jackson反序列化error handling

我的问题很简单：我有以下简单的类：

public class Foo { private int id = -1; public void setId(int _id){ this.id = _id; } public int getId(){ return this.id; } } 

我正在尝试处理以下JSON：

 { "id": "blah" } 

显然，这里有一个问题（“blah”无法解析为int）

以前，Jackson会抛出类似org.codehaus.jackson.map.JsonMappingException的东西：无法从String值’blah’构造java.lang.Integer的实例：不是有效的Integer值

我同意这一点，但我想在某处注册一些允许忽略这种映射错误的东西。 我尝试使用注册的DeserializationProblemHandler（参见此处 ），但它似乎只适用于未知属性而不是反序列化问题。

你对这个问题有什么线索吗？

Answer 1

感谢Jackson ML的Tatu ，我成功地解决了我的问题。

我必须为jackson处理的每种原始类型使用自定义非阻塞反序列化器。 像这个工厂的东西：

 public class JacksonNonBlockingObjectMapperFactory { /** * Deserializer that won't block if value parsing doesn't match with target type * @param  Handled type */ private static class NonBlockingDeserializer extends JsonDeserializer { private StdDeserializer delegate; public NonBlockingDeserializer(StdDeserializer _delegate){ this.delegate = _delegate; } @Override public T deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException, JsonProcessingException { try { return delegate.deserialize(jp, ctxt); }catch (JsonMappingException e){ // If a JSON Mapping occurs, simply returning null instead of blocking things return null; } } } private List jsonDeserializers = new ArrayList(); public ObjectMapper createObjectMapper(){ ObjectMapper objectMapper = new ObjectMapper(); SimpleModule customJacksonModule = new SimpleModule("customJacksonModule", new Version(1, 0, 0, null)); for(StdDeserializer jsonDeserializer : jsonDeserializers){ // Wrapping given deserializers with NonBlockingDeserializer customJacksonModule.addDeserializer(jsonDeserializer.getValueClass(), new NonBlockingDeserializer(jsonDeserializer)); } objectMapper.registerModule(customJacksonModule); return objectMapper; } public JacksonNonBlockingObjectMapperFactory setJsonDeserializers(List _jsonDeserializers){ this.jsonDeserializers = _jsonDeserializers; return this; } } 

然后像这样调用它（仅作为反序列化器传递给你想要非阻塞的那些）：

 JacksonNonBlockingObjectMapperFactory factory = new JacksonNonBlockingObjectMapperFactory(); factory.setJsonDeserializers(Arrays.asList(new StdDeserializer[]{ // StdDeserializer, here, comes from Jackson (org.codehaus.jackson.map.deser.StdDeserializer) new StdDeserializer.ShortDeserializer(Short.class, null), new StdDeserializer.IntegerDeserializer(Integer.class, null), new StdDeserializer.CharacterDeserializer(Character.class, null), new StdDeserializer.LongDeserializer(Long.class, null), new StdDeserializer.FloatDeserializer(Float.class, null), new StdDeserializer.DoubleDeserializer(Double.class, null), new StdDeserializer.NumberDeserializer(), new StdDeserializer.BigDecimalDeserializer(), new StdDeserializer.BigIntegerDeserializer(), new StdDeserializer.CalendarDeserializer() })); ObjectMapper om = factory.createObjectMapper(); 

Answer 2

您可能希望让控制器通过添加处理此特定exception的方法来处理问题

 @ExceptionHandler(HttpMessageNotReadableException.class) @ResponseBody public String handleHttpMessageNotReadableException(HttpMessageNotReadableException ex) { JsonMappingException jme = (JsonMappingException) ex.getCause(); return jme.getPath().get(0).getFieldName() + " invalid"; } 

当然，行

  JsonMappingException jme = (JsonMappingException) ex.getCause(); 

某些情况下可能会抛出一个类强制转换exception，但我还没有遇到它们。

Answer 3

创建一个简单的Mapper：

 @Provider @Produces(MediaType.APPLICATION_JSON) public class JSONProcessingErroMapper implements ExceptionMapper { @Override public Response toResponse(InvalidFormatException ex) { return Response.status(400) .entity(new ClientError("[User friendly message]")) .type(MediaType.APPLICATION_JSON) .build(); } 

}