相关疑难解决方法(0)

根据文档,hibernate 3.6应该支持java.util.UUID类型.但是当我将其映射为:

@Id protected UUID uuid;

我得到以下异常:

Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [test-applicationContext.xml]: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: persistenceUnit] Unable to build EntityManagerFactory
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1420) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:519) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:456) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory$1.getObject(AbstractBeanFactory.java:291) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:288) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:190) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor.findDefaultEntityManagerFactory(PersistenceAnnotationBeanPostProcessor.java:529) ~[spring-orm-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor.findEntityManagerFactory(PersistenceAnnotationBeanPostProcessor.java:495) ~[spring-orm-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor$PersistenceElement.resolveEntityManager(PersistenceAnnotationBeanPostProcessor.java:656) ~[spring-orm-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor$PersistenceElement.getResourceToInject(PersistenceAnnotationBeanPostProcessor.java:629) ~[spring-orm-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.annotation.InjectionMetadata$InjectedElement.inject(InjectionMetadata.java:147) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.beans.factory.annotation.InjectionMetadata.inject(InjectionMetadata.java:84) ~[spring-beans-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    at org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor.postProcessPropertyValues(PersistenceAnnotationBeanPostProcessor.java:338) ~[spring-orm-3.0.5.RELEASE.jar:3.0.5.RELEASE]
    ... …

我需要将在.NET中生成的Guid传递给Java应用程序.我用Guid.ToByteArray()它将它作为a存储在磁盘上byte[],然后将其读入Java并将其转换为UUID.为此,我复制了UUID的(私有)构造函数的实现,该构造函数采用byte[]:

private UUID(byte[] data) {
    long msb = 0;
    long lsb = 0;
    assert data.length == 16;
    for (int i=0; i<8; i++)
        msb = (msb << 8) | (data[i] & 0xff);
    for (int i=8; i<16; i++)
        lsb = (lsb << 8) | (data[i] & 0xff);
    this.mostSigBits = msb;
    this.leastSigBits = lsb;
}

但是,当我使用UUID检查时toString(),Java UUID与.NET Guid不同.

例如,.NET Guid

888794c2-65ce-4de1-aa15-75a11342bc63

变成Java UUID

c2948788-ce65-e14d-aa15-75a11342bc63

似乎前三组的字节顺序是相反的,而后两组中的顺序是相同的.

既然我希望toString()Guid和UUID都能产生相同的结果,有谁知道我应该如何正确地将.NET Guid读入Java UUID？

编辑:澄清一下,实施不是我自己的.它是类的私有构造函数,java.util.UUID它采用a byte[] …

我的数据库条目具有值的UUID(使用Microsoft SQL Server Management Studio提取)

CDF86F27-AFF4-2E47-BABB - 2F46B079E98B

将其加载到我的Scala应用程序后,toString方法会生成此值

276ff8cd-f4af-472e降低-BABB - 2f46b079e98b

这是怎么发生的？当我手边只有裸字符串CDF86F27-AFF4-2E47-BABB-2F46B079E98B时,如何以编程方式创建UUID实例？

相关的Slick代码(前者:表定义,后者:数据库访问对象)

class ChannelTable(tag: Tag) extends Table[ChannelTuple](tag, "Channel") {
  def id = column[UUID]("Id", O.PrimaryKey)
  def channelId = column[Int]("Channel_Id", O.NotNull)
  def timer = column[UUID]("Timer_Id", O.NotNull)
  def from = column[Timestamp]("FromTime", O.NotNull)
  def to = column[Timestamp]("ToTime", O.NotNull)
  def mon = column[Boolean]("Mon", O.NotNull)
  def tues = column[Boolean]("Tues", O.NotNull)
  def wed = column[Boolean]("Wed", O.NotNull)
  def thu = column[Boolean]("Thu", O.NotNull)
  def fri = column[Boolean]("Fri", O.NotNull)
  def sat = column[Boolean]("Sat", O.NotNull) …

我有一个应用程序，我想在 PostgreSQL 和 SQL Server 上运行。我想使用 java.util.UUID 作为 ID。

我已将 SQL Server 中的列定义为

id  UNIQUEIDENTIFIER ROWGUIDCOL NOT NULL UNIQUE

我已将 PostgreSQL 中的列定义为

id  UUID NOT NULL

这些列在我的 JPA 实体中定义为

@Id
@Column(name = "id")
public UUID getId() {
    return id;
}

这适用于 PostgreSQL，因为它将 UUID 传递给 PostgreSQL JDBC 驱动程序。这种方式适用于 SQL Server，因为 Hibernate 在将 UUID 发送到 SQL Server 之前将其转换为其二进制形式。不幸的是，二进制格式略有不同，导致 GUID 的字符串表示（例如，使用 SSMS 查看它们时）不同，这至少是令人困惑的。

这可以在 SQL Server 中通过将列的类型更改为 uuid-char 来解决

@Id
@Type(type = "uuid-char")
@Column(name = "id")
public UUID getId() {
    return …