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Question: Zipping two Lists into a List of Objects #623
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Okay, so first we want to get the columnHeaders as an array of names:
Then, for each row, we take the $headers as entries (if this doesn't mean anything to you, refer to the
Then we put it all together: wrapping it on a filter is left as an exercise for the reader.
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Brilliant! 👍 😄 |
I added this to wiki, because I can't get enough of good JQ example usage. https:/stedolan/jq/wiki/jq-Cookbook/ba821a5760a7925e4a52e682e4b0a6248848d732 |
That's so cool! I didn't know the wiki had this. I'll have a go at |
Probably a simple question, but given the following JSON:
How can I convert this into a form like:
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