My question is kind of similar to Prevent GSON from serializing JSON string but the solution there uses GSON library and I am restricted to using Jackson (fasterxml).
I have an entity class as follows:
package com.dawson.model;
import com.dawson.model.audit.BaseLongEntity;
import lombok.extern.log4j.Log4j;
import javax.persistence.*;
#Table(name = "queue", schema = "dawson")
#Entity
#Log4j
public class Queue extends BaseLongEntity {
protected String requestType;
protected String body;
protected Queue() {
}
public Queue(String requestType, String body) {
this.requestType = requestType;
this.body = body;
}
#Column(name = "request_type")
public String getRequestType() {
return requestType;
}
public void setRequestType(String requestType) {
this.requestType = requestType;
}
#Column(name = "body")
#Lob
public String getBody() {
return body;
}
public void setBody(String body) {
this.body = body;
}
}
I want to populate the body field with the json string representation of a map and then send this as part of the ResponseEntity. Something as follows:
ObjectMapper mapper = new ObjectMapper();
mapper.configure(SerializationFeature.FAIL_ON_EMPTY_BEANS, false);
mapper.disable(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS);
mapper.writer().withDefaultPrettyPrinter();
HashMap<String, String> map = new HashMap<>(5);
map.put("inquiry", "How Can I solve the problem with Jackson double serialization of strings?");
map.put("phone", "+12345677890");
Queue queue = null;
try {
queue = new Queue("General Inquiry", mapper.writeValueAsString(map));
} catch (JsonProcessingException e) {
e.printStackTrace();
}
String test = mapper.writeValueAsString(map)
System.out.println(test);
Expected Output: "{"requestType": "General Inquiry","body": "{"inquiry":"How Can I solve the problem with Jackson double serialization of strings?","phone":"+12345677890"}"}"
Actual Output:"{"requestType": "General Inquiry","body": "{\"inquiry\":\"How Can I solve the problem with Jackson double serialization of strings?\",\"phone\":\"+12345677890\"}"}"
I am using
Jackson Core v2.8.2
I tried playing with
#JsonIgnore
and
#JsonProperty
tags but that doesn't help because my field is already serialized from the map when writing to the Entity.
Add the #JsonRawValue annotation to the body property. This makes Jackson treat the contents of the property as a literal JSON value, that should not be processed.
Be aware that Jackson doesn't do any validation of the field's contents, which makes it dangerously easy to produce invalid JSON.
Related
In my Spring project I have several objects that should be serialized to a specific JSON format.
public class Person {
private Integer id;
private String name;
private String height;
private Address address;
}
and
public class Address {
private String street;
private String city;
private String phone;
}
Let assume that Person.height and Address.phone should not appear in the JSON.
The resulting JSON should look like
{
"attributes": ["id", "name", "street", "city"],
"values": [12345, "Mr. Smith", "Main street", "Chicago"]
}
I can create create a standard JSON with an ObjectMapper and some annotations like #JsonProperty and #JsonUnwrapped where I disable some SerializationFeatures. But at the moment I'm not able to create such a JSON.
Is there an easy way to create this JSON? And how would the way back (deserialization) look like?
There are good reasons Jackson doesn't serializes maps in this format. It's less readable and also harder to deserialize properly.
But if you just create another POJO it's very easy to achieve what you want to do:
public class AttributeList {
public static AttributeList from(Object o) {
return from(new ObjectMapper().convertValue(o, new TypeReference<Map<String, Object>>() {}));
}
public static AttributeList from(Map<String, Object> attributes) {
return new AttributeList(attributes);
}
private final List<String> attributes;
private final List<Object> values;
private AttributeList(Map<String, Object> o) {
attributes = new ArrayList<>(o.keySet());
values = new ArrayList<>(o.values());
}
}
At the deserialization process (which as I understand is the process of converting JSON data into a Java Object), how can I tell Jackson that when it reads a object that contains no data, it should be ignored?
I'm using Jackson 2.6.6 and Spring 4.2.6
The JSON data received by my controller is as follows:
{
"id": 2,
"description": "A description",
"containedObject": {}
}
The problem is that the object "containedObject" is interpreted as is and it's being instantiated. Therefore, as soon as my controller reads this JSON data, it produces an instance of the ContainedObject object type but I need this to be null instead.
The easiest and fastest solution would be that in the JSON data received, this value be null like this:
{
"id": 2,
"description": "A description",
"containedObject": null
}
But this isn't possible since I'm not in control of the JSON data that is sent to me.
Is there an annotation (like this explained here) that works for the deserialization process and could be helpfull in my situation?
I leave a representation of my classes for more information:
My entity class is as follows:
public class Entity {
private long id;
private String description;
private ContainedObject containedObject;
//Contructor, getters and setters omitted
}
And my contained object class as follows:
public class ContainedObject {
private long contObjId;
private String aString;
//Contructor, getters and setters omitted
}
I would use a JsonDeserializer. Inspect the field in question, determine, if it is emtpy and return null, so your ContainedObject would be null.
Something like this (semi-pseudo):
public class MyDes extends JsonDeserializer<ContainedObject> {
#Override
public String deserialize(JsonParser jsonParser, DeserializationContext context) throws IOException, JsonProcessingException {
//read the JsonNode and determine if it is empty JSON object
//and if so return null
if (node is empty....) {
return null;
}
return node;
}
}
then in your model:
public class Entity {
private long id;
private String description;
#JsonDeserialize(using = MyDes.class)
private ContainedObject containedObject;
//Contructor, getters and setters omitted
}
Hope this helps!
You can implement a custom deserializer as follows:
public class Entity {
private long id;
private String description;
#JsonDeserialize(using = EmptyToNullObject.class)
private ContainedObject containedObject;
//Contructor, getters and setters omitted
}
public class EmptyToNullObject extends JsonDeserializer<ContainedObject> {
public ContainedObject deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException, JsonProcessingException {
JsonNode node = jp.getCodec().readTree(jp);
long contObjId = (Long) ((LongNode) node.get("contObjId")).numberValue();
String aString = node.get("aString").asText();
if(aString.equals("") && contObjId == 0L) {
return null;
} else {
return new ContainedObject(contObjId, aString);
}
}
}
Approach 1 : This is mostly used. #JsonInclude is used to exclude properties with empty/null/default values.Use #JsonInclude(JsonInclude.Include.NON_NULL) or #JsonInclude(JsonInclude.Include.NON_EMPTY) as per your requirement.
#JsonInclude(JsonInclude.Include.NON_NULL)
public class Employee {
private String empId;
private String firstName;
#JsonInclude(JsonInclude.Include.NON_NULL)
private String lastName;
private String address;
private String emailId;
}
More info about the jackson annotations : https://github.com/FasterXML/jackson-annotations/wiki/Jackson-Annotations
Approach 2 : GSON
use GSON (https://code.google.com/p/google-gson/)
I have a controller that receives a JSON RequestBody like the one below:
{
"status":"pending",
"status1":"pending",
"status2":"pending",
"status3":"pending",
"specs":{
"width":1000,
"height":1507,
........some other fields, any number of fields..
},
}
I have an Entity
#Entity
public class MyBean implements Serializable {
#Id
#GeneratedValue
Long id;
public String status;
public String status1;
public String status2;
public String status3;
}
and my Controller class:
#RequestMapping(value = "/insert", method = RequestMethod.POST)
public void insert(#RequestBody MyBean myBean) {
return myService.save(myBean);
}
My problem is that I want to store the value of specs (which is a JSON document) as a String in a Lob field and I don't know how to do that.
Declare specs as public Map<String,Integer> specs;
and to convert Object to json use Jackson fasterxml api as below
MyBean bean=new MyBean();
bean.setId(new Long(1));
bean.setStatus("pending");
bean.setStatus1("pending");
bean.setStatus2("pending");
bean.setStatus3("pending");
Map<String, Integer> temp=new HashMap<String, Integer>();
temp.put("width",1000);
temp.put("height",1507);
bean.setSpecs(temp);
//Object to json
StringWriter sw=new StringWriter();
ObjectMapper mapper=new ObjectMapper();
mapper.writeValue(sw,bean);
System.out.println(sw.toString());
//json to object
MyBean newBean=mapper.readValue(sw.toString(), MyBean.class);
System.out.println(newBean.toString());
If you want serialize only specs before saving then use same mapper.writeValue() function to convert specs to json string
for more information refer this
I have a json object like this
{
"id":23 ,
"key": "AKEY",
"description": {
"plain": {
"value": "This is an example",
"representation": "plain"
}
}
}
I'd like to map it to this object
public class JsonResponse {
private int id;
private String key;
private String name;
private String type;
private String description;
/*usual getters and setters*/
}
I use the JSONSerialiser like this
JSONObject jsonObject = (JSONObject) JSONSerializer.toJSON(responseEntity);
But how do I map the "description.plain.value" to "JsonResponse.description"?
Can this be done using jackson annotations?
thanks for your help
I found this post Binding JSON child object property into Java object field in Jackson that partially solved my problem.
I wrote two setDescription() methods, one used by myself in my code, and one that's been called by jacskon
#JsonProperty(value = "description")
public void setDescription(Map<String, Map<String,String>> description) {
this.description = description.get("plain").get("value");
}
public void setDescription(String description) {
this.description = description;
}
It looks like the JsonProperty annotation is required to make jackson use the right setter.
Still I'm ok with this as long as it is a "short nested" property, but I guess Beri response is more acceptable with complex JSON responses.
We have a big table with a lot of columns. After we moved to MySQL Cluster, the table cannot be created because of:
ERROR 1118 (42000): Row size too large. The maximum row size for the used table type, not counting BLOBs, is 14000. This includes storage overhead, check the manual. You have to change some columns to TEXT or BLOBs
As an example:
#Entity #Table (name = "appconfigs", schema = "myproject")
public class AppConfig implements Serializable
{
#Id #Column (name = "id", nullable = false)
#GeneratedValue (strategy = GenerationType.IDENTITY)
private int id;
#OneToOne #JoinColumn (name = "app_id")
private App app;
#Column(name = "param_a")
private ParamA parama;
#Column(name = "param_b")
private ParamB paramb;
}
It's a table for storing configuration parameters. I was thinking that we can combine some columns into one and store it as JSON object and convert it to some Java object.
For example:
#Entity #Table (name = "appconfigs", schema = "myproject")
public class AppConfig implements Serializable
{
#Id #Column (name = "id", nullable = false)
#GeneratedValue (strategy = GenerationType.IDENTITY)
private int id;
#OneToOne #JoinColumn (name = "app_id")
private App app;
#Column(name = "params")
//How to specify that this should be mapped to JSON object?
private Params params;
}
Where we have defined:
public class Params implements Serializable
{
private ParamA parama;
private ParamB paramb;
}
By using this we can combine all columns into one and create our table. Or we can split the whole table into several tables. Personally I prefer the first solution.
Anyway my question is how to map the Params column which is text and contains JSON string of a Java object?
You can use a JPA converter to map your Entity to the database.
Just add an annotation similar to this one to your params field:
#Convert(converter = JpaConverterJson.class)
and then create the class in a similar way (this converts a generic Object, you may want to specialize it):
#Converter(autoApply = true)
public class JpaConverterJson implements AttributeConverter<Object, String> {
private final static ObjectMapper objectMapper = new ObjectMapper();
#Override
public String convertToDatabaseColumn(Object meta) {
try {
return objectMapper.writeValueAsString(meta);
} catch (JsonProcessingException ex) {
return null;
// or throw an error
}
}
#Override
public Object convertToEntityAttribute(String dbData) {
try {
return objectMapper.readValue(dbData, Object.class);
} catch (IOException ex) {
// logger.error("Unexpected IOEx decoding json from database: " + dbData);
return null;
}
}
}
That's it: you can use this class to serialize any object to json in the table.
The JPA AttributeConverter is way too limited to map JSON object types, especially if you want to save them as JSON binary.
You don’t have to create a custom Hibernate Type to get JSON support, All you need to do is use the Hibernate Types OSS project.
For instance, if you're using Hibernate 5.2 or newer versions, then you need to add the following dependency in your Maven pom.xml configuration file:
<dependency>
<groupId>com.vladmihalcea</groupId>
<artifactId>hibernate-types-52</artifactId>
<version>${hibernate-types.version}</version>
</dependency>
Now, you need to declare the new type either at the entity attribute level or, even better, at the class level in a base class using #MappedSuperclass:
#TypeDef(name = "json", typeClass = JsonType.class)
And the entity mapping will look like this:
#Type(type = "json")
#Column(columnDefinition = "json")
private Location location;
If you're using Hibernate 5.2 or later, then the JSON type is registered automatically by MySQL57Dialect.
Otherwise, you need to register it yourself:
public class MySQLJsonDialect extends MySQL55Dialect {
public MySQLJsonDialect() {
super();
this.registerColumnType(Types.JAVA_OBJECT, "json");
}
}
And, set the hibernate.dialect Hibernate property to use the fully-qualified class name of the MySQLJsonDialect class you have just created.
If you need to map json type property to json format when responding to the client (e.g. rest API response), add #JsonRawValue as the following:
#Column(name = "params", columnDefinition = "json")
#JsonRawValue
private String params;
This might not do the DTO mapping for server-side use, but the client will get the property properly formatted as json.
It is simple
#Column(name = "json_input", columnDefinition = "json")
private String field;
and in mysql database your column 'json_input' json type
There is a workaround for those don't want write too much code.
Frontend -> Encode your JSON Object to string base64 in POST method, decode it to json in GET method
In POST Method
data.components = btoa(JSON.stringify(data.components));
In GET
data.components = JSON.parse(atob(data.components))
Backend -> In your JPA code, change the column to String or BLOB, no need Convert.
#Column(name = "components", columnDefinition = "json")
private String components;
In this newer version of spring boot and MySQL below code is enough
#Column( columnDefinition = "json" )
private String string;
I was facing quotes issue so I commented below line in my project
#spring.jpa.properties.hibernate.globally_quoted_identifiers=true
I had a similar problem, and solved it by using #Externalizer annotation and Jackson to serialize/deserialize data (#Externalizer is OpenJPA-specific annotation, so you have to check with your JPA implementation similar possibility).
#Persistent
#Column(name = "params")
#Externalizer("toJSON")
private Params params;
Params class implementation:
public class Params {
private static final ObjectMapper mapper = new ObjectMapper();
private Map<String, Object> map;
public Params () {
this.map = new HashMap<String, Object>();
}
public Params (Params another) {
this.map = new HashMap<String, Object>();
this.map.putAll(anotherHolder.map);
}
public Params(String string) {
try {
TypeReference<Map<String, Object>> typeRef = new TypeReference<Map<String, Object>>() {
};
if (string == null) {
this.map = new HashMap<String, Object>();
} else {
this.map = mapper.readValue(string, typeRef);
}
} catch (IOException e) {
throw new PersistenceException(e);
}
}
public String toJSON() throws PersistenceException {
try {
return mapper.writeValueAsString(this.map);
} catch (IOException e) {
throw new PersistenceException(e);
}
}
public boolean containsKey(String key) {
return this.map.containsKey(key);
}
// Hash map methods
public Object get(String key) {
return this.map.get(key);
}
public Object put(String key, Object value) {
return this.map.put(key, value);
}
public void remove(String key) {
this.map.remove(key);
}
public Object size() {
return map.size();
}
}
HTH
If you are using JPA version 2.1 or higher you can go with this case.
Link Persist Json Object
public class HashMapConverter implements AttributeConverter<Map<String, Object>, String> {
#Override
public String convertToDatabaseColumn(Map<String, Object> customerInfo) {
String customerInfoJson = null;
try {
customerInfoJson = objectMapper.writeValueAsString(customerInfo);
} catch (final JsonProcessingException e) {
logger.error("JSON writing error", e);
}
return customerInfoJson;
}
#Override
public Map<String, Object> convertToEntityAttribute(String customerInfoJSON) {
Map<String, Object> customerInfo = null;
try {
customerInfo = objectMapper.readValue(customerInfoJSON,
new TypeReference<HashMap<String, Object>>() {});
} catch (final IOException e) {
logger.error("JSON reading error", e);
}
return customerInfo;
}
}
A standard JSON object would represent those attributes as a HashMap:
#Convert(converter = HashMapConverter.class)
private Map<String, Object> entityAttributes;