Build a JSON string with Bash variables

问题:

I need to read these bash variables into my JSON string and I am not familiar with bash. any help is appreciated.

#!/bin/sh

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

JSON_STRING='{"bucketname":"$BUCKET_NAME"","objectname":"$OBJECT_NAME","targetlocation":"$TARGET_LOCATION"}'


echo $JSON_STRING 

回答1:

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}' )

回答2:

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

回答3:

A possibility:

JSON_STRING='{"bucketname":"'"$BUCKET_NAME"'","objectname":"'"$OBJECT_NAME"'","targetlocation":"'"$TARGET_LOCATION"'"}'

回答4:

First, don't use ALL_CAPS_VARNAMES: it's too easy to accidentally overwrite a crucial shell variable (like PATH)

Mixing single and double quotes in shell strings can be a hassle. In this case, I'd use printf:

bucket_name=testbucket
object_name=testworkflow-2.0.1.jar
target_location=/opt/test/testworkflow-2.0.1.jar
template='{"bucketname":"%s","objectname":"%s","targetlocation":"%s"}'

json_string=$(printf "$template" "$BUCKET_NAME" "$OBJECT_NAME" "$TARGET_LOCATION")

echo "$json_string"

For homework, read this page carefully: Security implications of forgetting to quote a variable in bash/POSIX shells

---

A note on creating JSON with string concatenation: there are edge cases. For example, if any of your strings contain double quotes, you can broken JSON:

$ bucket_name='a "string with quotes"'
$ printf '{"bucket":"%s"}\n' "$bucket_name"
{"bucket":"a "string with quotes""}

Do do this more safely with bash, we need to escape that string's double quotes:

$ printf '{"bucket":"%s"}\n' "${bucket_name//\"/\\\"}"
{"bucket":"a \"string with quotes\""}

回答5:

If you need to build a JSON representation where members mapped to undefined or empty variables should be ommited, then jo can help.

#!/bin/bash

BUCKET_NAME=testbucket
OBJECT_NAME=""

JO_OPTS=()

if [[ ! "${BUCKET_NAME}x" = "x" ]] ; then
        JO_OPTS+=("bucketname=${BUCKET_NAME}")
fi

if [[ ! "${OBJECT_NAME}x" = "x" ]] ; then
        JO_OPTS+=("objectname=${OBJECT_NAME}")
fi

if [[ ! "${TARGET_LOCATION}x" = "x" ]] ; then
        JO_OPTS+=("targetlocation=${TARGET_LOCATION}")
fi

jo "${JO_OPTS[@]}"

回答6:

can be done following way:

JSON_STRING='{"bucketname":"'$BUCKET_NAME'","objectname":"'$OBJECT_NAME'","targetlocation":"'$TARGET_LOCATION'"}'

回答7:

For Node.js Developer, or if you have node environment installed, you can try this:

JSON_STRING=$(node -e "console.log(JSON.stringify({bucketname: $BUCKET_NAME, objectname: $OBJECT_NAME, targetlocation: $TARGET_LOCATION}))")

Advantage of this method is you can easily convert very complicated JSON Object (like object contains array, or if you need int value instead of string) to JSON String without worrying about invalid json error.

Disadvantage is it's relying on Node.js environment.