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"
    }
  }
]

我想从嵌套中提取键/值的子集properties,同时保持外部对象的其他属性不变,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和之外的所有键NAME_LONG.

我确信必须有一个相当简单的方法来实现这一点与jq.帮助赞赏.

Answer 1

您可以使用此过滤器:

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

这会将properties对象过滤的数组中的每个项映射到仅包含NAME和NAME_LONG属性.

Answer 2

map(.properties |= {NAME, NAME_LONG}) 更直接，更容易理解。

我会将此作为评论添加到杰夫的答案中，但是关于评论的 SO 规则很愚蠢，所以它作为答案代替。