使用Java在JPA中以AM / PM插入并获取时间查询

Question

This is the question

OR

SELECT TIME_FORMAT( `locStart`, '%h:%i %p' ) FROM locationtimes.

我如何使用JPA实现它

这些是我的数据库表条目

“>

我的Java实体类

package models;

package models;

import util.MyConverter;

import javax.persistence.*;
import java.sql.Time;
import java.sql.Timestamp;

/**
 * Created by abhinav on 11/2/15.
 */

@Entity
@Table(name = "locationtimes")
//@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id", scope = Locationtimes.class)
public class Locationtimes {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Long id;

    @Column
    public Long locationId;

    @Column
    public String weekday;

    @Column(insertable = false, updatable = false)
    @Temporal(TemporalType.TIME)
    public Time locStart;

    @Column(insertable = false, updatable = false)
    public Time locEnd;

    @Column(insertable = false, updatable = false)
    public Time todayStart;

    @Column(insertable = false, updatable = false)
    public Time futureStart;

    @Column(insertable = false, updatable = false)
    public Time futureEnd;

    //@Transient
    @Column(name = "locStart")
    @Convert(converter = MyConverter.class)
    public String locStartString;

    //@Transient
    @Column(name = "locEnd")
    @Convert(converter = MyConverter.class)
    public String locEndString;

    @Column(name = "todayStart")
    @Convert(converter = MyConverter.class)
    public String todayStartString;

    //@Transient
    @Column(name = "futureStart")
    @Convert(converter = MyConverter.class)
    public String futureStartString;

    // @Transient
    @Column(name = "futureEnd")
    @Convert(converter = MyConverter.class)
    public String futureEndString;

    @Column
    public Long modifiedBy;

    @Column
    public Timestamp modified;
}

这是问题，或者从定位时间中选择SELECT TIME_FORMAT（`locStart`，'％h：％i％p'）。我如何使用JPA实现它？这是我的数据库表条目我的java Entity类包模型； ...

Answer 1

我想你的意思是你有时间用格式-

Answer 2

我得到了解决方案。这将通过此过程来实现。

@@ Farrellmr感谢您的支持