Flatten array of key-value pairs into object

Question

I've got very little experience with JSON and jq and am struggling to reformat my JSON objects. Each object contains a couple of key/value pairs and an array of objects. I'm trying to resturcture each object so that it's just a list of key/value pairs (i.e. extract the objects in the array).

[
   {
      "key1":"value1",
      "key2":"value2",
      "key 3":[
         {
            "a":"key_to_be_extracted_1",
            "v":"value_to_be_extracted_1"
         },
         {
            "a":"key_to_be_extracted_2",
            "v":"value_to_be_extracted_2"
         }
      ]
   },
   {
      "key1":"value1",
      "key2":"value2",
      "key 3":[
         {
            "a":"key_to_be_extracted_1",
            "v":"value_to_be_extracted_1"
         },
         {
            "a":"key_to_be_extracted_2",
            "v":"value_to_be_extracted_2"
         }
      ]
   }
]

What I'm trying to get to is:

[
   {
      "key1":"value1",
      "key2":"value2",
      "key_to_be_extracted_1":"value_to_be_extracted_1",
      "key_to_be_extracted_2":"value_to_be_extracted_2"
   },
   {
      "key1":"value1",
      "key2":"value2",
      "key_to_be_extracted_1":"value_to_be_extracted_1",
      "key_to_be_extracted_2":"value_to_be_extracted_2"
   }
]

Any help with this would be amazing!!

Botje · Accepted Answer · 2019-06-21 08:44:33Z

1

jq has a from_entries function that expects an array of {key, value} objects and produces an object. Thus, you can transform your "key 3" value into such an array and produce an object like so:

."key 3" | map({key: .a, value: .v} | from_entries

Putting that together with a map that just copies the two other keys:

map({key1, key2} + (."key 3" | map({key: .a, value: .v}) | from_entries))

answered Jun 21, 2019 at 8:44

Botje

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Comments

peak · Accepted Answer · 2019-06-21 09:55:52Z

0

If you want to avoid having to specify the keys to retain, you could go with:

map( .
     + (."key 3" | map( {(.a): .v} ) | add )
     | del(."key 3") )

That is, add in the new key-value pairs, and then delete "key 3". Notice also that add has been used here instead of from_entries, making for a more succinct and perhaps more straightforward solution.

edited Jun 21, 2019 at 9:55

answered Jun 21, 2019 at 9:48

peak

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