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jackson 常见问题

org.codehaus.jackson.map.JsonMappingException: No suitable constructor found for type

 

org.codehaus.jackson.map.JsonMappingException: No suitable constructor found for type [simple type, class test.jackson.Employee]: can not instantiate from JSON object (need to add/enable type information?)
 at [Source: c:\temp\employee.json; line: 1, column: 2]
    at org.codehaus.jackson.map.JsonMappingException.from(JsonMappingException.java:163)
    at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:483)
    at org.codehaus.jackson.map.deser.BeanDeserializer.deserialize(BeanDeserializer.java:350)
    at org.codehaus.jackson.map.ObjectMapper._readMapAndClose(ObjectMapper.java:2395)
    at org.codehaus.jackson.map.ObjectMapper.readValue(ObjectMapper.java:1549)
    at test.jackson.JSONToJavaExample.main(JSONToJavaExample.java:19)

一般来说,解决上面问题从下面几个方面入手:

1、是否缺少默认构造函数

2、是否是类的访问修饰符问题,即jackson访问不到。

 

jackson 的UnrecognizedPropertyException错误

java代码如下:

  1. public class Student implements Serializable{ 
  2.     private static final long serialVersionUID = 685922460589405829L; 
  3.  
  4.     private String name; 
  5.     private String age; 
  6.  
  7.    /*get set略*/ 

json字符串如下:

"{\"address\":\"hunan\",\"name\":\"china\",\"age\":\"5000\"}"

 

这种情况下,使用jackson从string转换为object的时候,会抛出下面异常:

  1. java.lang.IllegalArgumentException: org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "address" (Class util.Student), not marked as ignorable 

意思是说Student类里没有address这个属性,所以无法正常转化,同时还指明了not marked as ignorable,即没有标明可忽略的特性

 

解决方案:

1、在类上添加忽略声明

@JsonIgnoreProperties(ignoreUnknown=true)

2、更改str2obj方法

   ObjectMapper objectMapper = new ObjectMapper();    objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);    projectVO = objectMapper.readValue(yourjsonstring, Test.class);

 

Jackson annotation - How to rename element names?

you can rename a property just adding

@JsonProperty("contractor")

And by default Jackson use the getter and setter to serialize and deserialize.

 

Using a custom JsonSerializer.

public class Response { private String status; private String error; @JsonProperty("p") @JsonSerialize(using = CustomSerializer.class) private Object data; // ...}public class CustomSerializer extends JsonSerializer<Object> { public void serialize(Object value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException { jgen.writeStartObject(); jgen.writeObjectField(value.getClass().getName(), value); jgen.writeEndObject(); }}

Could not read JSON: Can not construct instance of java.util.Date from String value ‘2012-07-21 12:11:12‘: not a valid representation

 

  1. "yyyy-MM-dd‘T‘HH:mm:ss.SSSZ"
  2. "yyyy-MM-dd‘T‘HH:mm:ss.SSS‘Z‘"
  3. "EEE, dd MMM yyyy HH:mm:ss zzz"
  4. "yyyy-MM-dd"

 

当尸实体中存在Date类型,但是json字符串中是字符串类型

只支持以上几种格式否则报错

 

  1.  
  2. /** 
  3. *  java日期对象经过Jackson库转换成JSON日期格式化自定义类 
  4. * @author godfox 
  5. * @date 2010-5-3 
  6. */ 
  7. public class CustomDateSerializer extends JsonSerializer<Date> { 
  8.  
  9.         @Override 
  10.         public void serialize(Date value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException { 
  11.                 SimpleDateFormat formatter = new SimpleDateFormat("yyyy-MM-dd"); 
  12.                 String formattedDate = formatter.format(value); 
  13.                 jgen.writeString(formattedDate); 
  14.         } 

 

然后在你的POJO上找到日期的get方法

Java代码  技术分享

  1. @JsonSerialize(using = CustomDateSerializer.class)  
  2.        public Date getCreateAt() {  
  3.                return createAt;  
  4.        }  

  1. @JsonSerialize(using = CustomDateSerializer.class
  2.        public Date getCreateAt() { 
  3.                return createAt; 
  4.        } 

 

 

  1. //null的JSON序列 
  2.     private class NullSerializer extends JsonSerializer<Object> { 
  3.         public void serialize(Object value, JsonGenerator jgen, 
  4.                 SerializerProvider provider) throws IOException, 
  5.                 JsonProcessingException { 
  6.             jgen.writeString(""); 
  7.         } 
  8.     } 

在spring的配置文件中定义这个自定义的Mapper

 

在spring的配置文件中定义这个自定义的Mapper

Xml代码 技术分享

  1. <!-- 启动 MVC注解 -->
  2. <mvc:annotation-driven>
  3. <mvc:message-converters>
  4. <bean class="org.springframework.http.converter.json.MappingJacksonHttpMessageConverter">
  5. <property name="objectMapper" ref="customObjectMapper" />
  6. </bean>
  7. </mvc:message-converters>
  8. </mvc:annotation-driven>
  9. <!-- 自定义的JSON ObjectMapper -->
  10. <bean id="customObjectMapper" class="com.xixi.web4j.util.CustomObjectMapper" />

就这么简单,不用再像每个POJO对像上去@JsonSerializer(using...)了。

以后对所有DATE类型字段序列化JSON,都会默认传为“yyyy-MM-dd HH:mm:ss”格式,对于字段为null的也会转为“”。如果你对null有别的处理。你可以在自定义的NullSerializer修改你想要的格式。

jackson 常见问题