杰克逊反序列化的错误处理

Question

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; } 
} 

，我试图以处理以下JSON：

{ 
    "id": "blah" 
} 

显然，这里有一个问题（“blah”不能被解析为int）

以前，Jackson抛出类似org.codehaus.jackson.map.JsonMappingException的东西：无法从String构造java.lang.Integer的实例值“胡说”：不是一个有效的整数值

我同意这一点，但我想注册某个地方的东西，允许忽略这种类型的映射错误的。 我尝试使用已注册的DeserializationProblemHandler（请参阅here），但似乎只适用于未知属性而不是反序列化问题。

你对这个问题有任何线索吗？

Answer 1

我成功解决了我的问题，这要感谢Tatu from Jackson ML。

我不得不使用定制的非阻塞解串器在处理杰克逊每基本类型。 事情是这样的工厂：

public class JacksonNonBlockingObjectMapperFactory { 

    /** 
    * Deserializer that won't block if value parsing doesn't match with target type 
    * @param <T> Handled type 
    */ 
    private static class NonBlockingDeserializer<T> extends JsonDeserializer<T> { 
     private StdDeserializer<T> delegate; 

     public NonBlockingDeserializer(StdDeserializer<T> _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<StdDeserializer> jsonDeserializers = new ArrayList<StdDeserializer>(); 

    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<StdDeserializer> _jsonDeserializers){ 
     this.jsonDeserializers = _jsonDeserializers; 
     return this; 
    } 
} 

然后调用它像这样（传为解串器，只有那些你想成为非阻塞）：

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

你可能想要让你的控制器处理问题通过添加处理这个特定的异常

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

当然的方法，该线

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

可能会抛出一些情况下，类转换异常，但我还没有遇到过他们呢。

Answer 3

创建一个简单的映射：

@Provider 
@Produces(MediaType.APPLICATION_JSON) 
public class JSONProcessingErroMapper implements ExceptionMapper<InvalidFormatException> { 

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

}