json - 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，因此:

[
  {
    "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 中提取键/值的子集，同时保持外部对象的其他属性完整无缺，产生如下内容:

[
  {
    "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 之外的所有键和 NAME_LONG .

我相信一定有一种相当简单的方法可以用 jq 来实现这一点。帮助表示赞赏。

Answer 1

您可以使用此过滤器:

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

这将映射数组中的每一项，其中 properties对象被过滤为仅包含 NAME和 NAME_LONG特性。