Question

此处填充了subThings数组

$ echo '[{"thing-1" : "meta1","thing-2" : "meta2","thing-n" : "metan","subThings":[{"subThing-1" : "subMeta1","subThing-2" : "subMeta2","subThing-n" : "subMetan"}]}]' | jq '.'
[
  {
    "subThings": [
      {
        "subThing-n": "subMetan",
        "subThing-2": "subMeta2",
        "subThing-1": "subMeta1"
      }
    ],
    "thing-n": "metan",
    "thing-2": "meta2",
    "thing-1": "meta1"
  }
]

您可以在此处看到丢失的键只是设置为null。

$ echo '[{"thing-1" : "meta1","thing-2" : "meta2","thing-n" : "metan","subThings":[{"subThing-1" : "subMeta1","subThing-2" : "subMeta2","subThing-n" : "subMetan"}]}]' |
 jq '[.[] | {"thing-1","thing-5","subThing-2":.subThings[]["subThing-2"],}]'
[
  {
    "subThing-2": "subMeta2",
    "thing-5": null,
    "thing-1": "meta1"
  }
]

此处数组subThings为空

$ echo '[{"thing-1" : "meta1","thing-2" : "meta2","thing-n" : "metan","subThings":[]}]' | jq '.'
[
  {
    "subThings": [],
    "thing-n": "metan",
    "thing-2": "meta2",
    "thing-1": "meta1"
  }
]

而不是subThing-2的null，它将清除所有数据。

$ echo '[{"thing-1" : "meta1","thing-2" : "meta2","thing-n" : "metan","subThings":[]}]' |
  jq '[.[] | {"thing-1","thing-5","subThing-2":.subThings[]["subThing-2"],}]'
[]
$

为什么？
是否有其他方法可以获取数据，而无需空数组清除所有内容？

Answer 1

您正在尝试根据subThings对象的存在来生成对象。实际上，这就是.subThings[]在{"thing-1","thing-5","subThing-2":.subThings[]["subThing-2"]}的上下文中所做的。

您需要分别评估subThings并在没有null的情况下替换为empty，否则将只是map({ "thing-1", "thing-5", "subThing-2": (.subThings[]."subThing-2" // null) })而不会产生任何结果（这就是您在此处看到的内容））。

{{1}}

Answer 2

我仍然不知道我得到的json是否错误，或者这是否是jq中的错误。我发现此解决方案使用sed来使数组不为空。

$ echo '[{"thing-1" : "meta1","thing-2" : "meta2","thing-n" : "metan","subThings":[]}]' | sed 's/\[ *\]/[null]/g' | jq '[.[] | {"thing-1","thing-5","subThing-2":.subThings[]["subThing-2"],}]'
[
  {
    "thing-1": "meta1",
    "thing-5": null,
    "subThing-2": null
  }
]
$

如果嵌套数组为空，则jq不返回任何内容

2 个答案: