I'd like to print the value of members and id from below jq output:

$ cat test_|jq -r '.[] | select(.name=="AAA") | .'
{ "name": "AAA", "members": 10, "profiles": 0, "templates": 0, "ldapGroups": 0, "ldapMembers": 0, "id": "20"
}

Unfortunately it works for one of each only:

$ cat test_|jq -r '.[] | select(.name=="AAA") | .members'
10

with +" "+ I get error:

$ cat test_|jq -r '.[] | select(.name=="AAA") | .members+" "+.id'
jq: error: number and string cannot be added

1

4 Answers

I recommend to use string interpolation. It will automatically cast input to a string if necessary:

jq -r '.[]|select(.name=="AAA")|"\(.members) \(.id)"' file.json

Convert number to string with tostring function:

jq -r '.[] | select(.name=="AAA") | (.members|tostring) +" "+ .id test'

You have not specified precisely how you want the two values to appear, so it is worth pointing out that you could write:

.... | (.members, .id)

or

 .... | [.members, .id]

or

.... | [.members, .id] | E

where E could be @csv or @tsv or etc. In jq 1.5, join/1 will also do the type conversion.

1

You could start by just projecting the members of interest. e.g

$ jq -Mc '.[] | select(.name=="AAA") | {members,id}' data.json
{"members":10,"id":"20"}

Now you can see .members is a number and .id is a string. You can't add them directly with + but you can choose any of the options explained in the other answers.

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