Using jq to parse and display multiple fields in a json serially

Question

I have this Json

{
    "users": [
        {
            "first": "Stevie",
            "last": "Wonder"
        },
        {
            "first": "Michael",
            "last": "Jackson"
        }
    ]
}

Using jq I'd like to display first and last name serially. Like so -

Stevie Wonder
Michael Jackson

This is how far I have gotten -

jq '.users[].first, .users[].last'

But it displays

"Stevie"
"Michael"
"Wonder"
"Jackson"

Notice the following:

1. The double quotes that I do not want.
2. The carriage return that I do not want.
3. It's jumbled up.My query displays all the first names first, and then all the last names. However, I want first-last, first-last pair.

Answer 1

I recommend using String Interpolation:

jq '.users[] | "\(.first) \(.last)"'

reference

Answer 2

You can use addition to concatenate strings.

Strings are added by being joined into a larger string.

jq '.users[] | .first + " " + .last'

The above works when both first and last are string. If you are extracting different datatypes(number and string), then we need to convert to equivalent types. Referring to solution on this question . For example.

jq '.users[] | .first + " " + (.number|tostring)'

Answer 3

jq '.users[]|.first,.last' | paste - -

Answer 4

While both of the above answers work well if key,value are strings, I had a situation to append a string and integer (jq errors using the above expressions)

Requirement: To construct a url out below json

pradeep@seleniumframework>curl http://192.168.99.103:8500/v1/catalog/service/apache-443 | jq .[0]
  % Total    % Received % Xferd  Average Speed   Time    Time     Time  Current
                                 Dload  Upload   Total   Spent    Left  Speed
100   251  100   251    0     0   155k      0 --:--:-- --:--:-- --:--:--  245k
{
  "Node": "myconsul",
  "Address": "192.168.99.103",
  "ServiceID": "4ce41e90ede4:compassionate_wozniak:443",
  "ServiceName": "apache-443",
  "ServiceTags": [],
  "ServiceAddress": "",
  "ServicePort": 1443,
  "ServiceEnableTagOverride": false,
  "CreateIndex": 45,
  "ModifyIndex": 45
}

Solution:

curl http://192.168.99.103:8500/v1/catalog/service/apache-443 |
jq '.[0] | "http://" + .Address + ":" + "\(.ServicePort)"'

Answer 5

This will produce an array of names

> jq '[ .users[] | (.first + " " + .last) ]' ~/test.json

[
  "Stevie Wonder",
  "Michael Jackson"
]

Answer 6

my approach will be (your json example is not well formed.. guess thats only a sample)

jq '.Front[] | [.Name,.Out,.In,.Groups] | join("|")'  front.json  > output.txt

returns something like this

"new.domain.com-80|8.8.8.8|192.168.2.2:80|192.168.3.29:80 192.168.3.30:80"
"new.domain.com -443|8.8.8.8|192.168.2.2:443|192.168.3.29:443 192.168.3.30:443"

and grep the output with regular expression.

Answer 7

I got pretty close to what I wanted by doing something like this

cat my.json | jq '.my.prefix[] | .primary_key + ":", (.sub.prefix[] | "    - " + .sub_key)' | tr -d '"' 

The output of which is close enough to yaml for me to usually import it into other tools without much problem. (I am still looking for a way to basicallt export a subset of the input json)

Answer 8

In addition to what others have suggested, I think that two options are worth mentioning.

Print as CSV/TSV

$ cat file.json | jq -r '.users[] | [.first, .last] | @tsv'
Stevie  Wonder
Michael Jackson

cat file.json | jq -r '.users[] | [.first, .last] | @csv'
"Stevie","Wonder"
"Michael","Jackson"

The first expression, .users[] , unnests the objects from the outer-most array as in the code given in the question. The next expression, [.first, .last] , creates a new array of the values for each input object, and the final expression uses the built-in functions @tsv and @csv to print all input arrays as tab-separated and comma-seperated values, respectively.

Print as JSON values

Similarly, it is possible to construct JSON values again, which is interesting if you just want to keep a subset of the fields:

$ cat file.json | jq -c '.users[] | {first}'
{"first":"Stevie"}
{"first":"Michael"}