Question

实际上，密钥必须在JSON对象中是唯一的（例如Does JSON syntax allow duplicate keys in an object?）。但是，假设我有一个包含以下内容的文件：

{
    "a" : "1",
    "b" : "2",
    "a" : "3"
}

是否有将重复键转换为数组的简单方法？这样文件就变成了：

{
    "a" : [ {"key": "1"}, {"key": "3"}],
    "b" : "2"
}

或类似的东西，但它将重复的键组合成一个数组（或查找和替代方法来提取重复的键值）。

以下是Java的解决方案：Convert JSON object with duplicate keys to JSON array

有没有办法用awk / bash / python做到这一点？

Answer 1

如果您的输入实际上是一个平面JSON对象，其基元为值，则应该可以：

jq -s --stream 'group_by(.[0]) | map({"key": .[0][0][0], "value": map(.[1])}) | from_entries'

{
  "a": [
    "1",
    "3"
  ],
  "b": [
    "2"
  ]
}

对于更复杂的输出，这需要实际了解--stream应该如何使用，这超出了我的范围。

Answer 2

使用-s --stream建立圣地亚哥的答案，以下过滤器一次一步地构建对象，从而保留键的顺序和特定键的值：

reduce (.[] | select(length==2)) as $kv ({};
      $kv[0][0] as $k
      |$kv[1] as $v
      | (.[$k]|type) as $t
      | if $t == "null" then .[$k] = $v
        elif $t == "array" then .[$k] += [$v]
        else .[$k] = [ .[$k], $v ]
        end)

对于给定的输入，结果是：

{
  "a": [
    "1",
    "3"
  ],
  "b": "2"
}

为了说明保留每个键的值的顺序，请考虑以下输入：

{
    "c" : "C",
    "a" : "1",
    "b" : "2",
    "a" : "3",
    "b" : "1"
}

上述过滤器产生的输出是：

{
  "c": "C",
  "a": [
    "1",
    "3"
  ],
  "b": [
    "2",
    "1"
  ]
}

Answer 3

在峰值答案的基础上，以下过滤器适用于多个对象输入，适用于嵌套对象，适用于无法使用slurp-Option （ -s ）。

这不是最初问题的答案，但因为这里的jq-FAQ链接可能对某些访问者有用

文件 jqmergekeys.txt

def consumestream($arr): # Reads stream elements from stdin until we have enough elements to build one object and returns them as array
input as $inp 
| if $inp|has(1) then consumestream($arr+[$inp]) # input=keyvalue pair => Add to array and consume more
  elif ($inp[0]|has(1)) then consumestream($arr) # input=closing subkey => Skip and consume more
  else $arr end; # input=closing root object => return array

def convert2obj($stream): # Converts an object in stream notation into an object, and merges the values of duplicate keys into arrays
reduce ($stream[]) as $kv ({}; # This function is based on http://stackoverflow.com/a/36974355/2606757
      $kv[0] as $k
      | $kv[1] as $v
      | (getpath($k)|type) as $t # type of existing value under the given key
      | if $t == "null" then setpath($k;$v) # value not existing => set value
        elif $t == "array" then setpath($k; getpath($k) + [$v] ) # value is already an array => add value to array
        else setpath($k; [getpath($k), $v ]) # single value => put existing and new value into an array
        end);

def mainloop(f):  (convert2obj(consumestream([input]))|f),mainloop(f); # Consumes streams forever, converts them into an object and applies the user provided filter
def mergeduplicates(f): try mainloop(f) catch if .=="break" then empty else error end; # Catches the "break" thrown by jq if there's no more input

#---------------- User code below --------------------------    

mergeduplicates(.) # merge duplicate keys in input, without any additional filters

#mergeduplicates(select(.layers)|.layers.frame) # merge duplicate keys in input and apply some filter afterwards

示例：

tshark -T ek | jq -nc --stream -f ./jqmergekeys.txt

Answer 4

这是一个可以很好地概括的简单选择：

reshape.jq

def augmentpath($path; $value):
  getpath($path) as $v
  | setpath($path; $v + [$value]);

reduce (inputs | select(length==2)) as $pv
  ({}; augmentpath($pv[0]; $pv[1]) )

用法

jq -n -f reshape.jq input.json

输出

使用给定的输入：

{
  "a": [
    "1",
    "3"
  ],
  "b": [
    "2"
  ]
}

后记

如果避免单例数组很重要，则可以修改augmentpath的def或添加后处理步骤。

JSON字段具有相同的名称

4 个答案:

reshape.jq

用法

输出

后记