Jackson和java.sql.Time序列化/反序列化

本文介绍了Jackson和java.sql.Time序列化/反序列化的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

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

我发布 JSON - 对象为 @RequestBody 到一个Spring Controller，它应该映射到实体的实例（pojo）中。 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;