jackson 常見問題

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 )

一般來說，解決上面問題從下面幾個方面入手：

1、是否缺少默認構造函數

2、是否是類的訪問修飾符問題，即jackson訪問不到。

 

jackson 的UnrecognizedPropertyException錯誤

java代碼如下：

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\"}"

 

這種情況下，使用jackson從string轉換為object的時候，會拋出下面異常：

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

意思是說Student類里沒有address這個屬性，所以無法正常轉化，同時還指明了not marked as ignorable，即沒有標明可忽略的特性

 

解決方案：

1、在類上添加忽略聲明

@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"

 

當屍實體中存在Date類型，但是json字符串中是字符串類型

只支持以上幾種格式否則報錯

 

/**
 *  java日期對象經過Jackson庫轉換成JSON日期格式化自定義類
 * @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;
       }