JQ combine array of json maps based on keys

Question

My json looks like the following

[
  {
    "100": "ONE",
    "200": "TWO"
  },
  {
    "100": "1",
    "200": "2"
  }
]

and I am expecting the following output

{
  "1": "ONE",
  "2": "TWO"
}

I found few answers here but all have static keys but in my case keys are dynamic

Another example

{
  "apiVersion": "v1",
  "data": {
    "bac6f56f-101c-26da-edfa-c08e6622a337": "1"
  },
  "kind": "ConfigMap",
  "metadata": {
    "annotations": {
      "bac6f56f-101c-26da-edfa-c08e6622a337": "restart"
    },
    "creationTimestamp": "2020-06-25T14:53:06Z",
    "uid": "7b1dfc3a-1357-400e-b750-a1ff98a204b9"
  }
}

and the expected output is

{"restart":"1"}

@Inian added another example and it is an output of kubectl get configmap — Sanchu Varkey
– Sanchu Varkey, Commented Jun 24, 2021 at 10:20

oguz ismail · Accepted Answer · 2021-06-24 11:03:48Z

2

Here is one way:

. as [$V, $K]
| reduce ($K | keys_unsorted)[] as $k ({};
  . + {($K[$k]): $V[$k]}
)

This iterates over the second object's keys, retrieves values associated with each key in both objects, and pairs them in a new one. And can be adapted for your second example by changing

. as [$V, $K]

to

. as {data: $V, metadata: {annotations: $K}}

^{Online demo}

edited Jun 24, 2021 at 11:03

answered Jun 24, 2021 at 10:07

oguz ismail

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Comments

peak · Accepted Answer · 2021-06-25 08:20:03Z

1

Here's a reduce-free solution to the first problem that hopefully also elucidates the general approach:

.[0] as $dict
| .[1]
| with_entries( {value: $dict[.key], key: (.value|tostring) } )

Solution to second problem

Using the above approach, we have only to tweak the first two lines:

.data as $dict
| .metadata.annotations
| with_entries( {value: $dict[.key], key: (.value|tostring) } )

edited Jun 25, 2021 at 8:20

answered Jun 24, 2021 at 21:16

peak

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6 Comments

Sanchu Varkey Over a year ago

if I want just a pipe-separated string as output like restart|1 for the second problem, how would you rewrite it. Is it possible?

peak Over a year ago

Yes, of course. One way would be to create the appropriate array(s) and then use join("|").

Sanchu Varkey Over a year ago

I piped this to your code | keys[] as $k | "\($k)|\(.[$k])" and it works

peak Over a year ago

Where there's one way, there's usually another :-)

Sanchu Varkey Over a year ago

How would you write your way create the appropriate array(s) and then use join("|") ?

|

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