How to retrieve "name" value from json using bash? - bash

Question

I am using the following curl command:
curl -s -v --user admin:orca --insecure -X GET https://insecure.registry.com/api/v0/repositories/authi-api/tags
Getting following output:
{
"name": "Dev_ReleaseRollout_Lane-3",
"inRegistry": true,
"hashMismatch": false,
"inNotary": false
},
{
"name": "Dev_ReleaseRollout_Lane-latest",
"inRegistry": true,
"hashMismatch": false,
"inNotary": false
},
{
"name": "Payments_Dev_Lane-267",
"inRegistry": true,
"hashMismatch": false,
"inNotary": false
}
I want to get only name values in a variable.
I need only Dev_ReleaseRollout_Lane-3 Dev_ReleaseRollout_Lane-latest Payments_Dev_Lane-267 in a variable

Answer 1

Assuming you actually have an array around the three objects:
$ curl ... | jq -r '.[].name'
Dev_ReleaseRollout_Lane-3
Dev_ReleaseRollout_Lane-latest
Payments_Dev_Lane-267
It's fairly simple, . is the array, [].name take name from each element in the array. -r is raw output.
--raw-output / -r:
With this option, if the filter’s result is a string then it will be written directly to standard output rather than being formatted as a JSON string with quotes. This can be useful for making jq filters talk to non-JSON-based systems.
If the cURL output is actually as mentioned above the following will work:
jq -rRs '"[\(.)]" | fromjson[].name' file.json
However I think there is a better way to wrap an array around input,
-R is raw input and -s is slurp. \(...) is string interpolation.
--slurp/-s:
Instead of running the filter for each JSON object in the input, read the entire input stream into a large array and run the filter just once.