JPA序列化
我正在 MySQL 5 上使用 JPA 2.0-EclipseLink-Hibernate 3。我的类结构如下所示(我对其进行了简化)
public class Status implements Serialization {
private String name = null;
public Status(string n) {
this.name = n;
}
@Override
public String toString() {
return this.name;
}
}
@Entity
@Table(name="ACCOUNT")
public class Account {
public static final Status ACTIVE = new Status("AC");
public static final Status DISABLE = new Status("DI");
private Status status = null;
private long id = 0;
public Account(long id, Status s) {
this.id = id;
this.status = s;
}
@Id
@Column(name="id")
public long getId() {
return this.id;
}
public setId(long id) {
this.id = id;
}
@Column(name="status")
public Status getStatus() {
return this.status;
}
public setStatus(Status s) {
this.status = s;
}
}
现在,我的问题是,当我使用 EntityManager.persist 插入新帐户时,程序会引发异常:
java.sql.SQLException:错误的字符串值:'\xAC\xED\x00\x05sr...' 对于第 1 行的“状态”列
,在我手动将新帐户添加到数据库中,然后选择它后,程序显示:
异常描述:无法从字节数组反序列化对象。 内部异常:java.io.EOFException 映射: org.eclipse.persistence.mappings.DirectToFieldMapping[状态-->ACCOUNT.status] 描述: RelationalDescriptor(net.serenco.serenekids.model.bean.account.Account --> [数据库表(帐户)])
请告诉我如何控制将哪些值写入数据库,顺便请帮助我修复这些错误。提前致谢!
---------------- 编辑----------------
我自己设法找到了解决方案。这是:
@Embeddable
public class Status implements Serialization {
...
protected String getName() {
return name;
}
protected void setName(String name) {
this.name = name;
}
}
@Entity
@Table(name="ACCOUNT")
public class Account {
...
@Embedded
@AttributeOverrides( {
@AttributeOverride(name="name", column = @Column(name="status") )
} )
public Status getStatus() {
return this.status;
}
...
}
I'm working with JPA 2.0-EclipseLink-Hibernate 3 on MySQL 5. My classes structure is like below (I simplified it)
public class Status implements Serialization {
private String name = null;
public Status(string n) {
this.name = n;
}
@Override
public String toString() {
return this.name;
}
}
@Entity
@Table(name="ACCOUNT")
public class Account {
public static final Status ACTIVE = new Status("AC");
public static final Status DISABLE = new Status("DI");
private Status status = null;
private long id = 0;
public Account(long id, Status s) {
this.id = id;
this.status = s;
}
@Id
@Column(name="id")
public long getId() {
return this.id;
}
public setId(long id) {
this.id = id;
}
@Column(name="status")
public Status getStatus() {
return this.status;
}
public setStatus(Status s) {
this.status = s;
}
}
Now, my question is, when I insert a new Account using EntityManager.persist, the program throws an exception:
java.sql.SQLException: Incorrect string value: '\xAC\xED\x00\x05sr...'
for column 'status' at row 1
And after I added a new Account manually into the database, then I select it, the program say:
Exception Description: Could not deserialize object from byte array.
Internal Exception: java.io.EOFException Mapping:
org.eclipse.persistence.mappings.DirectToFieldMapping[status-->ACCOUNT.status]
Descriptor:
RelationalDescriptor(net.serenco.serenekids.model.bean.account.Account
--> [DatabaseTable(ACCOUNT)])
Please tell me how to control what value will be written to DB, and by the way, please help me fix these errors. Thanks in advance!
---------------- EDIT ----------------
I have managed to find out the solution myself. It is:
@Embeddable
public class Status implements Serialization {
...
protected String getName() {
return name;
}
protected void setName(String name) {
this.name = name;
}
}
@Entity
@Table(name="ACCOUNT")
public class Account {
...
@Embedded
@AttributeOverrides( {
@AttributeOverride(name="name", column = @Column(name="status") )
} )
public Status getStatus() {
return this.status;
}
...
}
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