使用 Bash 变量构建 JSON 字符串

发布于01月27日

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

推荐答案

如果您事先不知道变量的内容是否被正确转义以包含在JSON中，那么最好使用像jq这样的程序来生成JSON.否则，您的麻烦只会以无效的JSON结束.

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}' )