You can go through this tutorial. Hope it will help you.

  1. How to convert Java object to / from JSON (Jackson)
  2. https://dzone.com/articles/deserializing-json-java-object
Answer from SkyWalker on Stack Overflow
🌐
GeeksforGeeks
geeksforgeeks.org › java › deserialize-json-array-to-a-single-java-object-with-jackson
Deserialize JSON Array to a Single Java Object with Jackson - GeeksforGeeks
July 23, 2025 - The Course class defines a POJO to map JSON data with id, courseName, and coursePrice properties. Getters and setters are provided for Jackson to populate objects. After creating POJO class, we will add logic of Deserialization of Java Array to Java Object in the main class. ... package org.example; import com.fasterxml.jackson.databind.ObjectMapper; public class Main { public static void main(String[] args) { // JSON data representing an array of Course objects String json = "[{\n" + " \"id\": 1,\n" + " \"courseName\": \"Introduction to Java\",\n" + " \"coursePrice\": 30000.99\n" + "}\n]"; //
Discussions

Deserialize JSON file into Java object - Stack Overflow
For example, I have a JSON file named simple.json which has the below code. More on stackoverflow.com
🌐 stackoverflow.com
Deserialization of JSON from an API
However, I cannot figure out a consistent way to deserialize it (using jackson databind) as the players are each an object/attribute of the main object instead of being items in an array. In the past I have made "template" classes of other APIs like this one. package org.example.Data; public class Adp { public Player player; public double value; public int overallRank; public int positionRank; } package org.example.Data; import java... More on reddit.com
🌐 r/learnjava
9
4
June 24, 2024
Having trouble with deserialization of JSON String to a Java object using Gson
But in my case, JSON fields such as String APP_DATA.ITEM_INTIAL; and String APP_DATA.ITEM_NUMBER; would not be valid java variable names. You can add an annotation to fields so that the JSON-serialized property name is different from the Java field name. If you check out the Gson user's guide , there's a section titled "JSON Field Naming Support" which explains how to do this. There are also lots of examples if you google "gson rename field". Some old versions of Gson had limitations on which characters were allowed in JSON keys, but if you're using the latest version you shouldn't have issues. More on reddit.com
🌐 r/learnprogramming
3
2
August 29, 2022
Why can't I deserialize JSON that I had serialized?
Please ensure that: Your code is properly formatted as code block - see the sidebar (About on mobile) for instructions You include any and all error messages in full You ask clear questions You demonstrate effort in solving your question/problem - plain posting your assignments is forbidden (and such posts will be removed) as is asking for or giving solutions. Trying to solve problems on your own is a very important skill. Also, see Learn to help yourself in the sidebar If any of the above points is not met, your post can and will be removed without further warning. Code is to be formatted as code block (old reddit: empty line before the code, each code line indented by 4 spaces, new reddit: https://i.imgur.com/EJ7tqek.png ) or linked via an external code hoster, like pastebin.com, github gist, github, bitbucket, gitlab, etc. Please, do not use triple backticks (```) as they will only render properly on new reddit, not on old reddit. Code blocks look like this: public class HelloWorld { public static void main(String[] args) { System.out.println("Hello World!"); } } You do not need to repost unless your post has been removed by a moderator. Just use the edit function of reddit to make sure your post complies with the above. If your post has remained in violation of these rules for a prolonged period of time (at least an hour), a moderator may remove it at their discretion. In this case, they will comment with an explanation on why it has been removed, and you will be required to resubmit the entire post following the proper procedures. To potential helpers Please, do not help if any of the above points are not met, rather report the post. We are trying to improve the quality of posts here. In helping people who can't be bothered to comply with the above points, you are doing the community a disservice. I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns. More on reddit.com
🌐 r/javahelp
6
1
October 30, 2025
🌐
Baeldung
baeldung.com › home › json › jackson › getting started with custom deserialization in jackson
Getting Started with Deserialization in Jackson | Baeldung
September 3, 2024 - Let’s implement a custom deserializer for this case. First, we need to implement the ContextualDeserializer interface to get the entity type inside the Wrapper. We’ll do this by overriding the createContextual() method. When this method is called, the context is resolved, and the actual content of the Wrapper can be retrieved via the BeanProperty argument. We also have to extend JsonDeserializer.
🌐
Medium
medium.com › @bubu.tripathy › json-serialization-and-deserialization-in-java-2a3f08266b70
JSON Serialization and Deserialization in Java | by Bubu Tripathy | Medium
April 9, 2023 - In this example, we create a JSON string containing the same data as in the previous example. We then call the `readValue()` method of the `ObjectMapper` class, passing in the JSON string and the class of the Java object we want to deserialize to.
🌐
Rip Tutorial
riptutorial.com › deserialize json collection to collection of objects using jackson
Java Language Tutorial => Deserialize JSON collection to collection...
CollectionType mapType = factory.constructMapLikeType(Map.class, String.class, Person.class); List<Person> list = mapper.readValue(jsonString, mapType);
🌐
ZetCode
zetcode.com › java › json
Java JSON - JSON serialization and deserialization in Java with JSON-Java
January 27, 2024 - JSONArray is an ordered sequence of values. Its external text form is a string wrapped in square brackets with commas separating the values. The constructor can convert a JSON text into a Java object.
🌐
Mosy
thepracticaldeveloper.com › java-and-json-jackson-serialization-with-objectmapper
Java and JSON – Jackson Serialization with ObjectMapper | Mosy
July 31, 2018 - As this example shows, you can’t get backward compatibility by default in this case since the unknown property causes an error. But this has a very easy fix, as you’ll see in the next section. @Test public void deserializePersonV2AsPersonIgnoringProperties() throws IOException { var mapper = new ObjectMapper(); mapper.registerModule(new JavaTimeModule()); mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false); var json = "{\"name\":\"Juan Garcia\",\"birthdate\":[1980,9,15]," + "\"hobbies\":[\"football\",\"squash\"]}"; PersonEC value = mapper.readValue(json, PersonEC.class); log.info("Deserializing a simple POJO ignoring unknown properties: {}", value); assertThat(value.getName()).isEqualTo("Juan Garcia"); assertThat(value.getBirthdate()) .isEqualTo(LocalDate.of(1980, 9, 15)); }
Find elsewhere
🌐
CodeMiner42
blog.codeminer42.com › home › posts › serialization and deserialization in java using jackson
Serialization and Deserialization in Java using Jackson - The Miners
May 29, 2023 - The name attribute is now being mapped to the fullname JSON key, and the cars attribute is being mapped to vehicles key. To deserialize this object we must pass the JSON string with the attributes exactly as they are above.
Top answer
1 of 3
2

You can use ObjectMapper for this:

ObjectMapper objectMapper = new ObjectMapper();

String jsonRecordArray =
        "[{ \"name\" : \"David\", \"age\" : 45 },\n" +
        " { \"name\" : \"Norma\", \"age\" : 44 }\n" +
        "]";
List<Record> listRecord = objectMapper.readValue(jsonRecordArray, new TypeReference<>() {
});

In maven the relevant dependency:

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.13.2.2</version>
</dependency>

Remember to handle JsonProcessingException, for example by adding it to the method signature.

2 of 3
1

First of all the answer from jwpol is correct and could be used as a solution. Just a bit of additional clarifications: The library that @jwpol suggests is called JSON-Jackson or also known as faster XML. Here is the main site for it and there are many others. This library is de-facto is JSON standard (used by default in Spring Boot). One popular alternative would be Gson - a Google implementation of JSON library. Here are 2 sites to read about it: https://github.com/google/gson, https://sites.google.com/site/gson/gson-user-guide.

Also, for simplistic use where you just need simple serialization/de-serialization capabilities I wrote my own simple feature that is a thin wrapper over Jackson Json library that gives you simplified use without directly importing Jackson library and configuring it. With my feature your code could look like

    String jsonRecordArray =
            "[{ \"name\" : \"David\", \"age\" : 45 },\n" +
            " { \"name\" : \"Norma\", \"age\" : 44 }\n" +
            "]";
List<Record> listRecord;
    try {
      listRecord = JsonUtils.readObjectFromJsonString(jsonRecordArray, List.class);
    } catch(IOException ioe) {
    ...
    }

BTW, in jwpol exaple exception handling is missing.

If you want to use my feature you need to import MgntUtils open source library written by me. You can get Maven artifacts here and or get it on Github (including source code and Javadoc). Here is Javadoc for JsonUtils class

🌐
Oracle
blogs.oracle.com › javamagazine › java-json-serialization-jackson
Efficient JSON serialization with Jackson and Java | javamagazine
With these hints, this piece of JSON will be correctly deserialized. { "first_name" : "Grant", "last_name" : "Hughes", "age" : 19 } ... @JsonCreator, which labels a constructor or factory method that will be used to create new Java objects from JSON
🌐
Medium
medium.com › @bectorhimanshu › how-to-use-jackson-library-to-serialize-java-objects-into-json-and-to-deserialize-json-into-java-6c3a6b6377e1
Using Jackson library to serialize Java objects into Json and to deserialize Json into Java objects | by bectorhimanshu | Medium
February 26, 2024 - import com.fasterxml.jackson.d... " + jsonString); // Deserialize JSON back to object Person deserializedPerson = deserializeJson(jsonString, Person.class); System.out.println("Deserialized Person: " + deserializedPerson); ...
🌐
YouTube
youtube.com › watch
Serialize and Deserialize JSON using Jackson - YouTube
Learn how to use Jackson to serialize and deserialize JSON. Learn the fundamentals of Jackson quickly and easily through examples. Jackson makes JSON with Ja...
Published: December 5, 2022
🌐
Baeldung
baeldung.com › home › json › jackson › intro to the jackson objectmapper
Intro to the Jackson ObjectMapper | Baeldung
December 9, 2025 - Also, on the writing side, we can use the writeValue API to serialize any Java object as JSON output. We’ll use the following Car class with two fields as the object to serialize or deserialize throughout this article: public class Car { private String color; private String type; // standard getters setters } Let’s see a first example of serializing a Java object into JSON using the writeValue method of the ObjectMapper class:
🌐
YouTube
youtube.com › brian fraser
Deserializing JSON: GSON and Java - YouTube
Use GSON library in IntelliJ with Java to deserialize a JSON file. Explicitly extracting values from the JSON file, which makes sense when the JSON format do...
Published: January 15, 2020
Views: 20K
🌐
TutorialsPoint
tutorialspoint.com › jackson_annotations › jackson_annotations_jsondeserialize.htm
Jackson Annotations - @JsonDeserialize
package com.tutorialspoint; import java.io.IOException; import java.text.ParseException; import java.util.Date; import com.fasterxml.jackson.databind.ObjectMapper; public class JacksonTester { public static void main(String args[]) throws ParseException{ ObjectMapper mapper = new ObjectMapper(); String jsonString = "{\"name\":\"Mark\",\"dateOfBirth\":\"20-12-1984\"}"; try { Student student = mapper .readerFor(Student.class) .readValue(jsonString); System.out.println(student.dateOfBirth); } catch (IOException e) { e.printStackTrace(); } } } class Student { public String name; public Date dateOf
🌐
TutorialsPoint
tutorialspoint.com › article › how-to-deserialize-a-json-into-an-existing-object-in-java
How to deserialize a JSON into an existing object in Java?
May 13, 2025 - Print the Person object. Use the Gson object to deserialize the JSON string into the existing object. Print the updated object. In the below example, we will create a class Person with fields for name and age.
🌐
Adobe Developer
developer.adobe.com › experience-manager › reference-materials › 6-5 › javadoc › com › fasterxml › jackson › databind › JsonDeserializer.html
JsonDeserializer (The Adobe AEM Quickstart and Web Application.)
public abstract class JsonDeserializer<T> extends java.lang.Object implements NullValueProvider · Abstract class that defines API used by ObjectMapper (and other chained JsonDeserializers too) to deserialize Objects of arbitrary types from JSON, using provided JsonParser.
🌐
Index.dev
index.dev › blog › deserialize-json-java-mapper
How to Deserialize JSON in Java Using a Mapper | Index.dev
January 28, 2025 - In this guide, we shall dive into the details of these methods, providing examples tailored to optimize workflows. Jackson’s ObjectMapper is a powerful utility that simplifies deserialization with minimal setup. It is feature-rich, supports advanced annotations, and is highly performant for most use cases. It acts as a bridge between JSON strings and Java classes, automating the parsing process while reducing boilerplate code.
🌐
Baeldung
baeldung.com › home › json › jackson › forcing jackson to deserialize to specific type
Forcing Jackson to Deserialize to Specific Type | Baeldung
October 15, 2025 - For our example, we’ll define a JSON structure with a field that can have multiple types of values. We’ll then create a Java class to represent the JSON structure and use Jackson to deserialize the ...
🌐
DZone
dzone.com › coding › languages › deserializing json to a java object using google’s gson library
Deserializing Json to a Java Object Using Google’s Gson Library
August 27, 2014 - of all the libraries mentioned ... . to have the json string deserialized, a java object must be created that has the same fields names with the fields in the json string....