escape variable in bash script curl request - linux

Question

I am trying to curl via bash script but unable to pass the value of var1 in curl request and getting error upon execution ...
#!/bin/bash
var1="some test message"
curl 'https://link' \
-H 'Content-Type: application/json' \
-d '
{"msgtype": "text",
"text": {
"content": $var1
}
}'

Answer 1

Your variable $var1 doesn't get expanded by the shell because it is inside single quote '.
You need to use double quote to let bash do the parameter expansion, and escape your json data:
#!/bin/bash
var1="some test message"
curl 'https://link' \
-H 'Content-Type: application/json' \
-d "
{
\"msgtype\": \"text\",
\"text\": {
\"content\": \"$var1\"
}
}"
Or you can use inline document (without the escaping hell, but the command becomes awkward):
#!/bin/bash
var1="some test message"
curl 'https://link' \
-H 'Content-Type: application/json' \
-d "$(cat <<EOT
{
"msgtype": "text",
"text": {
"content": "$var1"
}
}
EOT
)"

Answer 2

In general, don't use parameter expansion to define JSON dynamically like this. There is no guarantee that your template, combined with the contents of the variable, produces well-formed JSON. (This is for the same reasons you don't use string interpolation to create dynamic SQL queries.) Use a tool like jq instead.
curl 'https://link' ... \
-d "$(jq --argjson x "$var" \
-n '{msgtype: "text", text: {content: $x}}')"

Answer 3

Use below script.
https://aaa.com' -H 'Content-Type: application/json' -d '{"msgtype": "text", "text": { "content": '"'$var1'"' }}'