Get object as JSON in IntelliJ Idea from debugger - json

Question

Is it possible to get the whole object from debugger as Json?
There is an option View text but can I somehow View JSON?

Answer 1

EDIT: as noted in the comments, this is not perfect, as for some variables you will get a "stackoverflow" response
As suggested by #Mr Han's answer, here's how you can do this:
Add a new way to view objects in Intellij debugger as json by
Going to File | Settings | Build, Execution, Deployment | Debugger | Data Views | Java Type Renderers
Click + to add new renderer
Call it JSON renderer
Supply java.lang.Object for Apply renderer to objects of type
Choose Use following expression: and supply an expression like so:
if (null == this || this instanceof String)
return this;
new com.google.gson.GsonBuilder().setPrettyPrinting().create().toJson(this);
Click OK
Now when you choose Copy Value on a variable, it will copy as json.

Answer 2

Alternatively, as seen here, you can use the following piece of code in your debug watcher:
new ObjectMapper()
.setSerializationInclusion(JsonInclude.Include.NON_NULL)
.writerWithDefaultPrettyPrinter()
.writeValueAsString( myObject )

Answer 3

You can try this code fragment into the Evaluate Expression(Alt + F8) on IntelliJ :
new com.fasterxml.jackson.databind.ObjectMapper() .registerModule(new com.fasterxml.jackson.datatype.jsr310.JavaTimeModule()) .disable(com.fasterxml.jackson.databind.SerializationFeature.WRITE_DATES_AS_TIMESTAMPS) .writerWithDefaultPrettyPrinter() .writeValueAsString( myObject );
image IntelliJ

Answer 4

You could use the Show as ... plugin for IntelliJ.
A small plugin to display formatted data out of the debugger and console.
Uses IntelliJ's build-in formatting capabilities.
No more need to copy values from debugger or console to a file to format them there. Following formats are supported: JSON, SQL, XML, Base64 encoded JSON, Base64 encoded text

Answer 5

If you have gson dependency in your project you can create a watch variable
new GsonBuilder().setPrettyPrinting().create().gson.toJson(myObject)
where myObject is your object.

Answer 6

Just follow it : File | Settings | Build, Execution, Deployment | Debugger | Data Views | Java Type Renderers, click + to add new render ,
copy is OK :) u can choose another jar to format it
And now , Apply, join it ~

Answer 7

Follow the instructions of #BradParks, and use the following expression.
For me it did not work without fully-qualified class names. I also added some modifications to the ObjectMapper. For some reason which I don't understand, even if I have Apply renderers to object of type set to java.lang.Object, I needed to typecast this as (Object)this when used as a parameter of the writeValueAsString() method.
if (this == null
|| this instanceof CharSequence
|| this instanceof Number
|| this instanceof Character
|| this instanceof Boolean
|| this instanceof Enum) {
// Here you may add more sophisticated test which types you want to exclude from the JSON conversion.
return this;
}
new com.fasterxml.jackson.databind.ObjectMapper()
.registerModule(new com.fasterxml.jackson.datatype.jsr310.JavaTimeModule())
.disable(com.fasterxml.jackson.databind.SerializationFeature.WRITE_DATES_AS_TIMESTAMPS)
.setVisibility(
com.fasterxml.jackson.annotation.PropertyAccessor.FIELD,
JsonAutoDetect.Visibility.ANY)
.setSerializationInclusion(com.fasterxml.jackson.annotation.JsonInclude.Include.NON_NULL)
.writerWithDefaultPrettyPrinter()
.writeValueAsString((Object)this);

Answer 8

In case someone is having hard time to make the renderers work for more complex objects - below you can find a combined JSON renderer from:
https://stackoverflow.com/a/60189764/3837638
https://stackoverflow.com/a/69802068/3837638
https://stackoverflow.com/a/63060407/3837638
The following renderer helped me to identify multiple fields with the same name in the class hierarchy, so I could change that.
Initially I was having IllegalArgumentException for serialization of a nested object that I wasn't able to analyse.
If there is an issue during serialization, with this renderer you can find the stack trace from the exception that you need to fix in the console.
Good luck!
if (null == this)
return "null";
if (this instanceof CharSequence
|| this instanceof Number
|| this instanceof Character
|| this instanceof Boolean
|| this instanceof Enum) {
// Here you may add more sophisticated test which types you want to exclude from the JSON conversion.
return this;
}
try {
String json = new GsonBuilder().setPrettyPrinting().create().toJson(this);
return json;
} catch (Exception e) {
e.printStackTrace();
}

Answer 9

Use Intellij plugin Debug Variable Extractor
More information -
https://plugins.jetbrains.com/plugin/16362-debug-variable-extractor