我使用以下技术堆栈:
用于连接池的
我的实际代码如下所示。
/// My trigger looks like this
CREATE OR REPLACE TRIGGER FILE_BRI
BEFORE INSERT
ON FILE
FOR EACH ROW
BEGIN
SELECT FILE_SEQ.NEXTVAL INTO :NEW.ID FROM DUAL;
END;
///
public class FILE implements Serializable {
@Id
@SequenceGenerator(
name = "FILE_SEQ",
sequenceName = "FILE_SEQ",
allocationSize = 1)
@GeneratedValue(
strategy = GenerationType.SEQUENCE,
generator = "FILE_SEQ"
)
private long id;
}
public class ServiceA () {
@Transactional(propagation = REQUIRES_NEW, isolation = READ_COMMITTED)
public File insertFile() {
// Below line returns the inserted File object with ID as '58496'
return fileRepository.save(file)
}
@Transactional(propagation = REQUIRES_NEW, isolation = READ_COMMITTED)
public AccessControl insertAccessControl() {
// Below line results in 'SQLIntegrityConstraintViolationException' (full error at the bottom of this post)
return accessControlRepository.save(accessControlFile)
}
}
Public class FileProcessor() {
ServiceA serviceA;
public void someMethod() {
// insert the file and get the inserted record
File insertedFile = serviceA.insertFile(file);
// get the ID from the inserted file and make another insert into another table
serviceA.insertAccessControl(insertedFile.getId()); // inserted file ID is '58496'
}
}
这是我的调查:
当我验证"FILE“表中插入记录的ID为'58497‘时,repository.save()
返回了一个不同的值。当我在FILE_ID
为'58496‘的表"ACCESS_CONTROL_FILE“上进行第二次插入时,会导致下面的错误,因为ID为'58496’的文件不存在。
原因: java.sql.SQLIntegrityConstraintViolationException: ORA-01400:无法在("DB_OWNER"."ACCESS_CONTROL_FILE"."FILE_ID")中插入NULL
我不明白为什么repository.save()
会返回一个与数据库中实际插入的ID(即ID=58496)不同的ID(即ID=58497)!
我研究了所有我能在互联网上找到的与“传播与隔离”相关的选项。
发布于 2018-12-14 01:09:53
正如评论中提到的,看起来是数据库触发器导致了这个问题。禁用触发器,让JPA管理ID生成。
https://stackoverflow.com/questions/53765290
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