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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 
932
First, your name property is a string, so you need to add double quotes to it in your json. Second, using single quotes, bash won't do variable expansion: it won't replace $r_name with the variable content (see Expansion of variable inside single quotes in a command in bash shell script for more information).
married = false; //convert object to json string var string = JSON. stringify(obj); //convert string to Json Object console. log(JSON. parse(string)); // this is your requirement.
To initialize a string, you directly start with the name of the variable followed by the assignment operator(=) and the actual value of the string enclosed in single or double quotes. Output: This simple example initializes a string and prints the value using the “$” operator.
json has been loaded into a JSONObject named jsonData. Using data. json as your guide, create String variables named name, publisher, and language and set them to the appropriate values from data.
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}' ) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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