使用jq展平JSON文档 [英] Flatten a JSON document using jq

问题描述

我正在考虑以下JSON对象数组:

I'm considering the following array of JSON objects:

[
  {
    "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"
      ]
    }
  }
]

我想把它弄平以获得:

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

我正在尝试使用jq，但是我无法展平"fields".我该怎么办?目前，我对命令行工具很感兴趣，但是我也欢迎其他建议.

I'm trying to work with jq but I fail to flatten the "fields". How should I do it? At the moment I'm interested in command-line tools, but I'm open to other suggestions as well.

推荐答案

这是一个棘手的问题.

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

我们将其分解并解释其内容

Let's break it down and explain the bits of it

1. 对于数组中的每个项目...

1. For every item in the array...

map(...)

创建一个新对象，其中包含除fields属性以外的所有值.

Create a new object containing the values for all except the fields property.

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

用...组合……

Combine that with...

+

每个fields都将每个值投影到每个数组的第一项

Each of the fields projecting each of the values to the first item of each array

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

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