
json[0].fields = {};
json[0].fields.country = [];
json[0].fields.country[0] = "DE";
json[0].fields.deviceID = [];
json[0].fields.deviceID[0] = "deviceID1";
json[0].fields.deviceOs = [];
json[0].fields.deviceOs[0] = "Android";
json[0].fields.type = [];
json[0].fields.type[0] = "type";
json[0].id = "id1";
json[0].index = "index1";
json[0].type = "type1";
json[1] = {};
json[1].fields = {};
json[1].fields.country = [];
json[1].fields.country[0] = "US";
json[1].fields.deviceID = [];
json[1].fields.deviceID[0] = "deviceID2";
json[1].fields.deviceOs = [];
json[1].fields.deviceOs[0] = "Android";
json[1].fields.type = [];
json[1].fields.type[0] = "type";
json[1].id = "id2";
json[1].index = "index2";
json[1].type = "type2";

Answer 3

您可以使用以下过滤器：

[.[] | {index: .index, type: .type, id: .id, deviceOs: .fields.deviceOs[],deviceID: .fields.deviceID[],country: .fields.country[]}]

您可以在这里测试https://jqplay.org

Answer 4

以下是一些变体，首先使用 + 将 .fields 合并到包含对象中，然后展平数组元素。首先我们处理 .fields

  .[]
| . + .fields
| del(.fields)

这给我们留下了看起来像的物体

{
  "index": "index1",
  "type": [
    "type"
  ],
  "id": "id1",
  "deviceOs": [
    "Android"
  ],
  "deviceID": [
    "deviceID1"
  ],
  "country": [
    "DE"
  ]
}

然后我们可以通过多种方式展平按键。一种方法是使用with_entries

| with_entries( .value = if .value|type == "array" then .value[0] else .value end )

另一种方法是使用reduce和setpath

| . as $v
| reduce keys[] as $k (
    {};
    setpath([$k]; if $v[$k]|type != "array" then $v[$k] else $v[$k][0] end)
  )