使用JQ交換嵌套對象中的鍵

Question

使用jq ，如何轉換：

{ "a": {"b": 0}, "c": {"d": 1}}

成：

{"b": {"a": 0}, "d": {"c": 1}}

不知道源中鍵的名稱？

（我知道這通常會丟失數據，但不會丟失我的數據）

Answer 1

這是使用with_entries的替代方法：

with_entries(.key as $parent
  | (.value|keys[0]) as $child
  | {
        key: $child,
        value: { ($parent): .value[$child] }
    }
)

Answer 2

def swapper:
  . as $in
  | reduce keys[] as $key
      ( {}; . + ( $in[$key] as $o
                  | ($o|keys[0]) as $innerkey
                  | { ($innerkey): { ($key): $o[$innerkey] } } ) ) ;

例：

{ "a": {"b": 0}, "c": {"d": 1}} | swapper

生產：

{"b":{"a":0},"d":{"c":1}}

Answer 3

這是使用jq流和變量的解決方案：

[
  . as $d
| keys[]
| $d[.] as $v
| ($v|keys[]) as $vkeys
| {
    ($vkeys): {
       (.): ($vkeys|$v[.])
    }
  }
] | add

很容易忘記最后的內容，因此更清楚地了解正在發生的事情是帶有其他注釋和變量的稍微擴展的版本。

[
  . as $d                    # $d:     {"a":{"b":0},"c":{"d": 1}}
| keys[] | . as $k           # $k:     "a", "c"
| $d[$k] as $v               # $v:     {"b": 0}, {"d": 1}
| ($v|keys[]) as $vkeys      # $vkeys: "b", "d"
| ($vkeys|$v[.]) as $vv      # $vv:    0, 1
| {
    ($vkeys): {              # "b": {     "d": {
       ($k): $vv             #   "a": 0      "c": 1
    }                        # }        , }
  }
] | add