Question 108818 in Json Question posted in
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Json – How to generate the following string?

Consider the following a snippet from my package.json

"dependencies": {
    "@types/lodash.camelcase": "^4.3.6",
    "@types/lodash.kebabcase": "^4.1.6",
    "@types/lodash.lowercase": "^4.3.6",
    "@types/lodash.upperfirst": "^4.3.6",
    "@types/mime": "^2.0.3",
    "@types/swagger-jsdoc": "^6.0.1",
    "async": "^3.2.2",
    "aws-arn": "^1.0.1",
}

From this, I want to generate the following string –

@types/lodash.camelcase@^4.3.6 @types/lodash.kebabcase@^4.1.6 @types/lodash.lowercase@^4.3.6 @types/lodash.upperfirst@^4.3.6 @types/mime@^2.0.3 @types/swagger-jsdoc@^6.0.1 async@^3.2.2 aws-arn@^1.0.1

Any idea how to accomplish this?

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2

Answers


  1. 0

    As jq is tagged: Make a valid JSON, then disassemble with to_entries, and join twice:

    jq -r '.dependencies | to_entries | map(join("@")) | join(" ")' package.json

    @types/lodash.camelcase@^4.3.6 @types/lodash.kebabcase@^4.1.6 @types/lodash.lowercase@^4.3.6 @types/lodash.upperfirst@^4.3.6 @types/mime@^2.0.3 @types/swagger-jsdoc@^6.0.1 async@^3.2.2 aws-arn@^1.0.1

    Demo

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  2. 0

    Assuming actually valid JSON, you can use string interpolation) and join:

    .dependencies | to_entries | map("(.key)@(.value)") | join(" ")

    String interpolation gives you a little more flexibility compared to using join twice. YMMV, and for sample cases join will suffice.

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