Jackson json property (string) to instance

Question

Assuming I have the following JSON:

{
  "property": "123:1234"
}

How do I use Jackson annotations to ensure that the string value of "property" is de-serialized to a self-defined class rather than a String object? I went through their documentation and I was unable to find this particular feature.

Thanks in advance.

do you want your property to be stored in some object with fields? Maybe delimeted by `:` ? — Michał Krzywański, May 16 '19 at 15:04
@michalk Yes. The string is used as an identifier within our system and I'd like to wrap it in a class dedicated to those identifiers. — Byebye, May 16 '19 at 15:57

score 0 · Answer 1 · answered May 16 '19 at 15:08

First thing first define your class that needs to be used:

@JsonInclude(JsonInclude.Include.NON_NULL)
public class JsonTest{
@JsonProperty("property")
private String property;

    //define your getters and setters for the field

Then you can use the ObjectMapper class from jackson:

  public static <T> T extractObjectFromJson(String jsonText, Class<T> type) {
    try {
        return new ObjectMapper().configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false).reader().forType(type)
                .readValue(jsonText);
    } catch (Exception e) {
        //Manage your exception here
    }
    return null;
}

So you can just call the method extractobjectFromJson(//Your JSON String, JsonTest.class) to get your JSON deserialized.

score 0 · Accepted Answer · answered May 16 '19 at 15:20

You could create custom deserializer for your field. Assuming you want to map it to SomeClass object :

public class SomeClass {

    @JsonDeserialize(using = CustomPropertyDeserializer.class)
    private Properties property;

    public Properties getProperty() {
        return property;
    }

    public void setProperty(Properties property) {
        this.property = property;
    }
}

You annotate your field that you want to deserialize customly with @JsonDeserialize annotation passing custom deserializer. Your deserializer could look like this :

public class CustomPropertyDeserializer extends StdDeserializer<Properties> {

    public CustomPropertyDeserializer() {
        super(Properties.class);
    }

    @Override
    public Properties deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        String valueAsString = p.getValueAsString();
        String[] split = valueAsString.split(":");

        return new Properties(split[0], split[1]);
    }
}

And custom property class :

public class Properties {
    private String first;
    private String second;

    public Properties(String first, String second) {
        this.first = first;
        this.second = second;
    }

    public String getFirst() {
        return first;
    }

    public void setFirst(String first) {
        this.first = first;
    }

    public String getSecond() {
        return second;
    }

    public void setSecond(String second) {
        this.second = second;
    }
}

For testing it :

    public static void main(String[] args) throws IOException {
        String s = Files.lines(Paths.get("src/main/resources/data.json")).collect(Collectors.joining());

        ObjectMapper objectMapper = new ObjectMapper();

        SomeClass someClass = objectMapper.readValue(s, SomeClass.class);

        System.out.println(someClass.getProperty().getFirst());
        System.out.println(someClass.getProperty().getSecond());

    }

The output is then :

123
1234

So all the custom logic how to map your String to some class that you define could be placed in deserialize method of your custom deserializer.

Jackson json property (string) to instance

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