使用jq从嵌套的JSON对象中提取所选属性

Question

给定一个如下的对象数组：

[
  {
    "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"
    }
  }
]

我希望从嵌套的属性(properties)中提取一部分键/值，同时保留外层对象的其他属性，生成类似以下内容:

[
  {
    "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"
    }
  }
]

即仅保留NAME和NAME_LONG键，删除所有其他键。我相信使用jq可以很容易实现这一点。感谢您的帮助。

- Rob Cowie

Answer 1

您可以使用此过滤器：

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

这将映射数组中每个项目，其中properties对象被过滤，只包括NAME和NAME_LONG属性。

- Jeff Mercado

---

有没有办法将键的名称作为变量传递给 jq？还是构建==右侧需要使用插值来完成？ - GcL

Answer 2

map(.properties |= {NAME, NAME_LONG}) 更加直观易懂。

我本来想将这个作为对 Jeff 回答的注释，但由于 Stack Overflow 对注释的规定太愚蠢了，所以只好将其作为回答发布。

- user4011114