How to Fix "Cannot Deserialize Instance of java.lang.String Out of START_ARRAY Token" Error in Java

What does the error "Cannot Deserialize Instance of java.lang.String Out of START_ARRAY Token" mean in Java?

The error "Cannot Deserialize Instance of java.lang.String Out of START_ARRAY Token" typically occurs when you're trying to deserialize JSON data into a Java object. This error indicates that the parser expected a single string value but encountered an array instead.

import com.fasterxml.jackson.databind.ObjectMapper;
import java.io.IOException;
import java.util.List;

public class Example {
    public static void main(String[] args) throws IOException {
        String json = "{ 'name': ['John', 'Doe'] }"; // JSON with an array
        ObjectMapper objectMapper = new ObjectMapper();
        Person person = objectMapper.readValue(json.replace("'", "\'"), Person.class);
    }
}

class Person {
    private List<String> name; // Change to List<String> if JSON is an array
    public List<String> getName() { return name; }
    public void setName(List<String> name) { this.name = name; }
}

Causes

• The JSON structure does not match the expected Java object structure.
• The Java object has a field of type `String`, but the corresponding JSON element is an array.
• Incorrect data types are specified in the Java class.

Solutions

• Check the JSON data to ensure it matches the expected structure.
• Change the Java field type from `String` to `List<String>` or `String[]` if the JSON data is indeed an array.
• Update the JSON data to provide a single string value that matches the expected type.

Mistake: Assuming the JSON data is always structured as a single object when it may contain arrays.

Solution: Always verify the structure of the incoming JSON data before deserialization.

Mistake: Not updating the Java class structure when changing the expected JSON format.

Solution: When the expected JSON format changes, ensure that the corresponding Java class is updated accordingly.