How can I Serialize/De-serialize a Boolean Value from FasterXML\Jackson as an Int?

Question

I'm writing a JSON Client for a Server that returns Boolean values as "0" and "1". When I try to run my JSON Client I currently get the following Exception:

HttpMessageNotReadableException: Could not read JSON: Can not construct instance of java.lang.Boolean from String value '0': only "true" or "false" recognized 

So how can I setup FasterXML\Jackson to correctly parse something like:

{    "SomeServerType" : {      "ID" : "12345",      "ThisIsABoolean" : "0",      "ThisIsABooleanToo" : "1"    } } 

Sample Pojo's:

@JsonInclude(JsonInclude.Include.NON_NULL) @JsonPropertyOrder({"someServerType"}) public class myPojo {    @JsonProperty("someServerType")    SomeServerType someServerType;     @JsonProperty("someServerType")    public SomeServerType getSomeServerType() { return someServerType; }     @JsonProperty("someServertype")    public void setSomeServerType(SomeServerType type)    { someServerType = type; } } 

---

@JsonInclude(JsonInclude.Include.NON_NULL) @JsonPropertyOrder({"someServerType"}) public class SomeServerType  {    @JsonProperty("ID")    Integer ID;     @JsonProperty("ThisIsABoolean")    Boolean bool;     @JsonProperty("ThisIsABooleanToo")    Boolean boolToo;     @JsonProperty("ID")    public Integer getID() { return ID; }     @JsonProperty("ID")    public void setID(Integer id)    { ID = id; }     @JsonProperty("ThisIsABoolean")    public Boolean getThisIsABoolean() { return bool; }     @JsonProperty("ThisIsABoolean")    public void setThisIsABoolean(Boolean b) { bool = b; }     @JsonProperty("ThisIsABooleanToo")    public Boolean getThisIsABooleanToo() { return boolToo; }     @JsonProperty("ThisIsABooleanToo")    public void setThisIsABooleanToo(Boolean b) { boolToo = b; } } 

Rest Client Line
Note 1: This is using Spring 3.2
Note 2: toJSONString() - is a helper method that uses Jackson to Serialize my Parameters Object
Note 3: The Exception happens on Reading IN the result object

DocInfoResponse result = restTemplate.getForObject(docInfoURI.toString()                                   + "/?input={input}",                                   DocInfoResponse.class,                                   toJSONString(params)); 

Answer 1

As Paulo Pedroso's answer mentioned and referenced, you will need to roll your own custom JsonSerializer and JsonDeserializer. Once created, you will need to add the @JsonSerialize and @JsonDeserialize annotations to your property; specifying the class to use for each.

I have provided a small (hopefully straightforward) example below. Neither the serializer nor deserializer implementations are super robust but this should get you started.

public static class SimplePojo {      @JsonProperty     @JsonSerialize(using=NumericBooleanSerializer.class)     @JsonDeserialize(using=NumericBooleanDeserializer.class)     Boolean bool; }  public static class NumericBooleanSerializer extends JsonSerializer<Boolean> {      @Override     public void serialize(Boolean bool, JsonGenerator generator, SerializerProvider provider) throws IOException, JsonProcessingException {         generator.writeString(bool ? "1" : "0");     }    }  public static class NumericBooleanDeserializer extends JsonDeserializer<Boolean> {      @Override     public Boolean deserialize(JsonParser parser, DeserializationContext context) throws IOException, JsonProcessingException {         return !"0".equals(parser.getText());     }        }  @Test public void readAndWrite() throws JsonParseException, JsonMappingException, IOException {     ObjectMapper mapper = new ObjectMapper();      // read it     SimplePojo sp = mapper.readValue("{\"bool\":\"0\"}", SimplePojo.class);     assertThat(sp.bool, is(false));      // write it     StringWriter writer = new StringWriter();     mapper.writeValue(writer, sp);     assertThat(writer.toString(), is("{\"bool\":\"0\"}")); }