Question

我正在使用springmvc和hibernate进行休息服务的示例，其中我以毫秒格式获取日期数据类型的日期响应，我通过谷歌搜索使用了下面的代码，但这些代码没有影响它的结果相同毫秒格式。下面是我使用的代码

controller.java

/* Getting List of objects in Json format in Spring Restful Services */  
 @RequestMapping(value = "/list", method = RequestMethod.GET)  
 public @ResponseBody  List getDatalist() { 
  List DataList = null;  
  try {  
      DataList = dataServices.getDataEntityList();  
  } catch (Exception e) {  
   e.printStackTrace();  
  }  
  return DataList;  
 }

DataValueTable.java

  @JsonAutoDetect
@Entity  
@Table(name = "DataValueTable")  
@JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})  
public class DataValueTable  implements Serializable {  

 private static final long serialVersionUID = 1L;  

 @Id  
 @GeneratedValue  
 @Column(name = "ID")  
 private long id;  

 @JsonSerialize(using=JsonDateSerializer.class)
 @Column(name = "Datatype") 
 private Date dataType;  

 @Column(name = "Datacategory")  
 private String dataCategory;  

 @Column(name = "DataValue")  
 private boolean dataValue;
     public Date getDataType() {
    return dataType;
}

public void setDataType(Date dataType) {
    this.dataType = dataType;
}

JsonDateSerializer .java

package com.beingjavaguys.model;

import java.io.IOException;
import java.text.SimpleDateFormat;
import java.util.Date;
import org.codehaus.jackson.JsonGenerator;
import org.codehaus.jackson.JsonProcessingException;
import org.codehaus.jackson.map.JsonSerializer;
import org.codehaus.jackson.map.SerializerProvider;
import org.springframework.stereotype.Component;
/**
 * Used to serialize Java.util.Date, which is not a common JSON
 * type, so we have to create a custom serialize method;.
 *
 * @author Loiane Groner
 * http://loianegroner.com (English)
 * http://loiane.com (Portuguese)
 */
@Component
public class JsonDateSerializer extends JsonSerializer<Date>{
private static final SimpleDateFormat dateFormat = new SimpleDateFormat("MM-dd-yyyy");
@Override
public void serialize(Date date, JsonGenerator gen, SerializerProvider provider)
throws IOException, JsonProcessingException {
String formattedDate = dateFormat.format(date);
gen.writeString(formattedDate);
}
}

弹簧-config.xml中

 <context:component-scan base-package="com.beingjavaguys.controller" />  
 <mvc:annotation-driven />  

 <bean id="dataSource"  
  class="org.springframework.jdbc.datasource.DriverManagerDataSource"> 
  <property name="driverClassName" value="com.microsoft.sqlserver.jdbc.SQLServerDriver" />  
  <property name="url" value="jdbc:sqlserver://localhost;database=Sample" />  
  <property name="username" value="sdsd" />  
  <property name="password" value="sdsd" /> 
    <property name="hibernate.dialect" value="com.beingjavaguys.model.SqlServerDialectWithNvarchar" /> 
 </bean>  


 <bean id="sessionFactory"  
 class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">  
 <property name="dataSource" ref="dataSource" />  
 <property name="annotatedClasses">  
 <list>  
   <value>com.beingjavaguys.model.Employee</value>   
      <value>com.beingjavaguys.model.DataValueTable</value> 
   </list>  
  </property>  
  <property name="hibernateProperties">  
   <props>  

    <prop key="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</prop>  
    <prop key="hibernate.show_sql">${hibernate.show_sql}</prop>  
   </props>  
  </property>  
 </bean>  


 <bean id="txManager"  
  class="org.springframework.orm.hibernate4.HibernateTransactionManager">  
  <property name="sessionFactory" ref="sessionFactory" />  
 </bean>  

 <bean id="persistenceExceptionTranslationPostProcessor"  
  class="org.springframework.dao.annotation.PersistenceExceptionTranslationPostProcessor" />  

 <bean id="dataDao" class="com.beingjavaguys.dao.DataDaoImpl"></bean>  
 <bean id="dataServices" class="com.beingjavaguys.services.DataServicesImpl"></bean>  
</beans>

现在我得到了响应＆＃34; 12312121000＆＃34;。我希望它以02-02-2015格式显示。欢迎任何建议和答案。

Answer 1

使用jackson，您可以注释实体，例如

@JsonFormat(shape=JsonFormat.Shape.STRING, pattern="yyyy-MM-dd HH:mm:ss.000 ", timezone="UTC")
public Timestamp getDate() {
    return date;
}

这是我用来取回模式中时间戳的测试用例的一个例子。

    ObjectMapper mapper = new ObjectMapper();
    JaxbAnnotationModule module = new JaxbAnnotationModule();
    mapper.registerModule(module);      
    mapper.setSerializationInclusion(Include.NON_NULL);
    // Object to JSON - entityBean I want to map to JSON
    String jsonOutput = mapper.writeValueAsString(entityBean);

请注意。您必须将带注释的JsonFormat分配给getter方法。

如何从springmvc和hibernate中的json响应中获取日期格式响应而不是毫秒

1 个答案: