我正在尝试像Hibernate User Guide (2.7.2)那样执行单向@OneTaMany关系,但是当我尝试将以下对象保存在MariaDB数据库中时:
Filter filter = new Filter("TLS_A320");
filter.addConstraint(new Constraint(ConstraintType.DEPARTURE, "TLS"));
filter.addConstraint(new Constraint(ConstraintType.AIRCRAFT, "A320"));
session.save(filter);
我遇到了这个例外:
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'Filter_idFilter' in 'field list'
以下是两个班级:
Filter.java
public class Filter {
@Id
@Column(name = "idFilter")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@Column(name = "name")
private String name;
@OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
private List<Constraint> constraintList;
//more code
};
Constraint.java
public class Constraint {
@Id
@Column(name = "idConstraint")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@Column(name = "type")
@Enumerated(EnumType.ORDINAL)
private ConstraintType constraintType;
@Column(name = "value")
private String value;
//more code
}
以下是表格定义:
CREATE TABLE `Constraint` (
idConstraint INT NOT NULL AUTO_INCREMENT,
type INT NOT NULL,
value VARCHAR(10) NOT NULL,
PRIMARY KEY (idConstraint)
);
CREATE TABLE Filter (
idFilter INT NOT NULL AUTO_INCREMENT,
name VARCHAR(50) UNIQUE NOT NULL,
PRIMARY KEY (idFilter)
);
CREATE TABLE Filter_Constraint (
idFilter INT UNIQUE NOT NULL,
idConstraint INT NOT NULL,
CONSTRAINT fk_Filter_Constraint_Filter FOREIGN KEY (idFilter) REFERENCES Filter(idFilter),
CONSTRAINT fk_Filter_Constraint_Constraint FOREIGN KEY (idConstraint) REFERENCES `Constraint`(idConstraint)
);
在我看来,Filter和Constraint插入很好,插入Filter_Constraint表时会发生异常:
DEBUG org.hibernate.SQL - insert into Filter (name) values (?)
DEBUG org.hibernate.id.IdentifierGeneratorHelper - Natively generated identity: 4
DEBUG org.hibernate.SQL - insert into `Constraint` (type, value) values (?, ?)
DEBUG org.hibernate.id.IdentifierGeneratorHelper - Natively generated identity: 5
DEBUG org.hibernate.SQL - insert into `Constraint` (type, value) values (?, ?)
DEBUG org.hibernate.id.IdentifierGeneratorHelper - Natively generated identity: 6
DEBUG org.hibernate.SQL - insert into `Filter_Constraint` (Filter_idFilter, `constraintList_idConstraint`) values (?, ?)
DEBUG org.hibernate.engine.jdbc.spi.SqlExceptionHelper - could not execute statement [n/a]
DEBUG com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'Filter_idFilter' in 'field list'
我对Hibernate很陌生,我无法弄清楚我做错了什么。
答案 0 :(得分:2)
您没有在constraintList
上定义任何连接列或反向连接列,因此Hibernate会自动生成最可能导致名称为Filter_idFilter
的列名。
实体名称将成为名称的一部分,因为Constraints
可以具有相同名称的ID字段,即idFilter
。当然你知道并非如此,但Hibernate没有(或者至少列名生成代码不是那么复杂)因此它必须假设可能就是这样。
连接表名称也是如此,因此您可能希望@JoinTable
使用constraintList
上的连接列和反连接列的定义。但是,由于关系为OneToMany
,为什么不向Constraint
添加反向引用,将过滤器的ID添加到表Constraint
并删除完全连接表?