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How to deserialize the generic response using Spring WebClient

I have a problem during the deserialization of a response. Let's suppose I have this response from third party using webclient .
Response :
{
"name":"FirstName",
"type":"Steel",
"Fee":{
"id":"1234",
"name":"FeeFirstName"
},
"address":"2nd Street"
}
This is how my pojo classes looks like
public class Fee{} //generic OR empty class
public class Foo{
private String name;
private String type;
private Fee fee;
private String address;
}
My webclient get response code :
#Autowired
private WebClient fooWebClient;
public Foo getFoo()
{
try{
return fooWebClient.get()
.uri(uriBuilder -> uriBuilder.path("/foo/fee").build("123"))
.header(HttpHeaders.CONTENT_TYPE,MediaType.APPLICATION_JSON_VALUE)
.accept(MediaType.APPLICATION_JSON)
.retrieve()
.bodyToMono(Foo.class)
.block();
}catch(Exception e){throw new ApiClientException(e.getMessage());}
}
The above webclient getFoo() code is not giving me the full response, the Fee is coming blank stating "Class has no fields". Rest of the values are coming properly in response.
Fee needs to be empty as any other object can also come.
Please let me know how to deserialize the whole response.

You don't need the Fee class, you can get rid of it entirely and use a Map instead:
public class Foo {
private String name;
private String type;
private Map<String, Object> fee;
private String address;
}

We cannot dynamically create POJO and hence we are left with two options.
Add necessary fields to the 'Fee' class (If you know Fee structure upfront)
If you are not sure about the 'Fee' structure go for Map.

Because spring integrates Jackson you can create a custom Jackson JSON Deserializer for the Fee class that gives you more control:
#JsonDeserialize(using = FeeDeserializer.class)
public class Fee {
private String id;
private String name;
public Fee(String id, String name) {
this.id = id;
this.name = name;
}
}
import com.fasterxml.jackson.*;
public class FeeDeserializer extends JsonDeserializer<Fee> {
#Override
public Fee deserialize(JsonParser jsonParser, DeserializationContext ctxt) throws IOException {
ObjectCodec codec = jsonParser.getCodec();
JsonNode tree = codec.readTree(jsonParser);
JsonNode id = tree.get("id");
JsonNode name = tree.get("name");
return (id != null && name != null) ? new Fee(id.asText(), name.asText()) : null;
}
}
For more details see
https://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#features.json.jackson.custom-serializers-and-deserializers
https://docs.spring.io/spring-framework/docs/current/reference/html/web-reactive.html#webflux-codecs-jackson

Set the default value of a property if null in springboot

I get a JSON string that I convert to an object. One of the property within the JSON sometimes would be null. If the property is null, I want the default value to be set to 0.
This is my class:
#Data
#NoArgsConstructor
#AllArgsConstructor
#Builder
public class Car {
#JsonProperty(value = "car_name")
private String carName;
#JsonProperty(value = "car_value")
private Long carValue;
#JsonProperty(value = "Specifications")
private Map<String, String> Specifications;
}
I use object mapper to convert the JSON string to the object
public List<Car> stringToCar(String json) throws JsonProcessingException {
ObjectMapper om = new ObjectMapper();
return om.readValue(json, new TypeReference<List<Car>>() {} );
}
carValue would sometimes have null value, if that happens I want it be set as 0. Is it possible to do in a efficient way rather than looping through the object and manually setting the value to 0

You have multiple ways to do this.
1) Setter that receives long instead
This is actually not straightforward but it works. If you define the setter as follows it will do what you need:
public void setCarValue(long carValue) {
this.carValue = carValue;
}
However, this feels like a hack to me, so I would not suggest you use it.
2) Custom deserializer
This one is more complex but also much easier to understand and explicit about your intent.
public class CustomLongDeserializer extends JsonDeserializer<Long> {
#Override
public Long deserialize(JsonParser jp, DeserializationContext ctxt)
throws IOException, JsonProcessingException {
String text = jp.getText();
if (text == null || text.isEmpty()) {
return 0;
} else {
return Long.valueOf(text);
}
}
}
Then you could apply the serializer on the attribute as follows:
#Data
#NoArgsConstructor
#AllArgsConstructor
#Builder
public class Car {
#JsonProperty(value = "car_name")
private String carName;
#JsonProperty(value = "car_value")
#JsonDeserialize(using = CustomLongDeserializer.class)
private Long carValue;
#JsonProperty(value = "Specifications")
private Map<String, String> Specifications;
}
Or apply it as a global deserializer to be used to deserialize every single Long:
ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addDeserializer(Long.class, new CustomLongDeserializer());
mapper.registerModule(module);

It's not a generic solution, but in your specific use case you could use the long primitive instead of Long which will coerce null to 0 when deserialized.

If you don't want to create a separate class like #João Dias mentioned, you can define the constructor yourself instead of using #AllArgsConstructor, something like :
public Car(String carName,
Long carValue,
Map<String, String> Specifications){
this.carName = carName;
this.Specifications = Specifications;
this.carValue = (carValue == null ? 0 : carValue);
}
The same can be used for #Builder annotation like :
builder().carValue(value == null ? 0 : value).build();
Again, if you have more variables to handle you should consider previous response.

How to convert embedded JSON String inside JSON to java object?

I have the below json, where the body key contains a value which is a string representation of a JSON object, how do I convert it to a Java Object ?
I can extract the body value by converting the JSON to a Map, but I don't know how I should proceed from there
input.json file
{
"body": "{\n\t\"username\": \"TestUser\",\n\t\"password\": \"TestPassword\"\n}"
}
The User POJO is as below,
class User {
private String username;
private String password;
... getters, setters and no-arg constructor
}
My code looks something like this, I need to implement convertToUser function
public static void main(String[] args) {
String jsonContent = readJsonFile("input.json");
String escapedJsonBody = getBody(s);
User user = convertToUser(escapedJsonBody, User.class);
}
I am already using jackson java library, any insights on doing this with jackson is highly appreciated.

One way to do it is to create DTOs and converter. Having DTOs like (i have nested the class declarations jsut to save space in answer):
#Getter #Setter
public class Input { // this level maps to the whole input.json
#JsonDeserialize(using = BodyDeserializer.class) // custom deserializer below
private Body body; // this is the body-attribute in JSON
#Getter #Setter
public static class Body {
private User user;
#Getter #Setter
public static class User {
private String username;
private String password;
}
}
}
the converter:
public class BodyDeserializer extends JsonDeserializer<Body> {
private ObjectMapper om = new ObjectMapper(); // this is to read the user from string
#Override
public Body deserialize(JsonParser p, DeserializationContext ctxt) throws IOException {
String embedded = p.readValueAs(String.class);
Body body = new Body();
body.setUser(om.readValue(embedded, User.class)); // here is the trick
return body;
}
}
Use like:
ObjectMapper om = new ObjectMapper();
String input = "{\"body\": \"{\\n\\t\\\"username\\\": \\\"TestUser\\\",\\n\\t\\\"password\\\": \\\"TestPassword\\\"\\n}\"}";
Input r = om.readValue(input, Input.class);
This way the conversion happens in generic way only con might be that you do not like to create DTOs and dig the user like Input.getBody().getUser();

To convert a JSON String to a java pojo you can use Jackson's ObjectMapper class that will assist you to do this.
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.readValue(inputJson, User.class);
More info can be found on Jackson's github page

Custom Deserialization using Jackson in Java

I am trying to implement a university project where I try to fetch values from two Json fields and map it to one pojo class.
Sample Json:
"event":[{"D17-32":0,"S10":"D"}]
Pojo class
public class Event {
#JsonDeserialize(using = SignalCustomDeserializer.class)
#JsonTypeInfo(use = JsonTypeInfo.Id.NONE)
#JsonProperty("S10")
private Signal S10;
#JsonProperty("D17-32")
private String D17_32;
//Getter and setter implementation
}
Class which I need to serialize the fields to
public class Signal{
private String value;
private String detectorId;
private int detectorValue; //this value has to be fetched from another json
//Getter and setter implementation
}
Custom deserializer class
public class SignalCustomDeserializer extends JsonDeserializer {
#Override
public Signal deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException, JsonProcessingException {
String signalId = jsonParser.getCurrentName();
String signalVal = jsonParser.getValueAsString();
String detectorVal = jsonParser.getValueAsString("D01-16");
Signal signal = new Signal();
signal.setValue(signalVal);
signal.setDetectorId(getDetectorId(signalId));
return signal;
}
}
I am able to get the signalId and signalValue but I am unable to get the value for the other field. I am unsure if its available in the JsonObject when the custom deserialization class is called.

Deserialize nested array as ArrayList with Jackson

I have a piece of JSON, that looks like this:
{
"authors": {
"author": [
{
"given-name": "Adrienne H.",
"surname": "Kovacs"
},
{
"given-name": "Philip",
"surname": "Moons"
}
]
}
}
I have created a class to store Author information:
public class Author {
#JsonProperty("given-name")
public String givenName;
public String surname;
}
And two wrapper classes:
public class Authors {
public List<Author> author;
}
public class Response {
public Authors authors;
}
This is working, but having two wrapper classes seems to be unnecessary. I want to find a way to remove Authors class and have a list as a property of Entry class. Is something like that is possible with Jackson?
Update
Solved that with custom deserializer:
public class AuthorArrayDeserializer extends JsonDeserializer<List<Author>> {
private static final String AUTHOR = "author";
private static final ObjectMapper mapper = new ObjectMapper();
private static final CollectionType collectionType =
TypeFactory
.defaultInstance()
.constructCollectionType(List.class, Author.class);
#Override
public List<Author> deserialize(JsonParser jsonParser, DeserializationContext deserializationContext)
throws IOException, JsonProcessingException {
ObjectNode objectNode = mapper.readTree(jsonParser);
JsonNode nodeAuthors = objectNode.get(AUTHOR);
if (null == nodeAuthors // if no author node could be found
|| !nodeAuthors.isArray() // or author node is not an array
|| !nodeAuthors.elements().hasNext()) // or author node doesn't contain any authors
return null;
return mapper.reader(collectionType).readValue(nodeAuthors);
}
}
And using it like this:
#JsonDeserialize(using = AuthorArrayDeserializer.class)
public void setAuthors(List<Author> authors) {
this.authors = authors;
}
Thanks #wassgren for the idea.

I see at least two approaches to do this if you want to get rid of wrapper classes. The first is to use the Jackson Tree Model (JsonNode) and the second is to use a deserialization feature called UNWRAP_ROOT_VALUE.
Alternative 1: Use JsonNode
When deserializing JSON using Jackson there are multiple ways to control what type of objects that are to be created. The ObjectMapper can deserialize the JSON to e.g. a Map, JsonNode (via the readTree-method) or a POJO.
If you combine the readTree-method with the POJO conversion the wrappers can be completely removed. Example:
// The author class (a bit cleaned up)
public class Author {
private final String givenName;
private final String surname;
#JsonCreator
public Author(
#JsonProperty("given-name") final String givenName,
#JsonProperty("surname") final String surname) {
this.givenName = givenName;
this.surname = surname;
}
public String getGivenName() {
return givenName;
}
public String getSurname() {
return surname;
}
}
The deserialization can then look something like this:
// The JSON
final String json = "{\"authors\":{\"author\":[{\"given-name\":\"AdrienneH.\",\"surname\":\"Kovacs\"},{\"given-name\":\"Philip\",\"surname\":\"Moons\"}]}}";
ObjectMapper mapper = new ObjectMapper();
// Read the response as a tree model
final JsonNode response = mapper.readTree(json).path("authors").path("author");
// Create the collection type (since it is a collection of Authors)
final CollectionType collectionType =
TypeFactory
.defaultInstance()
.constructCollectionType(List.class, Author.class);
// Convert the tree model to the collection (of Author-objects)
List<Author> authors = mapper.reader(collectionType).readValue(response);
// Now the authors-list is ready to use...
If you use this Tree Model-approach the wrapper classes can be completely removed.
Alternative 2: remove one of the wrappers and unwrap the root value
The second approach is to remove only one of the wrappers. Assume that you remove the Authors class but keep the Response-wrapper. If you add the a #JsonRootName-annotation you can later unwrap the top-level name.
#JsonRootName("authors") // This is new compared to your example
public class Response {
private final List<Author> authors;
#JsonCreator
public Response(#JsonProperty("author") final List<Author> authors) {
this.authors = authors;
}
#JsonProperty("author")
public List<Author> getAuthors() {
return authors;
}
}
Then, for your mapper simply use:
ObjectMapper mapper = new ObjectMapper();
// Unwrap the root value i.e. the "authors"
mapper.configure(DeserializationFeature.UNWRAP_ROOT_VALUE, true);
final Response responsePojo = mapper.readValue(json, Response.class);
The second approach only removes one of the wrapper classes but instead the parsing function is quite pretty.