Build a JSON string with Bash variables

You can use printf:

JSON_FMT='{"bucketname":"%s","objectname":"%s","targetlocation":"%s"}\n'
printf "$JSON_FMT" "$BUCKET_NAME" "$OBJECT_NAME" "$TARGET_LOCATION"

much clear and simpler

You are better off using a program like jq to generate the JSON, if you don't know ahead of time if the contents of the variables are properly escaped for inclusion in JSON. Otherwise, you will just end up with invalid JSON for your trouble.

BUCKET_NAME=testbucket
OBJECT_NAME=testworkflow-2.0.1.jar
TARGET_LOCATION=/opt/test/testworkflow-2.0.1.jar

JSON_STRING=$( jq -n \
                  --arg bn "$BUCKET_NAME" \
                  --arg on "$OBJECT_NAME" \
                  --arg tl "$TARGET_LOCATION" \
                  '{bucketname: $bn, objectname: $on, targetlocation: $tl}' )