小编典典

使用jq从嵌套的JSON对象中提取所选属性

json

因此,给定对象的JSON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nested中提取键/值的子集properties,同时保持外部对象的其他属性不变,产生如下结果:

[
  {
    "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和以外的所有键NAME_LONG

我确信必须有一种使用jq实现此目标的合理简便的方法。帮助表示赞赏。


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2020-07-27

共1个答案

小编典典

您可以使用以下过滤器:

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

这会映射数组中properties过滤对象的每个项目,仅包括NAMENAME_LONG属性。

2020-07-27