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) 

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)