How do I deserialize a generic object using fromJson?

How do I deserialize a generic object using the Cobalt infrastructure utility library's JSONHelper.fromJson method? For example, if I have the following (ignoring JSON annotations):

class Foo {
String bar;
}

I can do this:

String json = "{\"bar\":\"abc\"}";
Foo f = JSONHelper.fromJson(json, Foo.class);

But suppose Foo is instead defined:

class Foo {
T bar;
}

How can I deserialize a Foo of T when I know T? E.g.:

Foo f = JSONHelper.fromJson(???)

I see that fromJsonMap and fromJsonList takes a TyperReference object. Do I just need to fall back directly to Jackson the way it's implemented in those methods?

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Accepted answers

Andrew Gerber

I'm thinking the best approach would to just create your own wrapper around Jackson's `ObjectMapper`. Though you could use a bit of trickery to get what you want out of the JSONHelper's `fromJsonMap` method:

class Foo {
T bar;
}

Map> fooo = JSONHelper.fromJsonMap("{\"foo\":{\"bar\":5}}", new TypeReference>>() {});

System.out.print(fooo.get("foo").bar.equals(5)); // true

Here is a link which shows how to use `TypeReference`:
[http://www.cowtowncoder.com/blog/archives/2009/01/entry_137.html
][1]

[1]: http://www.cowtowncoder.com/blog/archives/2009/01/entry_137.html

All comments

Samuel Slotsky

What vertical are you working in? If this is Website code, we typically use `JsonDotNetSerializer.CreateFromJson(string json, bool ignoreMissingMembers)`. Perhaps that would work better for you.

Ryan Morlok

This is in the WebContent module, in Java land.

Samuel Slotsky

In that case, I've not a clue

Andrew Gerber

Ryan Morlok

This is what I ended up doing only I went straight to Jackson rather than creating a wrapper.

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