Jackson with JSON: Unrecognized field, not marked as ignorable

Question

I need to convert a certain JSON string to a Java object. I am using Jackson for JSON handling. I have no control over the input JSON (I read from a web service). This is my input JSON:

{"wrapper":[{"id":"13","name":"Fred"}]}

Here is a simplified use case:

private void tryReading() {
    String jsonStr = "{\"wrapper\"\:[{\"id\":\"13\",\"name\":\"Fred\"}]}";
    ObjectMapper mapper = new ObjectMapper();  
    Wrapper wrapper = null;
    try {
        wrapper = mapper.readValue(jsonStr , Wrapper.class);
    } catch (Exception e) {
        e.printStackTrace();
    }
    System.out.println("wrapper = " + wrapper);
}

My entity class is:

public Class Student { 
    private String name;
    private String id;
    //getters & setters for name & id here
}

My Wrapper class is basically a container object to get my list of students:

public Class Wrapper {
    private List<Student> students;
    //getters & setters here
}

I keep getting this error and "wrapper" returns null. I am not sure what's missing. Can someone help please?

org.codehaus.jackson.map.exc.UnrecognizedPropertyException: 
    Unrecognized field "wrapper" (Class Wrapper), not marked as ignorable
 at [Source: java.io.StringReader@1198891; line: 1, column: 13] 
    (through reference chain: Wrapper["wrapper"])
 at org.codehaus.jackson.map.exc.UnrecognizedPropertyException
    .from(UnrecognizedPropertyException.java:53)

Waste

I found this useful to avoid creating a wrapper class: Map dummy<String,Student> = myClientResponse.getEntity(new GenericType<Map<String, Student>>(){}); and then Student myStudent = dummy.get("wrapper"); – Ellissa

JSON should looks like: String jsonStr = "{\"students\"\:[{\"id\":\"13\",\"name\":\"Fred\"}]}"; if you are expecting Wrapper object in your REST POST request – Cryptanalysis

Related (but different) question: Ignoring new fields on JSON objects using Jackson – Huberman

And incidentally, most answers to this question actually answer a different question, namely one similar to the one I linke above. – Huberman

majority of answers help brush problem under rug rather than actually solving it :( – Rhombic

It is so funny that why so many answers in this questions answer the things that are not related to the questions ? Apparently , the OP does not want to ignore deserialising the "wrapper" property ........ – Abbreviated

@Rhombic sometimes you can only fix what's under your control, and the rest you have to ignore – Wretched

Answer 1

You can use Jackson's class-level annotation:

import com.fasterxml.jackson.annotation.JsonIgnoreProperties

@JsonIgnoreProperties
class { ... }

It will ignore every property you haven't defined in your POJO. Very useful when you are just looking for a couple of properties in the JSON and don't want to write the whole mapping. More info at Jackson's website. If you want to ignore any non declared property, you should write:

@JsonIgnoreProperties(ignoreUnknown = true)

Dunkle

Ariel, is there any way to declare this external to the class? – Ulrica

I haven't done it but I believe that you can get somewhere in the annotations processing code and add the behavior programatically, although I can't think why you would like to do it. Can you give me an example? – Dunkle

I'm serializing classes that I do not own (cannot modify). In one view, I'd like to serialize with a certain set of fields. In another view, I want a different set of fields serialized (or perhaps rename the properties in the JSON). – Ulrica

i must add that you do need the (ignoreUnknown = true) when annotating your class otherwise it won't work – Reciprocate

Julián, this is not the correct answer to the question of the OP. However, I suspect that people come here because they google how to ignore properties not defined in POJO and this is the first result, so they end up up-voting this and Suresh's response (thats what happened to me). Although the original question has nothing to do with wanting to ignore undefined properties. – Autonomous

this is a very stupid default setting imho, if you add a property to your api, the whole serialization fails – Phobos

Why on earth did this answer get so many updates. This does not solve the problem, it just ignores it!!! – Proselytize

I don't understand the difference between these 2 syntaxes above - can someone explain? – Stotinka

@ycomp, these are not 2 syntaxes, @JsonIgnoreProperties is the annotation and @JsonIgnoreProperties(ignoreUnknown = true) a possible way to use it. You can also specify which fields to ignore as stated on Jackson's documentation (e.g. @JsonIgnoreProperties({ "extra", "uselessValue" })) – Dunkle

I got to this stackoverflow post because of this error message in mongodb jackson library. In my case I did not need to add this @JsonIgnoreProperties annotation. I was just trying to make "_id" property work here as told in jackson tutorial mongojack.org/tutorial.html, but I coudn't. Since @Objectd @JsonProperty("_id") public String getId(){return this._id} do not work, I had to replace it with @ObjectId @JsonProperty("_id") public String _id;. Now the runtime error is gone and I am not using @JsonIgnoreProperties. The version is 1.4.2 – Calbert

Downvoted as it does not answer the OP's question. Ignoring unknown properties doesn't solve his problem, but leaves him with a Wrapper instance where the students list is null or empty. – Accost

If you don't want your application to respond to unknown props defaultly by throwing a 500 error and barfing out all your property names, then this is the correct answer. I really don't know why most would desire this default behavior. – Yap

@Ariel Kogan ,so how does it behave just using @ JsonIgnoreProperties, I find it's ok to compile – Edwinedwina

@yuxh, ignoreUnknown defaults to false. You can check the documentation. – Dunkle

FWIW after almost a decade, regarding @JonLorusso's question about annotating classes you do not own, take a look at Jackson Mixin (and the function addMixInAnnotations). It's meant exactly for that. – Adjutant

In my case I got this response because the resource anwers with 500 error. I setted with @JsonIgnoreProperties and then get null. I knew the problem was then REST. – Lida

Answer 2

You can use

ObjectMapper objectMapper = getObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

It will ignore all the properties that are not declared.

Shurlock

This didn't work for me, it still fails on unknown properties – Bibliology

Could u please paste atleast a piece of code what exactly u are doing, You might have missed something there..Either by doing this or by using "@JsonIgnoreProperties(ignoreUnknown = true) " Your problem should be resolved. Anyways good luck. – Shurlock

I run this code: pastebin.com/rdBNezEn. However, I have realized that what this question asks is a bit different. Your answer is valid – Bibliology

FWIW -- I had to import this slightly different enum: org.codehaus.jackson.map.DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES – Unifoliate

^Above is for Jackson versions prior to 2 – Unopened

You can also chain these calls like: getObjectMapper().disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES) – Shel

Downvoted as it does not answer the OP's question. Ignoring unknown properties doesn't solve his problem, but leaves him with a Wrapper instance where the students list is null or empty. – Accost

mapper.configure(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false); worked for me – Sampler

Answer 3

The first answer is almost correct, but what is needed is to change getter method, NOT field -- field is private (and not auto-detected); further, getters have precedence over fields if both are visible.(There are ways to make private fields visible, too, but if you want to have getter there's not much point)

So getter should either be named getWrapper(), or annotated with:

@JsonProperty("wrapper")

If you prefer getter method name as is.

Mecke

Please elaborate - which getter needs to be changed or annotated? – Accost

you mean annotated with @JsonGetter("wrapper"), right? – Saadi

@pedrambashiri No, I mean @JsonProperty. While @JsonGetter is a legal alias, it is rarely used as @JsonProperty works for getters, setters and fields; setters and getters can be distinguished by signature (one takes no arguments, returns non-void; other takes one argument). – Mecke

This is the answer I was looking for! Sounds like Jackson has trouble mapping the source JSON to your POJO, but you can guarantee matches by tagging the getters. Thanks! – Weathertight

Answer 4

using Jackson 2.6.0, this worked for me:

private static final ObjectMapper objectMapper = 
    new ObjectMapper()
        .configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

or with the setting:

@JsonIgnoreProperties(ignoreUnknown = true)

Phylissphyll

With that config annotation is unnecessary – Lambkin

Do you need to configure both ObjectMapper and Annotation on class? I guess objectMapper will fix for all, without a need to annotate each class. I use the annotation though. – Dyslexia

You do not need both settings in the same class. You might also use DI to get a global singleton instance of the ObjectMapper, to set the FAIL_ON_UNKNOWN_PROPERTIES property globally. – Bulbiferous

You don't need both, you can choose one or the other. – Lyallpur

Answer 5

it can be achieved 2 ways:

1. Mark the POJO to ignore unknown properties

   @JsonIgnoreProperties(ignoreUnknown = true)
   

2. Configure ObjectMapper that serializes/De-serializes the POJO/json as below:

   ObjectMapper mapper =new ObjectMapper();            
   // for Jackson version 1.X        
   mapper.configure(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false);
   // for Jackson version 2.X
   mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false) 
   

Disjoint

Why should this be the accepted answer? This just tells to ignore unknown properties, whereas the question was to find a way to get the json wrapped into an object which this solution clearly says to ignore. – Taimi

Nice answer by simply using the first option. – Scorcher

Answer 6

Adding setters and getters solved the problem, what I felt is the actual issue was how to solve it but not how to suppress/ignore the error. I got the error "Unrecognized field.. not marked as ignorable.."

Though I use the below annotation on top of the class it was not able to parse the json object and give me the input

@JsonIgnoreProperties(ignoreUnknown = true)

Then I realized that I did not add setters and getters, after adding setters and getters to the "Wrapper" and to the "Student" it worked like a charm.

Exordium

This appears to be the only answer that actually answers the question. All the other answers are just marking unknown properties as ignored, but 'wrapper' is not an unknown property, it is what we are trying to parse. – Otte

Answer 7

This just perfectly worked for me

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(
    DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false);

@JsonIgnoreProperties(ignoreUnknown = true) annotation did not.

Depository

Downvoted as it does not answer the OP's question. Ignoring unknown properties doesn't solve his problem, but leaves him with a Wrapper instance where the students list is null or empty. – Accost

Answer 8

This works better than All please refer to this property.

import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.ObjectMapper;

    ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
    projectVO = objectMapper.readValue(yourjsonstring, Test.class);

Chromous

Yeah, this is the one that solved my issue - which matched the title of this post. – Armindaarming

Answers work well for me and it's very easy.Thank you – Upbeat

after this my projectVO data is null.yourjsonstring having value but projectVO fields are null. Any help guys ? – Tennietenniel

this really works and the POJO also doesnt need any modifications! – Rhoda

Answer 9

If you are using Jackson 2.0

ObjectMapper mapper = new ObjectMapper();
mapper.disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);

Isotonic

why this config has no effect for me? – Leadsman

@Leadsman It depends on Jackson version – Isotonic

Answer 10

According to the doc you can ignore selected fields or all uknown fields:

 // to prevent specified fields from being serialized or deserialized
 // (i.e. not include in JSON output; or being set even if they were included)
 @JsonIgnoreProperties({ "internalId", "secretKey" })

 // To ignore any unknown properties in JSON input without exception:
 @JsonIgnoreProperties(ignoreUnknown=true)

Tubate

Answer 11

It worked for me with the following code:

ObjectMapper mapper =new ObjectMapper();    
mapper.configure(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false);

Numeral

Answer 12

Jackson is complaining because it can't find a field in your class Wrapper that's called "wrapper". It's doing this because your JSON object has a property called "wrapper".

I think the fix is to rename your Wrapper class's field to "wrapper" instead of "students".

Azarria

Thanks Jim. I tried that and it did not fix the problem. I am wondering if I am missing some annotation.. – Waste

Hmm, what happens when you create the equivalent data in Java and then use Jackson to write it out in JSON. Any difference between that JSON and the JSON above should be a clue to what's going wrong. – Azarria

This answer worked for me, with the example from the question. – Huberman

Answer 13

I have tried the below method and it works for such JSON format reading with Jackson. Use the already suggested solution of: annotating getter with @JsonProperty("wrapper")

Your wrapper class

public Class Wrapper{ 
  private List<Student> students;
  //getters & setters here 
} 

My Suggestion of wrapper class

public Class Wrapper{ 

  private StudentHelper students; 

  //getters & setters here 
  // Annotate getter
  @JsonProperty("wrapper")
  StudentHelper getStudents() {
    return students;
  }  
} 


public class StudentHelper {

  @JsonProperty("Student")
  public List<Student> students; 

  //CTOR, getters and setters
  //NOTE: If students is private annotate getter with the annotation @JsonProperty("Student")
}

This would however give you the output of the format:

{"wrapper":{"student":[{"id":13,"name":Fred}]}}

Also for more information refer to https://github.com/FasterXML/jackson-annotations

Cheops

Answer 14

If you want to apply @JsonIgnoreProperties to all class in you application then the best way it is override Spring boot default jackson object.

In you application config file define a bean to create jackson object mapper like this.

@Bean
    public ObjectMapper getObjectMapper() {
        ObjectMapper mapper = new ObjectMapper();
        mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
        return mapper;
    }

Now, you don't need to mark every class and it will ignore all unknown properties.

Frag

Why not just create a static method that does the same? What is the significance of creating a bean? – Madcap

Answer 15

This solution is generic when reading json streams and need to get only some fields while fields not mapped correctly in your Domain Classes can be ignored:

import org.codehaus.jackson.annotate.JsonIgnoreProperties;
@JsonIgnoreProperties(ignoreUnknown = true)

A detailed solution would be to use a tool such as jsonschema2pojo to autogenerate the required Domain Classes such as Student from the Schema of the json Response. You can do the latter by any online json to schema converter.

Idette

Answer 16

Simple solution for Java 11 and newer:

var mapper = new ObjectMapper()
        .registerModule(new JavaTimeModule())
        .disable(FAIL_ON_UNKNOWN_PROPERTIES)
        .disable(WRITE_DATES_AS_TIMESTAMPS)
        .enable(ACCEPT_EMPTY_ARRAY_AS_NULL_OBJECT);

Important for ignoring is disabling 'FAIL_ON_UNKNOWN_PROPERTIES'

Meso

Note that in org.codehaus.jackson.map.ObjectMapper, the method disable takes other parameters; you are referring here to com.fasterxml.jackson.databind.ObjectMapper – Categorize

Answer 17

Problem is your property in your JSON is called "wrapper" and your property in Wrapper.class is called "students".

So either...

1. Correct the name of the property in either the class or JSON.
2. Annotate your property variable as per StaxMan's comment.
3. Annotate the setter (if you have one)

Earthaearthborn

Answer 18

Annotate the field students as below since there is mismatch in names of json property and java property

public Class Wrapper {
    @JsonProperty("wrapper")
    private List<Student> students;
    //getters & setters here
}

Walli

Answer 19

Somehow after 45 posts and 10 years, no one has posted the correct answer for my case.

@Data //Lombok
public class MyClass {
    private int foo;
    private int bar;

    @JsonIgnore
    public int getFoobar() {
      return foo + bar;
    }
}

In my case, we have a method called getFoobar(), but no foobar property (because it's computed from other properties). @JsonIgnoreProperties on the class does not work.

The solution is to annotate the method with @JsonIgnore

Salters

What you should actually do here is ask the specific question for the issue you have had and then answer your own question with your solution. What you have added here is not a solution to what the original question asks. You will help a lot more people if you pose your specific issue as a question. – Lexielexigraphy

@Lexielexigraphy The primary purpose of Stackoverflow is not (just) to answer single-use questions, but to be a repository of useful questions and answers for future googlers. This is the first result on Google, hence the various answers. – Salters

Jeff Atwood would disagree with you. stackoverflow.blog/2011/07/01/…. – Lexielexigraphy

Upvoted because this helped me out as well. After mucking around with @JsonIgnoreProperties and adding dummy members, I found this and it did exactly what I needed. Thanks. – Stovall

Answer 20

I fixed this problem by simply changing the signatures of my setter and getter methods of my POJO class. All I had to do was change the getObject method to match what the mapper was looking for. In my case I had a getImageUrl originally, but the JSON data had image_url which was throwing the mapper off. I changed both my setter and getters to getImage_url and setImage_url.

Habergeon

you are right apparently: if the name you want is xxx_yyy The way to call it would be getXxx_yyy and setXxx_yyy. This is very strange but it works. – Rebuke

Answer 21

One other possibility is this property in the application.properties spring.jackson.deserialization.fail-on-unknown-properties=false, which won't need any other code change in your application. And when you believe the contract is stable, you can remove this property or mark it true.

Simpatico

Answer 22

This may not be the same problem that the OP had but in case someone got here with the same mistake I had then this will help them solve their problem. I got the same error as the OP when I used an ObjectMapper from a different dependency as the JsonProperty annotation.

This works:

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.annotation.JsonProperty;

Does NOT work:

import org.codehaus.jackson.map.ObjectMapper; //org.codehaus.jackson:jackson-mapper-asl:1.8.8
import com.fasterxml.jackson.annotation.JsonProperty; //com.fasterxml.jackson.core:jackson-databind:2.2.3

Nancienancy

thanks! import com.fasterxml.jackson.annotation.JsonProperty worked for me instead the other:-) – Toby

omfg this was killing me! ty – Bekelja

This!! Wasted 1 hr trying to figure out what was wrong. – Peaslee

Answer 23

What worked for me, was to make the property public.

Twinkling

It helps! Also class is better to do public. – Adduct

Answer 24

Either Change

public Class Wrapper {
    private List<Student> students;
    //getters & setters here
}

to

public Class Wrapper {
    private List<Student> wrapper;
    //getters & setters here
}

---- or ----

Change your JSON string to

{"students":[{"id":"13","name":"Fred"}]}

Calise

Answer 25

For my part, the only line

@JsonIgnoreProperties(ignoreUnknown = true)

didn't work too.

Just add

@JsonInclude(Include.NON_EMPTY)

Jackson 2.4.0

Fore

Answer 26

This worked perfectly for me

objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

Chaschase

This seems to be just a repeat of many of the existing answers. – Neuberger

Answer 27

Your input

{"wrapper":[{"id":"13","name":"Fred"}]}

indicates that it is an Object, with a field named "wrapper", which is a Collection of Students. So my recommendation would be,

Wrapper = mapper.readValue(jsonStr , Wrapper.class);

where Wrapper is defined as

class Wrapper {
    List<Student> wrapper;
}

Gaye

Answer 28

The new Firebase Android introduced some huge changes ; below the copy of the doc :

[https://firebase.google.com/support/guides/firebase-android] :

Update your Java model objects

As with the 2.x SDK, Firebase Database will automatically convert Java objects that you pass to DatabaseReference.setValue() into JSON and can read JSON into Java objects using DataSnapshot.getValue().

In the new SDK, when reading JSON into a Java object with DataSnapshot.getValue(), unknown properties in the JSON are now ignored by default so you no longer need @JsonIgnoreExtraProperties(ignoreUnknown=true).

To exclude fields/getters when writing a Java object to JSON, the annotation is now called @Exclude instead of @JsonIgnore.

BEFORE

@JsonIgnoreExtraProperties(ignoreUnknown=true)
public class ChatMessage {
   public String name;
   public String message;
   @JsonIgnore
   public String ignoreThisField;
}

dataSnapshot.getValue(ChatMessage.class)

---

AFTER

public class ChatMessage {
   public String name;
   public String message;
   @Exclude
   public String ignoreThisField;
}

dataSnapshot.getValue(ChatMessage.class)

If there is an extra property in your JSON that is not in your Java class, you will see this warning in the log files:

W/ClassMapper: No setter/field for ignoreThisProperty found on class com.firebase.migrationguide.ChatMessage

You can get rid of this warning by putting an @IgnoreExtraProperties annotation on your class. If you want Firebase Database to behave as it did in the 2.x SDK and throw an exception if there are unknown properties, you can put a @ThrowOnExtraProperties annotation on your class.

Autorotation

Answer 29

set public your class fields not private.

public Class Student { 
    public String name;
    public String id;
    //getters & setters for name & id here
}

Nguyen

poor practice - this breaks the encapsulation. – Rooky

Your class is in risk when user uses it because the internal state could be mutated through these fields. – Rooky

Answer 30

If for some reason you cannot add the @JsonIgnoreProperties annotations to your class and you are inside a web server/container such as Jetty. You can create and customize the ObjectMapper inside a custom provider

import javax.ws.rs.ext.ContextResolver;
import javax.ws.rs.ext.Provider;

import com.fasterxml.jackson.annotation.JsonInclude.Include;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.ObjectMapper;

@Provider
public class CustomObjectMapperProvider implements ContextResolver<ObjectMapper> {

    private ObjectMapper objectMapper;

    @Override
    public ObjectMapper getContext(final Class<?> cls) {
        return getObjectMapper();
    }

    private synchronized ObjectMapper getObjectMapper() {
        if(objectMapper == null) {
            objectMapper = new ObjectMapper();
            objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
        }
        return objectMapper;
    }
}

Muskrat

Answer 31

FAIL_ON_UNKNOWN_PROPERTIES option is true by default:

FAIL_ON_UNKNOWN_PROPERTIES (default: true)
Used to control whether encountering of unknown properties (one for which there is no setter; and there is no fallback "any setter" method defined using @JsonAnySetter annotation) should result in a JsonMappingException (when enabled), or just quietly ignored (when disabled)

Chiquia

Answer 32

The POJO should be defined as

Response class

public class Response {
    private List<Wrapper> wrappers;
    // getter and setter
}

Wrapper class

public class Wrapper {
    private String id;
    private String name;
    // getters and setters
}

and mapper to read value

Response response = mapper.readValue(jsonStr , Response.class);

Sams

Almost. Not wrappers, but wrapper. – Accost

@Accost Haha, thank you for catching the error. I naturally use plural for a collection. But per OP's question, it should be singular. – Sams

Answer 33

This may be a very late response, but just changing the POJO to this should solve the json string provided in the problem (since, the input string is not in your control as you said):

public class Wrapper {
    private List<Student> wrapper;
    //getters & setters here
}

Taimi

Answer 34

Google brought me here and i was surprised to see the answers... all suggested bypassing the error ( which always bites back 4 folds later in developement ) rather than solving it until this gentleman restored by faith in SO!

objectMapper.readValue(responseBody, TargetClass.class)

is used to convert a json String to an class object, whats missing is that the TargetClass should have public getter / setters. Same is missing in OP's question snippet too! :)

via lombok your class as below should work!!

@Data
@Builder
public class TargetClass {
    private String a;
}

Rhombic

Answer 35

import com.fasterxml.jackson.annotation.JsonIgnoreProperties;

@JsonIgnoreProperties

Seymourseys

Maybe some further explanations or doc would be nice – Matamoros

@JsonIgnoreProperties(ignoreUnknown = true) works fine, thanks – Jailer

Answer 36

When we are generating getters and setters, specially which starts with 'is' keyword, IDE generally removes the 'is'. e.g.

private boolean isActive;

public void setActive(boolean active) {
   isActive = active;
}

public isActive(){
   return isActive;
}

In my case, i just changed the getter and setter.

private boolean isActive;

public void setIsActive(boolean active) {
   isActive = active;
}

public getIsActive(){
   return isActive;
}

And it was able to recognize the field.

Pursley

Answer 37

I faced similar issue, only difference is i was working with YAML file instead of JSON file. So i was using YamlFactory() when creating an ObjectMapper. Also, the properties of my Java POJO were priviate. However, the solution mentioned in all these answers did not help. I mean, it stopped throwing error and started returning Java object but the properties of the java object would be null.

I just did the following: -

objectMapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);

And without the need of: -

objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

Now, i am able to deserialize data from Yaml file into Java object. All properties are not null now. This solution should work for JSON approach also.

Aulic

Answer 38

In my case it was simple: the REST-service JSON Object was updated (a property was added), but the REST-client JSON Object wasn't. As soon as i've updated JSON client object the 'Unrecognized field ...' exception has vanished.

Repand

This should be a comment – Pigeonhole

Answer 39

In my case a I have to add public getters and setters to leave fields private.

ObjectMapper mapper = new ObjectMapper();
Application application = mapper.readValue(input, Application.class);

I use jackson-databind 2.10.0.pr3 .

Alphonsa

Answer 40

The shortest solution without setter/getter is to add @JsonProperty to a class field:

public class Wrapper {
    @JsonProperty
    private List<Student> wrapper;
}

public class Student {
    @JsonProperty
    private String name;
    @JsonProperty
    private String id;
}

Also, you called students list "wrapper" in your json, so Jackson expects a class with a field called "wrapper".

Derna

Answer 41

Just in case anyone else is using force-rest-api like me, here is how I used this discussion to solve it (Kotlin):

var result = forceApi.getSObject("Account", "idhere")
result.jsonMapper.configure( DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES,  false)
val account: Account = result.`as`(Account::class.java)

It looks like force-rest-api is using an old version of jackson's.

Evapotranspiration

Answer 42

You should just change the field of List from "students" to "wrapper" just the json file and the mapper will look it up.

Bagman

Answer 43

Your json string is not inline with the mapped class. Change the input string

String jsonStr = "{\"students\"\:[{\"id\":\"13\",\"name\":\"Fred\"}]}";

Or change your mapped class

public class Wrapper {
    private List<Student> wrapper;
    //getters & setters here
}

Nauplius

Answer 44

In my case error came due to following reason

• Initially it was working fine,then i renamed one variable,made the changes in code and it gave me this error.

• Then i applied jackson ignorant property also but it did not work.

• Finally after re defining my getters and setters methods according to name of my variable this error was resolved

So make sure to redifine getters and setters also.

Unsocial

Answer 45

Change Wrapper class to

public Class Wrapper {

          @JsonProperty("wrapper")  // add this line
          private List<Student> students;
}

What this does is to recognise the students field as wrapper key of the json object.

Also I personally prefer using Lombok Annotations for Getters and Setters as

@Getter
@Setter
public Class Wrapper {

          @JsonProperty("wrapper")  // add this line
          private List<Student> students;
}

Since I have not tested the above code with Lombok and @JsonProperty together, I'll suggest you to add the following code to Wrapper class as well to override Lombok's default getter and setter.

public List<Student> getWrapper(){
     return students;
}

public void setWrapper(List<Student> students){
     this.students = students;
}

Also check this out to deserialise lists using Jackson.

Meingoldas

Answer 46

As per this documentation, you can use the Jackson2ObjectMapperBuilder to build your ObjectMapper:

@Autowired
Jackson2ObjectMapperBuilder objectBuilder;

ObjectMapper mapper = objectBuilder.build();
String json = "{\"id\": 1001}";

By default, Jackson2ObjectMapperBuilder disables the error unrecognizedpropertyexception.

Regal

Answer 47

Json field:

 "blog_host_url": "some.site.com"

Kotlin field

var blogHostUrl: String = "https://google.com"

In my case i needed just to use @JsonProperty annotation in my dataClass.

Example:

data class DataBlogModel(
       @JsonProperty("blog_host_url") var blogHostUrl: String = "https://google.com"
    )

Here's article: https://www.baeldung.com/jackson-name-of-property

Ania

Answer 48

ObjectMapper objectMapper = new ObjectMapper()
.configure(DeserializationFeature.ACCEPT_EMPTY_ARRAY_AS_NULL_OBJECT, true);

Musquash

Hi there, thanks for answering. Would be great if you could add some detail to the code you posted. What are you trying to achieve, and what is the reasoning behind it? Posting code is great but a bit of description could significantly help the people who is asking. – Zayas

Answer 49

I ran into this once when my JSON payload included a property that was not recognised by the API. The solution was to rename/remove the offending property.

Sitsang

Answer 50

Below is what I tried to suppress the unknown props which did not want to parse. Sharing Kotlin code

val myResponse = jacksonObjectMapper().configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
            .readValue(serverResponse, FooResponse::class.java)

Scansion

Answer 51

You need to verify all fields of the class you are parsing in, make it the same as in your original JSONObject. It helped me and in my case.

@JsonIgnoreProperties(ignoreUnknown = true) didn't help at all.

Miscegenation

it did not help? what you mean? @JsonIgnoreProperties(ignoreUnknown = true) definitely ignores properties which are undefined in java object, but are present in JSON – Dyslexia