Jackson: Deserialize JSON array from JSON object into Java List

Question

I have been stumbled by this for a while. I have a Spring application and would like to parse the following JSON:

{
  "metadata": {...}
  "response": {
    "objects": [
      {
        "name": "someName",
        "properties": [<array_of_properties>]
      },
      ...
    ]
  }
}

into a list of the following Java objects:

public class MyClass {
  String name;
  List<CustomProperties> customProperties;
}

Meaning, I want to extract only the objects array and parse only that. I have tried using a custom deserializer and that works, but I had to do:

@JsonDeserialize(using=MyDeserializer.class)
public class MyClassList extends ArrayList<MyClass>{}

and then:

ObjectMapper objectMapper = new ObjectMapper();
List<MyClass> list = objectMapper.readValue(json, MyClassList.class)

Is there anyway to avoid extending ArrayList, since currently I am doing that in order to be able to access the .class property.

shahaf · Accepted Answer · 2019-10-14 14:26:50Z

1

you can define your json structure with a couple of classes

public class MyJson {
  private MyResponse response;
  ...
}

public class MyResponse {
  private List<MyClass> objects;
  ...
}

public class MyClass {
  String name;
  List<CustomProperty> customProperties;
  ...
}

than you can use Jackson to parse the json string to MyJson class, no special @JsonDeserialize is needed

ObjectMapper objectMapper = new ObjectMapper();
MyJson myJson = objectMapper.readValue(json, MyJson.class);
List<MyClass> list =  myJson.getResponse().getObjects();

Keep in mind, this code is only a draft, all classes should have setters (and getters) and some null checks are required

answered Oct 14, 2019 at 14:26

shahaf

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2 Comments

Nikola Toshev Over a year ago

Thank you, shahaf! However, I was hoping to avoid the need for wrapper classes. I am not against it, and if it's the only way I will go with it, but is there no direct way to tell Jackson: Hey, I have this JSON string and this deserializer which returns a List of objects. Use it!

shahaf Over a year ago

@NikolaToshev, sry I don't familiar with one, because JAVA is statically type, a parsing of dynamic strings like JSON, is more construct, and one need to define the type and structure, the other option is to use JsonObject or JsonNode and traverse the tree to the specific key, more error prone...

pvpkiran · Accepted Answer · 2019-10-14 14:30:46Z

You can do something like this. I feel this would be cleaner

@JsonIgnoreProperties(ignoreUnknown = true)
class Wrapper{
  private Response response;
  //setters, getters
}

@JsonIgnoreProperties(ignoreUnknown = true)
class Response{
  private List<MyClass> objects;
  //setters, getters
}

@JsonIgnoreProperties(ignoreUnknown = true)
public class MyClass {
  String name;
  List<CustomProperties> customProperties;
  //setters, getters
}

ObjectMapper objectMapper = new ObjectMapper();
Wrapper wrapper = objectMapper.readValue(json, Wrapper.class)

You can extrat objects and consequently CustomProperties by traversing the list. You can declare only fields which you are interested in and ignore others by @JsonIgnoreProperties(ignoreUnknown = true)(for example i have not included metadata)

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