json - 使用 jq 合并 2 个 JSON 对象

Question

我在 2 个变量 V1 和 V2 中有 2 个 json 对象。
V1 是

{
  "Parameters: {
     "a": "value-of-a",
     "b": "value-of-b"
  }
}

V2是

{
  "Parameters": {
     "a": "new-value-of-a",
     "c": "value-of-c"
  }
}

我想用 V2 的值覆盖 V1 的值。我想生成一个新的 JSON，如下所示:
预期

{
   "Parameters": {
      "a": "new-value-of-a",
      "b": "value-of-b"
   }
}

我尝试了以下方法:

(
echo '{ "Parameters": { "a": "value-of-a", "b": "value-of-b" } }'\
'{ "Parameters": { "a": "new-value-of-a", "c": "value-of-c" } }'\
| jq --slurp 'reduce .[] as $item ({}; . * $item)'
)

但这会产生
实际

{
   "Parameters": {
      "a": "new-value-of-a",
      "b": "value-of-b",
      "c": "value-of-c"
   }
}

问题是我不希望 V1 中不存在的 V2 中的节点出现在结果中。谁能帮我？

Answer 1

这是一个概念上简单且可扩展的无减少解决方案:

jq -n '
   # Emit the object with key-value pairs of $o2 
   # for keys that appear in both . and $o2
   def commonKeys($o2):
     keys_unsorted as $k1
     | ($o2|keys_unsorted) as $k2
     | [($k1 - ($k1-$k2))[] | {(.): $o2[.]}] | add
   ;
   input.Parameters as $dict
   | input
   | .Parameters |= (. + commonKeys($dict))

' v2.json v1.json