BASH How to treat JSON inside a variable to remove only a specific part of the text

Question

I have a big JSON inside a var and I need to remove only from that specific comma ( and I have incontable number of others comma before that ) until the penultimate Curly Brackets.. in short, Only the BOLD text..... ( text between ** and next ** )

edit

originally there is no ** in json, I put it in the code just to show where it starts and ends what I want to remove ##################################################

              }
        ]
    }**,
    "meta": {
        "timeout": 0,
        "priority": "LOW_PRIORITY",
        "validationType": "SAME_FINGERS",
        "labelFilters": [],
        "externalIDs": [
            {
                "name": "chaveProcesso",
                "key": "01025.2021.0002170"
            }
        ]
    }**
}

William Pursell · Accepted Answer · 2021-08-26 18:38:34Z

2

It would help if you showed more context, but basically you want something like:

jq 'del(.meta)'

or:

jq 'with_entries(select(.key != "meta"))'

eg:

#!/bin/sh

json='{
    "foo": 5,
    "meta": {
        "timeout": 0,
        "priority": "LOW_PRIORITY",
        "validationType": "SAME_FINGERS",
        "labelFilters": [],
        "externalIDs": [
            {
                "name": "chaveProcesso",
                "key": "01025.2021.0002170"
            }
        ]
    }
}'

echo "$json" | jq 'del(.meta)'

edited Aug 26, 2021 at 18:38

answered May 12, 2021 at 14:01

William Pursell

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3 Comments

oguz ismail Over a year ago

You know del(.meta) does the same thing, right?

William Pursell Over a year ago

@oguzismail No, I'm still learning jq. Thanks for that!

Charles Duffy Over a year ago

(should be able to remove the . | from the start as well).

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