Question

Example JSON (note that the string has trailing spaces):

{ "aNumber": 0, "aString": "string   " }

Ideally, the deserialised instance would have an aString property with a value of “string” (i.e. without trailing spaces). This seems like something that is probably supported but I can’t find it (e.g. in DeserializationConfig.Feature).

We’re using Spring MVC 3.x so a Spring-based solution would also be fine.

I tried configuring Spring’s WebDataBinder based on a suggestion in a forum post but it does not seem to work when using a Jackson message converter:

@InitBinder
public void initBinder( WebDataBinder binder )
{
    binder.registerCustomEditor( String.class, new StringTrimmerEditor( " \t\r\n\f", true ) );
}

Answer 1

With a custom deserializer, you could do the following:

 <your bean>
 @JsonDeserialize(using=WhiteSpaceRemovalSerializer.class)
 public void setAString(String aString) {
    // body
 }

 <somewhere>
 public class WhiteSpaceRemovalDeserializer extends JsonDeserializer<String> {
     @Override
     public String deserialize(JsonParser jp, DeserializationContext ctxt) {
         // This is where you can deserialize your value the way you want.
         // Don't know if the following expression is correct, this is just an idea.
         return jp.getCurrentToken().asText().trim();
     }
 }

This solution does imply that this bean attribute will always be serialized this way, and you will have to annotate every attribute that you want to be deserialized this way.