I need to convert a certain JSON string to a Java object. I am using Jackson for JSON handling. Here is my Java class:
public class RequestClass {
String email_id;
String password;
public String getEmailId() {
return email_id;
}
public String getPassword() {
return password;
}
@Override
public String toString(){
return email_id+" "+password;
}
}
Here is the web service code:
@POST
@Consumes(MediaType.APPLICATION_JSON)
@Produces(MediaType.APPLICATION_JSON)
@Path("/dbconnect3")
public String connectToDbTest3(RequestClass rc) {
System.out.println("connectToDbTest3");
String email_id = rc.getEmailId();
String password = rc.getPassword();
System.out.println(email_id + " " + password);
}
This throws exception UnrecognizedPropertyException
with message "Unrecognized field "email_id" (Class jaxrs.RequestClass), not marked as ignorable"
.
In case i am not using the annotation @JsonIgnoreProperties(ignoreUnknown = true)
in my Java class, the output I am getting on line 09 is:
null myPassword
So I don't want to ignore Unrecognized field instead I want to get the value of email_id
.
Please tell why It shows email_id
as Unrecognized field while password is fetched successfully.