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*/ 
8. } 

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.         } 
15. } 

 

然后在你的POJO上找到日期的get方法

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.        }