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I am trying to send multi-line comment in the curl body from bash script. Below is my curl invocation.

#!/bin/bash

temp="This is sample data: 2019/05/21 03:33:04
      This is 2nd sample data: #2 Sample_Data"

response=$(curl -sS --location "${HEADERS[@]}" -w "%{http_code}\n" -X POST "$url" --header 'Content-Type: application/json' \
                        --data-raw  "{
                                \"id\" : \"111\",
                                \"status\" : {
                                .
                                .
                                .
                                \"details\" : [ \"$temp\" ]
                                }
                        }")
echo "$response"

In the actual script the variable temp is stdout. So, it will be a multi-line output. If I try above script I get below error:

{"exceptionClass":"xxx.exception.MessageNotReadableException","errorCode":"xxx.body.notReadable","message":"The given request body is not well formed"}400

Can anyone please let me know what is the issue and how to fix it?

Thanks in advance

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  • Since this seems to be JSON, there's no technical need for the line breaks in the data. A JSON parser doesn't care. Commented Apr 3, 2023 at 14:46
  • Anyway, the web server is quite specific in that your request body is not well formed. So we can't help you here - you are omitting a significant part of the request, so that nobody can verify the well-formedness of your JSON. Commented Apr 3, 2023 at 14:48
  • @MarcusMüller thanks for the update. But what is the general way of sending multi-line data in body? Do you have any examples for this? In my case the issue is only with variable \"details\" : [ \"$temp\" ] . If temp is single line then it's working fine.. Commented Apr 3, 2023 at 15:00
  • You're currently sending multiline data. Your problem has nothing to do with that. Again, JSON doesn't care about line breaks, either. Commented Apr 3, 2023 at 15:14
  • @MarcusMüller Those newline characters in the $temp string needs to be JSON encoded, so the multiline data is the issue. I'm pretty sure there is a duplicate question somewhere. Commented Apr 3, 2023 at 15:28

1 Answer 1

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A JSON string can't contain literal newlines. You can't take a shell variable and inject it into a JSON document without first properly JSON-encoding the variable's value.

One way to encode the string is by using jq:

status_details=$( jq -n --arg string "$temp" '$string' )

Note that $string in the command above is an internal jq variable, not a shell variable. The jq utility would encode the value given to $string on the command line and output it as a JSON-encoded string.

Given your example code, this would set the status_details variable to the following literal string, including the quotes:

"This is sample data: 2019/05/21 03:33:04\n      This is 2nd sample data: #2 Sample_Data"

You may then use this in your call to curl:

curl -s -S -L -w '%{http_code}\n' -X POST \
    --data-raw '{ "id": "111", "status": { "details": [ '"$status_details"' ] } }' \
    "$url"

If the details array is for storing each line from $temp as individual elements, you may split $temp into an array with jq like so:

status_details=$( jq -n --arg string "$temp" '$string | split("\n")' )

This would give you the following literal sting in $status_details:

[
  "This is sample data: 2019/05/21 03:33:04",
  "      This is 2nd sample data: #2 Sample_Data"
]

You could then use that with curl in pretty much the same way as shown above, but without surrounding $status_details with square brackets (... "details": '"$status_details"' ...).

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  • Hello @Kusalananda thanks for your answer. Your solution looks promising. But I still get "The given request body is not well formed"}400 error. I think this is becasue of quotes "" in $status_details. Is it possible to trim them? Tried jq -r option $( jq -r -n --arg string "$temp" '$string' ) Commented Apr 3, 2023 at 17:06
  • Update: Adding | tr -d '"' worked. Commented Apr 3, 2023 at 17:30
  • @Preeti You definitely should not need to trim the double quotes away! I'm assuming you are trying to insert $status_details in place of $temp in your code while ignoring the code that I used in my answer which was tailored to the way $status_details was created. Commented Apr 3, 2023 at 18:51

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