CODEWITHSUNDEEP
jsonjq
x3nr0s
x3nr0s

Reputation: 2156

jq - How do I print a parent value of an object when I am already deep into the object's children?

Say I have the following JSON, stored in my variable jsonVariable.

{
    "id": 1,
    "details": {
        "username": "jamesbrown",
        "name": "James Brown"
    }
}

I parse this JSON with jq using the following:

echo $jsonVariable | jq '.details.name | select(.name == "James Brown")'

This would give me the output

James Brown

But what if I want to get the id of this person as well? Now, I'm aware this is a rough and simple example - the program I'm working with at the moment is 5 or 6 levels deep with many different JQ functions other than select. I need a way to select a parent's field when I am already 5 or 6 layers deep after carrying out various methods of filtering.

Can anyone help? Is there any way of 'going in reverse', back up to the parent? (Not sure if I'm making sense!)

Upvotes: 73

Views: 42350

Answers (3)

user3899165
user3899165

Reputation:

For a more generic approach, save the value of the "parent" element at the detail level you want, then pipe it at the end of your filter:

jq '. as $parent | .details.name | select(. == "James Brown") | $parent'

Of course, for the trivial case you expose, you could omit this entirely:

jq 'select(.details.name == "James Brown")'

Also, consider that if your selecting filters return many matches for a single parent object, you will receive a copy of the parent object for each match. You may wish to make sure your select filters only return one element at the parent level by wrapping all matches below parent level into an array, or to deduplicate the final result with unique.

Upvotes: 109

Jeff Mercado
Jeff Mercado

Reputation: 134891

Rather than querying up to the value you're testing for, query up to the root object that contains the value you're querying on and the values you wish to select.

You need the object that contains both the id and the name.

$ jq --arg name 'James Brown' 'select(.details.name == $name).id' input.json

Upvotes: 4

mrtig
mrtig

Reputation: 2267

Give this a shot: 

echo $jsonVariable | jq '{Name: .details.name, Id: .Id}  | select(.name == "James Brown")'

Upvotes: 4

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