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：

public class Student implements Serializable{
    private static final long serialVersionUID = 685922460589405829L;

    private String name;
    private String age;

   /*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：

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 .

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

 

/**
 *  java日期对象经过Jackson库转换成JSON日期格式化自定义类
 * @author godfox
 * @date 2010-5-3
 */
public class CustomDateSerializer extends JsonSerializer<Date> {

        @Override
        public void serialize(Date value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
                SimpleDateFormat formatter = new SimpleDateFormat("yyyy-MM-dd");
                String formattedDate = formatter.format(value);
                jgen.writeString(formattedDate);
        }
}

 

然后在你的POJO上找到日期的get方法

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

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