使用hibernate注释进行软删除时出错

Question

我有以下JPA实体

@SQLDelete(sql="UPDATE service SET date_deletion =  CURRENT_DATE() WHERE id = ?")
@Where(clause="date_deletion IS NULL ")
public class Service {
...
}

选择工作确定所有通知date_deletion的元素都不显示，但是当我尝试删除....

16:38:26,836  DEBUG SQL:111 - UPDATE service SET date_deletion =  CURRENT_DATE() WHERE id = ?
16:38:26,836  DEBUG AbstractBatcher:418 - about to close PreparedStatement (open PreparedStatements: 1, globally: 1)
16:38:26,836  DEBUG JDBCExceptionReporter:225 - could not delete: [com.foo.domain.Service#1] [UPDATE service SET date_deletion =  CURRENT_DATE() WHERE id = ?]
java.sql.SQLException: Parameter index out of range (2 > number of parameters, which is 1).

SQL有什么问题？看起来像尝试处理CURRENT_DATE（）作为参数并期望2个参数而不是1 ...

Answer 1

固定。 我正在使用Spring Roo在内部处理作为参数发送的“版本”字段，正确的anotation是：

@SQLDelete(sql="UPDATE service SET date_deletion=CURRENT_DATE WHERE id=? and version=? ")
@Where(clause="date_deletion IS NULL ")
public class Service {
...