Jackson和java.sql.Time序列化/反序列化

问题描述

    @JsonFormat（pattern =HH：mm）
 @Column（name =start_time）
 private java.sql.Time startTime; 
  

我发布 c> Jackson >

Jackson 显然不能将时间字符串反序列化为 java.sql。 Time ，因为我得到这个异常：

  .wsmsDefaultHandlerExceptionResolver：无法读取HTTP消息： 
 org.springframework.http.converter.HttpMessageNotReadableException：
无法读取文档：无法构造java.sql.Time实例，
问题：null 
  

如何指示 Jackson 了解要做什么？

解决方案

解决方案是推出自己的反序列化器：

  import java.io.IOException; 
 import java.sql.Time; 
 import com.fasterxml.jackson.core.JsonParser; 
 import com.fasterxml.jackson.databind.DeserializationContext; 
 import com.fasterxml.jackson.databind.JsonDeserializer; 
 
公共类SqlTimeDeserializer扩展了JsonDeserializer< Time> {
 $ b $ @Override 
 public Time deserialize（JsonParser jp，DeserializationContext ctxt）throws IOException {
 return Time.valueOf（jp.getValueAsString（）+：00）; 
 
 
  

然后在实体中：

  @JsonFormat（pattern =HH：mm）
 @JsonDeserialize（using = SqlTimeDeserializer.class）
 @Column（ name =start_time）
私人时间startTime; 
  

Consider this property in an Hibernate-managed entity:

@JsonFormat(pattern = "HH:mm")
@Column(name = "start_time")
private java.sql.Time startTime;

I post a JSON-object as @RequestBody to a Spring Controller which Jackson is supposed to map into an instance of the entity (pojo).

Jackson does apparently not manage to de-serialize the time-string into a java.sql.Time, because I am getting this Exception:

.w.s.m.s.DefaultHandlerExceptionResolver : Failed to read HTTP message:
org.springframework.http.converter.HttpMessageNotReadableException:
Could not read document: Can not construct instance of java.sql.Time,
problem: null

How can I instruct Jackson to understand what to do?

解决方案

The solution is to roll your own deserializer:

import java.io.IOException;
import java.sql.Time;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonDeserializer;

public class SqlTimeDeserializer extends JsonDeserializer<Time> {

    @Override
    public Time deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException {
        return Time.valueOf(jp.getValueAsString() + ":00");
    }
}

And then in the entity:

@JsonFormat(pattern = "HH:mm")
@JsonDeserialize(using = SqlTimeDeserializer.class)
@Column(name = "start_time")
private Time                startTime;

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