jq：根据对象值条件递归删除对象的最简单方法

Question

我想使用 jq 删除 JSON "object" 中的所有词典（我通常使用该术语来指代数组或词典）

a) 包含一个名为 "delete_me" 的键，并且 b) 键 "delete_me" 满足某些预定条件（空、非零、真等）

基本上，我要实现的逻辑是：遍历输入，在每个节点处，如果该节点不是数组或对象，则保留它并继续前进，否则保留它但从中移除条件 a) 或 b) 失败的任何子字典。

有什么建议吗？

示例输入：

{
  "a": { "foo": "bar" },
  "b": {
    "i": {
      "A": {
        "i": [
          {
            "foo": {},
            "bar": {
              "delete_if_this_is_null": false,
              "an_array": [],
              "another_array": [
                {
                    "delete_if_this_is_null": null,
                    "foo": "bar"
                }
              ],
              "etc": ""
            },
            "foo2": "s"
          },
          {
            "foo": {
              "an_array": [
                {
                  "delete_if_this_is_null": "ok",
                  "foo":"bar",
                  "another_object": { "a":1 }
                },
                {
                  "delete_if_this_is_null": null,
                  "foo2":"bar2",
                  "another_object": { "a":1 },
                  "name": null
                }
              ],
              "an_object": {
                "delete_if_this_is_null":null,
                "foo3":"bar3"
              }
            },
            "zero": 0,
            "b": "b"
          }
        ]
      }
    }
  }
}

应该屈服，如果"delete_me"键是delete_if_this_is_null并且预定条件是delete_if_this_is_null == null：

{
  "a": { "foo": "bar" },
  "b": {
    "i": {
      "A": {
        "i": [
          {
            "foo": {},
            "bar": {
              "delete_if_this_is_null": false,
              "an_array": [],
              "another_array": [],
              "etc": ""
            },
            "foo2": "s"
          },
          {
            "foo": {
              "an_array": [
                {
                  "delete_if_this_is_null": "ok",
                  "foo":"bar",
                  "another_object": { "a":1 }
                }
              ]
            },
            "zero": 0,
            "b": "b"
          }
        ]
      }
    }
  }
}

更新：这是解决方案：假设输入在文件 'input.json':

中

jq 'def walk(f):
  . as $in
  | if type == "object" then
      reduce keys[] as $key
        ( {}; . + { ($key):  ($in[$key] | walk(f)) } ) | f
  elif type == "array" then map( walk(f) ) | f
  else f
  end;

def mapper(f):
  if type == "array" then map(f)
  elif type == "object" then
  . as $in
  | reduce keys[] as $key
      ({};
       [$in[$key] | f ] as $value
       | if $value | length == 0 then .
         else . + {($key): $value[0]} end)
  else .
  end;

walk( mapper(select((type == "object" and .delete_if_this_is_null == null) | not)) )' < input.json

Answer 1

我不确定你在问题中到底想完成什么，但我假设你想递归搜索 json 响应并删除 json 满足某些条件的对象条件。

您可以借助 walk 过滤器轻松完成此操作，该过滤器将在 jq 的未来版本中出现，请参阅 source.

中的实现

# Apply f to composite entities recursively, and to atoms
def walk(f):
  . as $in
  | if type == "object" then
      reduce keys[] as $key
        ( {}; . + { ($key):  ($in[$key] | walk(f)) } ) | f
  elif type == "array" then map( walk(f) ) | f
  else f
  end;

有了它，您可以像这样过滤掉它们：

def filter_objects(predicate): # removes objects that satisfies some predicate
    walk(
        if (type == "object") and (predicate) then
            empty
        else
            .
        end
    )
    ;
filter_objects(.delete_me) # remove objects that has a truthy property "delete_me"

Answer 2

Jeff 的解决方案可能耗电太多。例如，使用：

def data: [1,2, {"hello": {"delete_me": true, "a":3 }, "there": 4} ]; ];

Jeff 的解决方案是空的（即什么都没有）。

因此，以下内容可能更接近您要查找的内容：

walk(if (type == "object" and .delete_me) then del(.) else . end )

对于 data，这会产生：

[1,2,{"hello":null,"there":4}]

备选方案

如果需要消除上述示例中的 "hello":null 的解决方案，则需要 jq 的 map_values/1 的变体。这是一种方法：

def mapper(f):
  if type == "array" then map(f)
  elif type == "object" then
  . as $in
  | reduce keys[] as $key
      ({};
       [$in[$key] | f ] as $value
       | if $value | length == 0 then . 
         else . + {($key): $value[0]} end)
  else .
  end;

data | walk( mapper(select((type == "object" and .delete_me) | not)) )

结果是：

[1,2,{"there":4}]

Answer 3

这是一个使用递归函数的解决方案：

def clean(condition):
  if type == "object" then
    if condition
    then empty
    else
    with_entries(
      if (.value|type) == "object" and (.value|condition)
      then empty
      else .value |= clean(condition)
      end
    )
    end
  elif type == "array" then
    map(
      if type == "object" and condition
      then empty
      else clean(condition)
      end
    )
  else .
  end
;

clean(
  has("delete_if_this_is_null") and (.delete_if_this_is_null == null)
)

jq：根据对象值条件递归删除对象的最简单方法

jq: easiest way to recursively remove objects based on object value condition

json

jq

备选方案