CODEWITHSUNDEEP
jsonjq
Leo Gallucci
Leo Gallucci

Reputation: 16722

Exclude column from jq json output

I would like to get rid of the timestamp field here using jq JSON processor.

[
  {
    "timestamp": 1448369447295,
    "group": "employees",
    "uid": "elgalu"
  },
  {
    "timestamp": 1448369447296,
    "group": "employees",
    "uid": "mike"
  },
  {
    "timestamp": 1448369786667,
    "group": "services",
    "uid": "pacts"
  }
]

White listing would also works for me, i.e. select uid, group

Ultimately what I would really like is a list with unique values like this:

employees,elgalu
employees,mike
services,pacts

Upvotes: 80

Views: 52275

Answers (3)

pedr0
pedr0

Reputation: 7

Sed is your best friend - I can't think of anything simpler. I've got here having the same problem as the question's author - but maybe this is a simpler answer to the same problem:

< file sed -e '/timestamp/d'

Upvotes: -4

peak
peak

Reputation: 116780

For the record, an alternative would be:

$ jq -r '.[] | "\(.uid),\(.group)"' input.json

(The white-listing approach makes it easy to rearrange the order, and this variant makes it easy to modify the spacing, etc.)

The following example may be of interest to anyone who wants safe CSV (i.e. even if the values have embedded commas or newline characters):

$ jq -r '.[] | [.uid, .group] | @csv' input.json
"elgalu","employees"
"mike","employees"
"pacts","services"

Upvotes: 7

hek2mgl
hek2mgl

Reputation: 158040

If you just want to delete the timestamps you can use the del() function:

jq 'del(.[].timestamp)' input.json

However to achieve the desired output, I would not use the del() function. Since you know which fields should appear in output, you can simply populate an array with group and id and then use the join() function:

jq -r '.[]|[.group,.uid]|join(",")' input.json

-r stands for raw ouput. jq will not print quotes around the values.

Output:

employees,elgalu
employees,mike
services,pacts

Upvotes: 140

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