Select elements before nth element with jq

Question

I'm stuck on the following json with jq. The elements below are sorted descending by some timestamp (not included in the json). I need to select the ids before id X. E.g. select ids before id 1 should return 2, 3 and 5.

[
  {
    "id": 2,
    "somekey": "somevalue"
  },
  {
    "id": 3,
    "somekey": "somevalue"
  },
  {
    "id": 5,
    "somekey": "somevalue"
  },
  {
    "id": 1,
    "somekey": "somevalue"
  },
  {
    "id": 4,
    "somekey": "somevalue"
  }
]

Any idea how to do this in a one-liner with jq? Specifically the "select elements before" part.

peak · Accepted Answer · 2017-12-30T21:14:18.753

2

Short and simple:

.[0: map(.id) | index(1)]

Fancy but fast:

label $top | .[] | if .id == 1 then break $top else . end

edited Dec 30 '17 at 21:14

answered Dec 30 '17 at 20:22

peak

105,803
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If you want to include the seeked object, just increment index: `.[0: map(.id) | index(1)+1]` – poka Oct 02 '20 at 21:38

score 0 · Answer 2 · answered Dec 30 '17 at 18:35

0

I figured out this query:

[.[].id] | to_entries | .[0:map(select(.value==1))[].key][].value

If you know a less verbose way to achieve the same, let me know!

answered Dec 30 '17 at 18:35

Bas Harenslak

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Select elements before nth element with jq

2 Answers2