使用 jq 从嵌套的 JSON 对象中提取选定的属性

Question

给定一个 JSON 对象数组：

[
  {
    "geometry": {
      "type": "Polygon",
      "coordinates": [[[-69.9969376289999, 12.577582098000036]]]
    },
    "type": "Feature",
    "properties": {
      "NAME": "Aruba",
      "WB_A2": "AW",
      "INCOME_GRP": "2. High income: nonOECD",
      "SOV_A3": "NL1",
      "CONTINENT": "North America",
      "NOTE_ADM0": "Neth.",
      "BRK_A3": "ABW",
      "TYPE": "Country",
      "NAME_LONG": "Aruba"
    }
  },
  {
    "geometry": {
      "type": "MultiPolygon",
      "coordinates": [[[-63.037668423999946, 18.212958075000028]]]
    },
    "type": "Feature",
    "properties": {
      "NAME": "Anguilla",
      "WB_A2": "-99",
      "INCOME_GRP": "3. Upper middle income",
      "SOV_A3": "GB1",
      "NOTE_ADM0": "U.K.",
      "BRK_A3": "AIA",
      "TYPE": "Dependency",
      "NAME_LONG": "Anguilla"
    }
  }
]

我想从嵌套的 properties 中提取 key/values 的一个子集，同时保持外部对象的其他属性完好无损，生成如下内容：

[
  {
    "geometry": {
      "type": "Polygon",
      "coordinates": [[[-69.9969376289999, 12.577582098000036]]]
    },
    "type": "Feature",
    "properties": {
      "NAME": "Aruba",
      "NAME_LONG": "Aruba"
    }
  },
  {
    "geometry": {
      "type": "MultiPolygon",
      "coordinates": [[[-63.037668423999946, 18.212958075000028]]]
    },
    "type": "Feature",
    "properties": {
      "NAME": "Anguilla",
      "NAME_LONG": "Anguilla"
    }
  }
]

即删除除 NAME 和 NAME_LONG.

之外的所有键

我相信一定有一种相当简单的方法可以用 jq 实现这一点。帮助表示赞赏。

Answer 1

您可以使用这个过滤器：

map(
    .properties |= with_entries(select(.key == ("NAME", "NAME_LONG")))
)

这会映射数组中的每个项目，其中 properties 对象被过滤为仅包含 NAME 和 NAME_LONG 属性。

Answer 2

map(.properties |= {NAME, NAME_LONG})更直接易懂。

我会将其添加为对 Jeff 的回答的评论，但关于评论的 SO 规则很愚蠢，所以它改为作为答案。

使用 jq 从嵌套的 JSON 对象中提取选定的属性

Extracting selected properties from a nested JSON object with jq

json

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