Ignore fields from Java object dynamically while sending as JSON from Spring MVC

Question

I have model class like this, for hibernate

@Entity
@Table(name = "user", catalog = "userdb")
@JsonIgnoreProperties(ignoreUnknown = true)
public class User implements java.io.Serializable {

    private Integer userId;
    private String userName;
    private String emailId;
    private String encryptedPwd;
    private String createdBy;
    private String updatedBy;

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "UserId", unique = true, nullable = false)
    public Integer getUserId() {
        return this.userId;
    }

    public void setUserId(Integer userId) {
        this.userId = userId;
    }

    @Column(name = "UserName", length = 100)
    public String getUserName() {
        return this.userName;
    }

    public void setUserName(String userName) {
        this.userName = userName;
    }

    @Column(name = "EmailId", nullable = false, length = 45)
    public String getEmailId() {
        return this.emailId;
    }

    public void setEmailId(String emailId) {
        this.emailId = emailId;
    }

    @Column(name = "EncryptedPwd", length = 100)
    public String getEncryptedPwd() {
        return this.encryptedPwd;
    }

    public void setEncryptedPwd(String encryptedPwd) {
        this.encryptedPwd = encryptedPwd;
    }

    public void setCreatedBy(String createdBy) {
        this.createdBy = createdBy;
    }

    @Column(name = "UpdatedBy", length = 100)
    public String getUpdatedBy() {
        return this.updatedBy;
    }

    public void setUpdatedBy(String updatedBy) {
        this.updatedBy = updatedBy;
    }
}

In Spring MVC controller, using DAO, I am able to get the object. and returning as JSON Object.

@Controller
public class UserController {

    @Autowired
    private UserService userService;

    @RequestMapping(value = "/getUser/{userId}", method = RequestMethod.GET)
    @ResponseBody
    public User getUser(@PathVariable Integer userId) throws Exception {

        User user = userService.get(userId);
        user.setCreatedBy(null);
        user.setUpdatedBy(null);
        return user;
    }
}

View part is done using AngularJS, so it will get JSON like this

{
  "userId" :2,
  "userName" : "john",
  "emailId" : "john@gmail.com",
  "encryptedPwd" : "Co7Fwd1fXYk=",
  "createdBy" : null,
  "updatedBy" : null
}

If I don't want to set encrypted Password, I will set that field also as null.

But I don't want like this, I dont want to send all fields to client side. If I dont want password, updatedby, createdby fields to send, My result JSON should be like

{
  "userId" :2,
  "userName" : "john",
  "emailId" : "john@gmail.com"
}

The list of fields which I don't want to send to client coming from other database table. So it will change based on the user who is logged in. How can I do that?

I hope You got my question.

Answer 1

Add the @JsonIgnoreProperties("fieldname") annotation to your POJO.

Or you can use @JsonIgnore before the name of the field you want to ignore while deserializing JSON. Example:

@JsonIgnore
@JsonProperty(value = "user_password")
public String getUserPassword() {
    return userPassword;
}

GitHub example

Answer 2

I know I'm a bit late to the party, but I actually ran into this as well a few months back. All of the available solutions weren't very appealing to me (mixins? ugh!), so I ended up creating a new library to make this process cleaner. It's available here if anyone would like to try it out: https://github.com/monitorjbl/spring-json-view.

The basic usage is pretty simple, you use the JsonView object in your controller methods like so:

import com.monitorjbl.json.JsonView;
import static com.monitorjbl.json.Match.match;

@RequestMapping(method = RequestMethod.GET, value = "/myObject")
@ResponseBody
public void getMyObjects() {
    //get a list of the objects
    List<MyObject> list = myObjectService.list();

    //exclude expensive field
    JsonView.with(list).onClass(MyObject.class, match().exclude("contains"));
}

You can also use it outside of Spring:

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.module.SimpleModule;
import static com.monitorjbl.json.Match.match;

ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addSerializer(JsonView.class, new JsonViewSerializer());
mapper.registerModule(module);

mapper.writeValueAsString(JsonView.with(list)
      .onClass(MyObject.class, match()
        .exclude("contains"))
      .onClass(MySmallObject.class, match()
        .exclude("id"));

Answer 3

Can I do it dynamically?

Create view class:

public class View {
    static class Public { }
    static class ExtendedPublic extends Public { }
    static class Internal extends ExtendedPublic { }
}

Annotate you model

@Document
public class User {

    @Id
    @JsonView(View.Public.class)
    private String id;

    @JsonView(View.Internal.class)
    private String email;

    @JsonView(View.Public.class)
    private String name;

    @JsonView(View.Public.class)
    private Instant createdAt = Instant.now();
    // getters/setters
}

Specify the view class in your controller

@RequestMapping("/user/{email}")
public class UserController {

    private final UserRepository userRepository;

    @Autowired
    UserController(UserRepository userRepository) {
        this.userRepository = userRepository;
    }

    @RequestMapping(method = RequestMethod.GET)
    @JsonView(View.Internal.class)
    public @ResponseBody Optional<User> get(@PathVariable String email) {
        return userRepository.findByEmail(email);
    }

}

Data example:

{"id":"5aa2496df863482dc4da2067","name":"test","createdAt":"2018-03-10T09:35:31.050353800Z"}

Answer 4

Yes, you can specify which fields are serialized as JSON response and which to ignore. This is what you need to do to implement Dynamically ignore properties.

1) First, you need to add @JsonFilter from com.fasterxml.jackson.annotation.JsonFilter on your entity class as.

import com.fasterxml.jackson.annotation.JsonFilter;

@JsonFilter("SomeBeanFilter")
public class SomeBean {

  private String field1;

  private String field2;

  private String field3;

  // getters/setters
}

2) Then in your controller, you have to add create the MappingJacksonValue object and set filters on it and in the end, you have to return this object.

import java.util.Arrays;
import java.util.List;

import org.springframework.http.converter.json.MappingJacksonValue;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RestController;

import com.fasterxml.jackson.databind.ser.FilterProvider;
import com.fasterxml.jackson.databind.ser.impl.SimpleBeanPropertyFilter;
import com.fasterxml.jackson.databind.ser.impl.SimpleFilterProvider;

@RestController
public class FilteringController {

  // Here i want to ignore all properties except field1,field2.
  @GetMapping("/ignoreProperties")
  public MappingJacksonValue retrieveSomeBean() {
    SomeBean someBean = new SomeBean("value1", "value2", "value3");

    SimpleBeanPropertyFilter filter = SimpleBeanPropertyFilter.filterOutAllExcept("field1", "field2");

    FilterProvider filters = new SimpleFilterProvider().addFilter("SomeBeanFilter", filter);

    MappingJacksonValue mapping = new MappingJacksonValue(someBean);

    mapping.setFilters(filters);

    return mapping;
  }
}

This is what you will get in response:

{
  field1:"value1",
  field2:"value2"
}

instead of this:

{
  field1:"value1",
  field2:"value2",
  field3:"value3"
}

Here you can see it ignores other properties(field3 in this case) in response except for property field1 and field2.

Hope this helps.

Answer 5

We can do this by setting access to JsonProperty.Access.WRITE_ONLY while declaring the property.

@JsonProperty( value = "password", access = JsonProperty.Access.WRITE_ONLY)
@SerializedName("password")
private String password;

Answer 6

Add @JsonInclude(JsonInclude.Include.NON_NULL) (forces Jackson to serialize null values) to the class as well as @JsonIgnore to the password field.

You could of course set @JsonIgnore on createdBy and updatedBy as well if you always want to ignore then and not just in this specific case.

UPDATE

In the event that you do not want to add the annotation to the POJO itself, a great option is Jackson's Mixin Annotations. Check out the documentation

Answer 7

If I were you and wanted to do so, I wouldn't use my User entity in Controller layer.Instead I create and use UserDto (Data transfer object) to communicate with business(Service) layer and Controller. You can use Apache BeanUtils(copyProperties method) to copy data from User entity to UserDto.

Answer 8

I have created a JsonUtil which can be used to ignore fields at runtime while giving a response.

Example Usage : First argument should be any POJO class (Student) and ignoreFields is comma seperated fields you want to ignore in response.

 Student st = new Student();
 createJsonIgnoreFields(st,"firstname,age");

import java.util.logging.Logger;

import org.codehaus.jackson.map.ObjectMapper;
import org.codehaus.jackson.map.ObjectWriter;
import org.codehaus.jackson.map.ser.FilterProvider;
import org.codehaus.jackson.map.ser.impl.SimpleBeanPropertyFilter;
import org.codehaus.jackson.map.ser.impl.SimpleFilterProvider;

public class JsonUtil {

  public static String createJsonIgnoreFields(Object object, String ignoreFields) {
     try {
         ObjectMapper mapper = new ObjectMapper();
         mapper.getSerializationConfig().addMixInAnnotations(Object.class, JsonPropertyFilterMixIn.class);
         String[] ignoreFieldsArray = ignoreFields.split(",");
         FilterProvider filters = new SimpleFilterProvider()
             .addFilter("filter properties by field names",
                 SimpleBeanPropertyFilter.serializeAllExcept(ignoreFieldsArray));
         ObjectWriter writer = mapper.writer().withFilters(filters);
         return writer.writeValueAsString(object);
     } catch (Exception e) {
         //handle exception here
     }
     return "";
   }

   public static String createJson(Object object) {
        try {
         ObjectMapper mapper = new ObjectMapper();
         ObjectWriter writer = mapper.writer().withDefaultPrettyPrinter();
         return writer.writeValueAsString(object);
        }catch (Exception e) {
         //handle exception here
        }
        return "";
   }
 }    

Answer 9

I've solved using only @JsonIgnore like @kryger has suggested. So your getter will become:

@JsonIgnore
public String getEncryptedPwd() {
    return this.encryptedPwd;
}

You can set @JsonIgnore of course on field, setter or getter like described here.

And, if you want to protect encrypted password only on serialization side (e.g. when you need to login your users), add this @JsonProperty annotation to your field:

@JsonProperty(access = Access.WRITE_ONLY)
private String encryptedPwd;

More info here.

Answer 10

I've found a solution for me with Spring and jackson

First specify the filter name in the entity

@Entity
@Table(name = "SECTEUR")
@JsonFilter(ModelJsonFilters.SECTEUR_FILTER)
public class Secteur implements Serializable {

/** Serial UID */
private static final long serialVersionUID = 5697181222899184767L;

/**
 * Unique ID
 */
@Id
@JsonView(View.SecteurWithoutChildrens.class)
@Column(name = "id")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;

@JsonView(View.SecteurWithoutChildrens.class)
@Column(name = "code", nullable = false, length = 35)
private String code;

/**
 * Identifiant du secteur parent
 */
@JsonView(View.SecteurWithoutChildrens.class)
@Column(name = "id_parent")
private Long idParent;

@OneToMany(fetch = FetchType.LAZY)
@JoinColumn(name = "id_parent")
private List<Secteur> secteursEnfants = new ArrayList<>(0);

}

Then you can see the constants filters names class with the default FilterProvider used in spring configuration

public class ModelJsonFilters {

public final static String SECTEUR_FILTER = "SecteurFilter";
public final static String APPLICATION_FILTER = "ApplicationFilter";
public final static String SERVICE_FILTER = "ServiceFilter";
public final static String UTILISATEUR_FILTER = "UtilisateurFilter";

public static SimpleFilterProvider getDefaultFilters() {
    SimpleBeanPropertyFilter theFilter = SimpleBeanPropertyFilter.serializeAll();
    return new SimpleFilterProvider().setDefaultFilter(theFilter);
}

}

Spring configuration :

@EnableWebMvc
@Configuration
@ComponentScan(basePackages = "fr.sodebo")

public class ApiRootConfiguration extends WebMvcConfigurerAdapter {

@Autowired
private EntityManagerFactory entityManagerFactory;


/**
 * config qui permet d'éviter les "Lazy loading Error" au moment de la
 * conversion json par jackson pour les retours des services REST<br>
 * on permet à jackson d'acceder à sessionFactory pour charger ce dont il a
 * besoin
 */
@Override
public void configureMessageConverters(List<HttpMessageConverter<?>> converters) {

    super.configureMessageConverters(converters);
    MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();
    ObjectMapper mapper = new ObjectMapper();

    // config d'hibernate pour la conversion json
    mapper.registerModule(getConfiguredHibernateModule());//

    // inscrit les filtres json
    subscribeFiltersInMapper(mapper);

    // config du comportement de json views
    mapper.configure(MapperFeature.DEFAULT_VIEW_INCLUSION, false);

    converter.setObjectMapper(mapper);
    converters.add(converter);
}

/**
 * config d'hibernate pour la conversion json
 * 
 * @return Hibernate5Module
 */
private Hibernate5Module getConfiguredHibernateModule() {
    SessionFactory sessionFactory = entityManagerFactory.unwrap(SessionFactory.class);
    Hibernate5Module module = new Hibernate5Module(sessionFactory);
    module.configure(Hibernate5Module.Feature.FORCE_LAZY_LOADING, true);

    return module;

}

/**
 * inscrit les filtres json
 * 
 * @param mapper
 */
private void subscribeFiltersInMapper(ObjectMapper mapper) {

    mapper.setFilterProvider(ModelJsonFilters.getDefaultFilters());

}

}

Endly I can specify a specific filter in restConstoller when i need....

@RequestMapping(value = "/{id}/droits/", method = RequestMethod.GET)
public MappingJacksonValue getListDroits(@PathVariable long id) {

    LOGGER.debug("Get all droits of user with id {}", id);

    List<Droit> droits = utilisateurService.findDroitsDeUtilisateur(id);

    MappingJacksonValue value;

    UtilisateurWithSecteurs utilisateurWithSecteurs = droitsUtilisateur.fillLists(droits).get(id);

    value = new MappingJacksonValue(utilisateurWithSecteurs);

    FilterProvider filters = ModelJsonFilters.getDefaultFilters().addFilter(ModelJsonFilters.SECTEUR_FILTER, SimpleBeanPropertyFilter.serializeAllExcept("secteursEnfants")).addFilter(ModelJsonFilters.APPLICATION_FILTER,
            SimpleBeanPropertyFilter.serializeAllExcept("services"));

    value.setFilters(filters);
    return value;

}

Answer 11

Place @JsonIgnore on the field or its getter, or create a custom dto

@JsonIgnore
private String encryptedPwd;

or as mentioned above by ceekay annotate it with @JsonProperty where access attribute is set to write only

@JsonProperty( value = "password", access = JsonProperty.Access.WRITE_ONLY)
private String encryptedPwd;

Answer 12

Would not creating a UserJsonResponse class and populating with the wanted fields be a cleaner solution?

Returning directly a JSON seems a great solution when you want to give all the model back. Otherwise it just gets messy.

In the future, for example you might want to have a JSON field that does not match any Model field and then you're in a bigger trouble.

Answer 13

This is a clean utility tool for the above answer :

@GetMapping(value = "/my-url")
public @ResponseBody
MappingJacksonValue getMyBean() {
    List<MyBean> myBeans = Service.findAll();
    MappingJacksonValue mappingValue = MappingFilterUtils.applyFilter(myBeans, MappingFilterUtils.JsonFilterMode.EXCLUDE_FIELD_MODE, "MyFilterName", "myBiggerObject.mySmallerObject.mySmallestObject");
    return mappingValue;
}

//AND THE UTILITY CLASS
public class MappingFilterUtils {

    public enum JsonFilterMode {
        INCLUDE_FIELD_MODE, EXCLUDE_FIELD_MODE
    }
    public static MappingJacksonValue applyFilter(Object object, final JsonFilterMode mode, final String filterName, final String... fields) {
        if (fields == null || fields.length == 0) {
            throw new IllegalArgumentException("You should pass at least one field");
        }
        return applyFilter(object, mode, filterName, new HashSet<>(Arrays.asList(fields)));
    }

    public static MappingJacksonValue applyFilter(Object object, final JsonFilterMode mode, final String filterName, final Set<String> fields) {
        if (fields == null || fields.isEmpty()) {
            throw new IllegalArgumentException("You should pass at least one field");
        }

        SimpleBeanPropertyFilter filter = null;
        switch (mode) {
            case EXCLUDE_FIELD_MODE:
                filter = SimpleBeanPropertyFilter.serializeAllExcept(fields);
                break;
            case INCLUDE_FIELD_MODE:
                filter = SimpleBeanPropertyFilter.filterOutAllExcept(fields);
                break;
        }

        FilterProvider filters = new SimpleFilterProvider().addFilter(filterName, filter);
        MappingJacksonValue mapping = new MappingJacksonValue(object);
        mapping.setFilters(filters);
        return mapping;
    }
}

Answer 14

To acheive dynamic filtering follow the link - https://iamvickyav.medium.com/spring-boot-dynamically-ignore-fields-while-converting-java-object-to-json-e8d642088f55

1. Add the @JsonFilter("Filter name") annotation to the model class.

2. Inside the controller function add the code:-

   SimpleBeanPropertyFilter simpleBeanPropertyFilter =
           SimpleBeanPropertyFilter.serializeAllExcept("id", "dob"); 
   
   FilterProvider filterProvider = new SimpleFilterProvider()
           .addFilter("Filter name", simpleBeanPropertyFilter);
   
   List<User> userList = userService.getAllUsers();
   MappingJacksonValue mappingJacksonValue = new MappingJacksonValue(userList);
   mappingJacksonValue.setFilters(filterProvider);
   
   return mappingJacksonValue;
   

3. make sure the return type is MappingJacksonValue.

Answer 15

In your entity class add @JsonInclude(JsonInclude.Include.NON_NULL) annotation to resolve the problem

it will look like

@Entity
@JsonInclude(JsonInclude.Include.NON_NULL)