在深度嵌套的JSON中为特定的键名称选择特定的所有键路径和值

debugcn 发表于 Dev

用户名

我有一个非常大的JSON模式，该模式是半结构化的，嵌套的单元结构用键“ ObjectType”标记，以指示一些有意义的信息。

我正在尝试使用jq选择所有这些ObjectType及其值的路径。

json的示例部分是：

{
    "ObjectType": "ClassZ",
    "LastModifiedBy": "janeroe",
    "Name": "Anonymous",
    "ArrayProps": [],
    "Logistics": [
      {
        "ObjectType": "ClassA",
        "Source": "Vendor",
        "UUID": "x868-dhibye9-7678-12",
        "EffectiveDate": "2020-01-01",
        "Active": true,
        "Preferred": 0
      }
    ],
    "IsVirtual": true,
    "Convention": 3,
    "CruiseParams": [
      {
        "ObjectType": "ClassB",
        "Destinaton": "Atlantis",
        "Value": "3"
      }
    ],
    "InvolvedParties": [],
    "PartyEvents": [
      {
        "ObjectType": "ClassC",
        "CreatedDate": "2020-01-01",
        "CreatedBy": "johndoe"
      }
    ],
    "FunFactors": [
      {
        "ObjectType": "ClassD",
        "Level": 1
      }
    ]
  }

尝试输出类似或接近的内容：

"ObjectType":                "ClassZ"
"Logistics/0/ObjectType":    "ClassA"
"CruiseParams/0/ObjectType": "ClassB"
"PartyEvents/0/ObjectType":  "ClassC"
"FunFactors/0/ObjectType":   "ClassD"

钢铁司机

基于如何使用jq获取找到的值的索引路径，这是我能想到的最好的方法。

$ jq -rc 'paths as $p | select($p[-1] == "ObjectType") | "\($p|@csv): \"\(getpath($p))\""' sample.json
"ObjectType": "ClassZ"
"Logistics",0,"ObjectType": "ClassA"
"CruiseParams",0,"ObjectType": "ClassB"
"PartyEvents",0,"ObjectType": "ClassC"
"FunFactors",0,"ObjectType": "ClassD"

引号和定界符与您要求的不完全相同-使用map(tostring)| join(“/“)注释中的建议，它变成了

$ jq -rc 'paths as $p | select($p[-1] == "ObjectType") | "\"\($p|map(tostring)|join("/"))\": \"\(getpath($p))\""' sample.json
"ObjectType": "ClassZ"
"Logistics/0/ObjectType": "ClassA"
"CruiseParams/0/ObjectType": "ClassB"
"PartyEvents/0/ObjectType": "ClassC"
"FunFactors/0/ObjectType": "ClassD"

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