Question 109922 in Json Question posted in
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Json – Can this jq map be simplified?

Given this JSON:

{
  "key": "/books/OL1000072M",
  "source_records": [
    "ia:daywithtroubadou00pern",
    "bwb:9780822519157",
    "marc:marc_loc_2016/BooksAll.2016.part25.utf8:103836014:1267"
  ]
}

Can the following jq code be simplified?

jq -r '.key as $olid | .source_records | map([$olid, .])[] | @tsv'

The use of variable assignment feels like cheating and I’m wondering if it can be eliminated. The goal is to map the key value onto each of the source_records values and output a two column TSV.

Tags:

2

Answers


  1. 0

    Instead of mapping into an array, and then iterating over it (map(…)[]) just create an array and collect its items ([…]). Also, you can get rid of the variable binding (as) by moving the second part into its own context using parens.

    jq -r '[.key] + (.source_records[] | [.]) | @tsv'

    Alternatively, instead of using @tsv you could build your tab-separated output string yourself. Either by concatenation (… + …) or by string interpolation ("(…)"):

    jq -r '.key + "t" + .source_records[]'

    jq -r '"(.key)t(.source_records[])"'

    Output:

    /books/OL1000072M   ia:daywithtroubadou00pern
/books/OL1000072M   bwb:9780822519157
/books/OL1000072M   marc:marc_loc_2016/BooksAll.2016.part25.utf8:103836014:1267
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  2. 0

    It’s not much shorter, but I think it’s clearer than the original and clearer than the other shorter answers.

    jq -r '.key as $olid | .source_records[] | [ $olid, . ] | @tsv'
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