Print json block from a json array on new line joined by '#'-CodePudding

I have a json array that looks as below -

[
  {
    "Code": "some code",
    "Name": "some name",
    "Country": "some country",
    "Cost": "some cost",
    "Type": "Some type"
  },{
    "Code": "some code",
    "Name": "some name",
    "Country": "some country",
    "Cost": "some cost",
    "Type": "Some type"
  },{
    "Code": "some code",
    "Name": "some name",
    "Country": "some country",
    "Cost": "some cost",
    "Type": "Some type"
  }
]

I am trying print each json block on a new line with fields separated by '#'

Expected output

some code#some name#some country#some cost#some type
some code#some name#some country#some cost#some type
some code#some name#some country#some cost#some type

I've tried the code below which separates the content with '#'. However, it merges all blocks into a single line like this - some code#some name#some country#some cost#some type#some code#some name#some country#some cost#some type#some code#some name#some country#some cost#some type

 cat contents.json | sed 's/^[ \t]*//;s/[ \t]*$//' | awk  -F ': *' 'BEGIN { RS=",\n\"|\n}," } { gsub(/[\n\]\[\}]/,"",$2); if ($2) { printf("%s#", $2); } }'

Please could someone guide me to et this right?

CodePudding user response：

Using jq, the join builtin, when applied to an object, will pull out the field values and concatenate them using a separator string. The -r option ensures that raw text is being output (rather than JSON).

jq -r '.[] | join("#")' contents.json

some code#some name#some country#some cost#Some type
some code#some name#some country#some cost#Some type
some code#some name#some country#some cost#Some type

Demo

Update: If in the input objects the keys cannot be guaranteed to always be in the same order, just specify their order explicitly using an array.

jq -r '.[] | [.Code,.Name,.Country,.Cost,.Type] | join("#")' contents.json

some code#some name#some country#some cost#Some type
some code#some name#some country#some cost#Some type
some code#some name#some country#some cost#Some type

Demo

CodePudding user response：

I won't recommend mlr over jq for dealing with JSON, but in this case its powerful features could become handy for further processing of the input:

mlr --ijson --ocsv --ofs '#' unsparsify file.json

Code#Name#Country#Cost#Type
some code#some name#some country#some cost#Some type
some code#some name#some country#some cost#Some type
some code#some name#some country#some cost#Some type

^{note: You can remove the header with the -N option}

The advantage of this solution is that unordered or missing keys in the JSON objects won't jinx the output. Yet, if the JSON has a more complex structure then mlr won't be able to convert it.