How do I remove all keys except one with jq?

Question

Given a list of objects, with many keys I don't want:

[{
    "name": "Alice",
    "group": "Admins",
    "created": "2014"
}, {
    "name": "Bob",
    "group": "Users",
    "created": "2014"
}]

How do I filter these objects to only include keys I want?

[{
    "name": "Alice"
}, {
    "name": "Bob"
}]

I've tried jq '.[].name' but that extracts the values, rather than preserving the objects.

Answer 1

you can use map with del if you know the keys you don't want:

jq 'map(del (.group) | del (.created))'

Answer 2

Another solution without the map function:

jq '[.[] | {name: .name}]'

Answer 3

The accepted answer (with map) and the equivalent answer by @mauricio-tranjano will, in effect, add the specified key to objects that don't already have it. If that's not the behavior you want, then consider using has(_), e.g.:

$ jq -c 'map( if has("a") then {a} else {} end )'

Input:

[{id:1,a:1}, {id:2}]

Output:

[{"a":1},{}]

Answer 4

You can use the map() function to filter any key:

jq 'map({name: .name})'

Update

Suggested by @WilfredHughes: The above filter can be abbreviated as follows:

jq 'map({name})'