Hibernate oracle标识符太长ORA-00972

  • Z时代
  • 2024-01-10
  • 分类：问答

我陷入了这个问题。数据库架构是由其他人提供的，因此我不能简单地更改名称。我尝试在各处添加适当的注释，也许我遗漏了一些（显而易见的）？

这是我的完整映射（很多类），我将省略getter / setter。

问题是当hibernate试图获得全部 List<ControlRuleAttrib> controlRuleAttribs

@Entity
@Table(name = "CONTROL_RULE")
public class ControlRule implements Serializable {
 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 @Column(name = "CONTROL_RULE_ID")
 private Long id;
 @ManyToOne(fetch = FetchType.LAZY)
 @Cascade(CascadeType.ALL)
 @JoinColumn(name = "CONTROL_RULE_TYPE_ID")
 @ForeignKey(name = "CONTROL_RULE_TYPE_ID")
 private ControlRuleType controlRuleType;
 @Column(name = "JOB_NM")
 private String jobname;
 @Column(name = "LIBRARY_NM")
 private String libraryname;
 @Column(name = "TABLE_NM")
 private String tablename;
 @Column(name = "COLUMN_NM")
 private String columnname;
 @OneToMany(fetch = FetchType.LAZY)
 @Cascade(CascadeType.ALL)
 @JoinTable(name = "CONTROL_RULE_ATTRIB", joinColumns = {
  @JoinColumn(name = "CONTROL_RULE_ID", nullable = false, updatable = false)
 })
 private List < ControlRuleAttrib > controlRuleAttribs;
}

@Table(name = "CONTROL_RULE_ATTRIB")
@Entity
public class ControlRuleAttrib {
 @EmbeddedId
 private ControlRuleAttribPK controlRuleAttribPK;
 @Column(name = "ATTRIBUTE_VALUE")
 private String attributeValue;
}

这里的问题是，是否有可能以某种方式得到实体ControlRuleAttribType的ControlRuleAttrib？如您所见，下面ControlRuleAttribTypeId是IDControleRuleAttribType。我想得到整个对象的等值整数。

@Embeddable
public class ControlRuleAttribPK implements Serializable {
 @Column(name = "CONTROL_RULE_ID")
 private Long controlRuleId;
 @Column(name = "ATTRIBUTE_SEQ_NUM")
 private Integer attributeSeqNum;
 @Column(name = "CONTROL_RULE_ATTRIB_TYPE_ID")
 private Integer controlRuleAttribTypeId;
}

@Entity
@Table(name = "CONTROL_RULE_ATTRIB_TYPE")
public class ControlRuleAttribType implements Serializable {
 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 @Column(name = "CONTROL_RULE_ATTRIB_TYPE_ID")
 private Integer id;
 @Column(name = "CONTROL_RULE_ATTRIB_TYPE_NM")
 private String typename;
 @Column(name = "CONTROL_RULE_ATTRIB_TYPE_DESC")
 private String typedesc;
 @ManyToOne(fetch = FetchType.LAZY)
 @Cascade(CascadeType.ALL)
 @JoinColumn(name = "CONTROL_RULE_TYPE_ID")
 @ForeignKey(name = "CONTROL_RULE_TYPE_ID")
 private ControlRuleType controlruletype;
}

@Entity
@Table(name = "CONTROL_RULE_TYPE")
public class ControlRuleType implements Serializable {
 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 @Column(name = "CONTROL_RULE_TYPE_ID")
 private Integer id;
 @Column(name = "CONTROL_RULE_TYPE_NM")
 private String typename;
 @Column(name = "CONTROL_RULE_TYPE_DESC")
 private String typedesc;
}

这是stacktrace：

https://gist.github.com/a30dd9ce534d96bb9a97

您会发现，它在这里失败：

在com.execon.controllers.main.MainPageController.getMainPage（MainPageController.java:33）上[类：]

就是这样：

List<ControlRule> list = SessionFactoryUtils.openSession(
    sessionFactory ).createQuery( "from ControlRule" ).list();
System.out.println( list );

我添加的映射的每个对象都有这样toString()声明的方法：

@Override
public String toString()
{
    String s = "ControlRule{";
    s += "id=" + id.toString();
    s += ", controlRuleType=" + controlRuleType;
    s += ", jobname='" + jobname + '\'';
    s += ", libraryname='" + libraryname + '\'';
    s += ", tablename='" + tablename + '\'';
    s += ", columnname='" + columnname + '\'';
    s += ", controlRuleAttribs=" + controlRuleAttribs;
    s += '}';
    return s;
}

并hibernate请求：

https://gist.github.com/c8584113522757a4e0d8/4f31dc03e7e842eef693fa7ba928e19d27b3ca26

请帮助 ：）

在阅读@Jens答案之后，我对代码做了一些更改。首先，我按照您的描述进行操作，但出现了错误：

org.hibernate.AnnotationException：从com.execon.models.controlrules.ControlRule引用com.execon.models.controlrules.ControlRuleAttrib的外键具有错误的列数。应该是3

我想这是正确的，因为我有复合主键。

然后我以这种方式尝试：

@OneToMany(fetch = FetchType.LAZY)
@Cascade(CascadeType.ALL)
@JoinTable(name = "CONTROL_RULE_ATTRIB",
        joinColumns = {
                @JoinColumn(name = "CONTROL_RULE_ID", nullable = false, updatable = false)
        },
        inverseJoinColumns = {
                @JoinColumn(name = "CONTROL_RULE_ID", nullable = false, updatable = false),
                @JoinColumn(name = "CONTROL_RULE_ATTRIB_TYPE_ID", nullable = false, updatable = false),
                @JoinColumn(name = "ATTRIBUTE_SEQ_NUM", nullable = false, updatable = false)
        })
private List<ControlRuleAttrib> controlRuleAttribs;

非常接近，但它给了我以下异常：

映射中重复的列用于收集。

所以最后我删除了

joinColumns = 
{
    @JoinColumn(name = "CONTROL_RULE_ID", nullable = false, updatable = false)
}

编译了所有东西，除了当我尝试到达集合时，Hibernate正在执行以下查询：

https://gist.github.com/c88684392f0b7a62bea5

最后一行是controlrul0_.CONTROL_RULE_CONTROL_RULE_ID=?应为controlrul0_.CONTROL_RULE_ID=?。

无论如何，我可以使它工作吗？：/

回答：

经过数小时的努力，我终于使它在我的项目中正常工作。我所做的是这样的：

@OneToMany(fetch = FetchType.LAZY, mappedBy = "controlRuleAttribPK.controlRuleId")
@Cascade(CascadeType.ALL)
private List<ControlRuleAttrib> controlRuleAttribs;

基本上指出集合应该使用复合主键中的controlRuleId。到目前为止，它的工作很棒！

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