Converting Java object to JSON with specific syntax

Question

I want to convert Java object to JSON string. Lets say the object looks like this:

public class User{
    private Long id;
    private String firstName;
}

And the json should look like this:

{
    "inputFields":[
        {
            "name":"id",
            "value":"123"
        },
        {
            "name":"firstName",
            "value":"George"
        }
    ]
}

Tried to use Jackson, but it looks like it doesn't provide such kind of serialization.

You Need Custom Serializer to Serialize the array of Class User Object. — Vikrant Kashyap
– Vikrant Kashyap, Commented Apr 1, 2016 at 8:58
This is what you need to do: baeldung.com/jackson-custom-serialization — dabadaba
– dabadaba, Commented Apr 1, 2016 at 8:59
Why do you need such an awkward JSON structure? Wouldn't it be better to have "id":"123","name":"George"? — Magnus
– Magnus, Commented Apr 1, 2016 at 9:00
You might want to have a look at this thread - How to serialize Object to JSON? — Viorel Florian
– Viorel Florian, Commented Apr 1, 2016 at 9:05
@BadCash, there always is a reason. For this case I am trying to integrate with external WebService, and it requires such structure. — Yuriy Kravets
– Yuriy Kravets, Commented Apr 1, 2016 at 9:07

dumitru · Accepted Answer · 2016-04-07 08:11:07Z

2

Jackson can use custom serializer where you can have control in generating the output. The following are the step to do this:

Annotate your POJO to use a custom serializer

@JsonSerialize(using = CustomSerializer.class)
static class User{
    public Long id;
    public String firstName;

    public User(Long id, String firstName) {
        this.id = id;
        this.firstName = firstName;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }
}

Declare the serialzer

static class CustomSerializer extends JsonSerializer{

    @Override
    public void serialize(Object value, JsonGenerator jgen, SerializerProvider serializers) throws IOException, JsonProcessingException {
        jgen.writeStartObject();
        jgen.writeArrayFieldStart("inputFields");

        Field[] fields = value.getClass().getDeclaredFields();

        for (Field field : fields) {
            try {
                field.setAccessible(true);
                jgen.writeStartObject();
                jgen.writeObjectField("name", field.getName());
                jgen.writeObjectField("value", field.get(value));
                jgen.writeEndObject();
            } catch (IllegalArgumentException | IllegalAccessException e) {
                e.printStackTrace();
            }

        }

        jgen.writeEndArray();
        jgen.writeEndObject();
    }
}

A simple test

public static void main(String[] args) throws JsonProcessingException {
    User user = new User(1L, "Mike");
    ObjectMapper om = new ObjectMapper();
    om.writeValueAsString(user);
    System.out.println(om.writeValueAsString(user));
}

And the output will be

{"inputFields":[{"name":"id","value":1},{"name":"firstName","value":"Mike"}]}

edited Apr 7, 2016 at 8:11

answered Apr 1, 2016 at 9:14

dumitru

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4 Comments

Yuriy Kravets Over a year ago

what version of jackson are you using ?

dumitru Over a year ago

in the example i used jackson 2.6.1

Yuriy Kravets Over a year ago

add field.setAccessible(true); to the loop and I will accept this answer as the most "on point" and elegant.

dumitru Over a year ago

true, to prevent accessibility using reflection one should set that flag. thanks

stepanian · Accepted Answer · 2016-04-01 09:18:48Z

0

Create an InputFields class and move the User object data to it before converting to json.

Edit: The InputFields class would obviously have the same structure as the desired json result.

edited Apr 1, 2016 at 9:18

answered Apr 1, 2016 at 9:11

stepanian

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3 Comments

dambros Over a year ago

And make the field a Collection inside the InputFields class. Or create a custom converter.

Yuriy Kravets Over a year ago

@stepanian actually, the inputFields is not a big problem, the problem is structure of name-value pairs

stepanian Over a year ago

I understand that. The InputFields class will have that structure.

Duoglas Du · Accepted Answer · 2016-04-01 09:37:57Z

Just a example of stepanian's answer, but i prefer dumitru's way to implement a CustomSerializer.

create a InputField class

public class InputField {

    private String name;
    private String value;

    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

}

parse User

public static List<InputField> parseInputFields(Object model) {
    List<InputField> inputFields = new ArrayList<InputField>();

    Field[] field = model.getClass().getDeclaredFields();
    try {
        for (int j = 0; j < field.length; j++) {

            InputField inputField = new InputField();

            inputField.setName(field[j].getName());
            inputField.setValue(String.valueOf(field[j].get(model));

            inputFields.add(inputField);
        }
    } catch (NoSuchMethodException e) {
        e.printStackTrace();
    } catch (SecurityException e) {
        e.printStackTrace();
    } catch (IllegalAccessException e) {
        e.printStackTrace();
    } catch (IllegalArgumentException e) {
        e.printStackTrace();
    } catch (InvocationTargetException e) {
        e.printStackTrace();
    }

    return inputFields;
}

test

public static void main(String[] args) throws JsonProcessingException {
    User user = new User(1L, "Mike");

    Map<String, Object> tmpMap=new HashedMap();
    tmpMap.put("inputFields", parseInputFields(user));

    ObjectMapper om = new ObjectMapper();
    System.out.println(om.writeValueAsString(tmpMap));
}

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