需要使用JQ查找键值对并替换JSON中的键值对

Question

我有这个 JSON

{
  "firstName": "Rajesh",
  "lastName": "Kumar",
  "gender": "man",
  "age": 24,
  "address": {
    "streetAddress": "126 Udhna",
    "city": "Surat",
    "state": "WB",
    "postalCode": "394221"
  },
  "phoneNumbers": [
    {
      "type": "home",
      "number": "7383627627"
    }
  ]
}

我需要使用 JQ 找到“状态”键的值并替换 JSON 中的值。我不想通过提供键的位置来获取它，喜欢

firstName=$(cat sample-json.json | jq -r '.firstName')

我的预期输出

{
  "firstName": "Rajesh",
  "lastName": "Kumar",
  "gender": "man",
  "age": 24,
  "address": {
    "streetAddress": "126 Udhna",
    "city": "Surat",
    "state": "Bihar",
    "postalCode": "394221"
  },
  "phoneNumbers": [
    {
      "type": "home",
      "number": "7383627627"
    }
  ]
}

Answer 1

如果您愿意指定 .address：

jq '.address.state = "Bihar"' sample-json.json

否则：

jq 'walk(if type == "object" and has("state") then .state = "Bihar" else . end)' sample-json.json

最后一个将替换所有.state 值。如果您只想替换第一个匹配项：

jq 'first(..|objects|select(has("state"))).state = "Bihar"' sample-json.json

等等。如果您能明确要求，这将真正帮助所有相关人员。