Joda的当前日期时间 - mysql时间戳的时间日期格式

时间:2013-12-17 03:00:20

标签: java mysql spring spring-mvc jodatime

我需要将当前日期和时间导入到格式为TimeStamp的MySQL数据库字段中。从检查样本数据来看,MySQL TimeStamp数据类型的格式似乎是“yyyy-mm-dd hh:mm:ss”。我在我的spring hibernate应用程序中使用Joda-Time格式化。如何以基础MySQL TimeStamp格式化字段接受的格式获取当前日期时间?

这是我当前的代码,由于eclipse说.parseDateTime()需要字符串参数而不是DateTime参数,因此无法编译:

public void setCreated(){
    DateTime now = new org.joda.time.DateTime();
    DateTimeFormatter fmt = DateTimeFormat.forPattern("yyyy-MM-dd hh:mm:ss");
    created = fmt.parseDateTime(now);
    System.out.println("'''''''''''''''''''''' created is: "+created);
}

我试图坚持的实体定义如下:

@Entity
@Table(name = "documents")
public class Document {
@Id
@GeneratedValue
@Column(name="id")
private Integer id;

@ManyToOne
@JoinColumn(name = "client_id")
private Patient patient;

@ManyToOne
@JoinColumn(name = "type_id")
private DocumentType type;

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

@Column(name="description")
private String description;

@Column(name="filename")
private String filename;

@Column(name="content")
@Lob
private Blob content;

@Column(name="content_type")
private String contentType;

@Column(name = "created")
private DateTime created;

public Integer getId(){return id;}
public void setId(Integer i){id=i;}

protected void setPatient(Patient patient) {this.patient = patient;}
public Patient getPatient(){return this.patient;}

public void setType(DocumentType type) {this.type = type;}
public DocumentType getType() {return this.type;}

public String getName(){return name;}
public void setName(String nm){name=nm;}

public String getDescription(){return description;}
public void setDescription(String desc){description=desc;}

public String getFileName(){return filename;}
public void setFileName(String fn){filename=fn;}

public Blob getContent(){return content;}
public void setContent(Blob ct){content=ct;}

public String getContentType(){return contentType;}
public void setContentType(String ctype){contentType=ctype;}

public void setCreated(){
    DateTime now = new org.joda.time.DateTime();
    DateTimeFormatter fmt = DateTimeFormat.forPattern("yyyy-MM-dd hh:mm:ss");
    created = fmt.parseDateTime(now);
System.out.println("''''''''''''''''''''''''''' created is: "+created);
}
public DateTime getCreated() {return this.created;}

@Override
public String toString() {return this.getName();}
public boolean isNew() {return (this.id == null);}

}

如何更改上述内容以便以正确的格式保存数据以插入MySQL TimeStamp字段?


编辑:

与Sotirios的建议相关,以下是我当前pom.xml的相关部分,供讨论之用:

<properties>
    <jodatime-hibernate.version>1.3</jodatime-hibernate.version>
    <jodatime-jsptags.version>1.1.1</jodatime-jsptags.version>
    <jodatime.version>2.3</jodatime.version>
    <jadira-usertype-core.version>3.1.0.CR8</jadira-usertype-core.version>
</properties>
<dependencies>
    <dependency>
        <groupId>org.jadira.usertype</groupId>
        <artifactId>usertype.core</artifactId>
        <version>${jadira-usertype-core.version}</version>
    </dependency>
    <dependency>
        <groupId>joda-time</groupId>
        <artifactId>joda-time</artifactId>
        <version>${jodatime.version}</version>
    </dependency>
    <dependency>
        <groupId>joda-time</groupId>
        <artifactId>joda-time-hibernate</artifactId>
        <version>${jodatime-hibernate.version}</version>
    </dependency>
    <dependency>
        <groupId>joda-time</groupId>
        <artifactId>joda-time-jsptags</artifactId>
        <version>${jodatime-jsptags.version}</version>
    </dependency>
</dependencies>

2 个答案:

答案 0 :(得分:4)

因为你似乎需要它:

我的实体类

@Entity
@Table(name = "time_fields")
public class TimeFields {
    @Id
    @GeneratedValue(strategy= GenerationType.IDENTITY)
    private Long timeId;

    @Column
    @Temporal(TemporalType.DATE)
    private Date date;

    @Column
    @Type(type="org.jadira.usertype.dateandtime.joda.PersistentDateTime")
    private DateTime dateTime;

使用适当的getter和setter。

客户端代码

AnnotationConfigApplicationContext context = new AnnotationConfigApplicationContext(PersistenceContext.class);
SessionFactory sessionFactory = context.getBean(SessionFactory.class);
Session session = sessionFactory.openSession();

TimeFields timeFields = new TimeFields();
Date date = new Date();
DateTime dateTime = new DateTime();
System.out.println(date);
System.out.println(dateTime);
timeFields.setDate(date);
timeFields.setDateTime(dateTime);

session.beginTransaction();
session.persist(timeFields);
session.getTransaction().commit();

System.out.println(timeFields.getTimeId());
System.out.println(timeFields.getDate());
System.out.println(timeFields.getDateTime());

打印

Tue Dec 17 00:22:35 EST 2013
2013-12-17T00:22:35.843-05:00
3
Tue Dec 17 00:22:35 EST 2013
2013-12-17T00:22:35.843-05:00

除了joda-time和hibernate,你还需要jadira libs

<dependency>
    <groupId>org.jadira.usertype</groupId>
    <artifactId>usertype.jodatime</artifactId>
    <version>2.0.1</version>
</dependency>

您应该阅读有关Hibernate UserType s。

的更多信息

答案 1 :(得分:1)

日期的字符串表示纯粹供人类使用。时间戳在内部表示为与时区无关的数值。由于yyyy-mm-dd hh:mm:ss不包含时区,因此尝试自行进行转换会让您遇到时区错误。

Hibernate将直接处理java.util.Datejava.sql.Date,因此您可以将数据类型更改为其中之一,然后执行映射到joda对象getter / setter所以它可以通过这种方式传递给调用者。

最好是IMO,你可以为Hibernate提供joda对象的映射。已经为此准备了一个图书馆:

https://github.com/JodaOrg/joda-time-hibernate

或通过maven:

http://mvnrepository.com/artifact/joda-time/joda-time-hibernate

注意:我个人的经验法则是,如果你看到代码操纵日期为字符串,它就会被破坏......不是任何科学指标,但我发现它远远好于50/50。 :)

编辑:文档在这里:http://www.joda.org/joda-time-hibernate/userguide.html

在类路径中使用它,您应该能够注释您的列,如:

@Column(name = "created")
@Type(type="org.joda.time.contrib.hibernate.PersistentDateTime")
private DateTime created;