休眠:参数索引超出范围(8>参数数量,即7) [英] Hibernate: Parameter index out of range (8 > number of parameters, which is 7)
问题描述
不是该问题的重复参数索引超出范围(8个参数数量,即7)
我的SaltTranDef实体类是
My SaltTranDef entity class is
@Id
@Column(name="salt_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer saltId;
@Column(name="tran_type")
private String transactionType;
@Column(name="user_id")
private String userId;
@Column(name="parent_system")
private String parentSystem;
@Column(name="parent_sys_ref_id")
private String parentSystemReference;
@Column(name="status")
private int status;
@OneToMany(mappedBy = "saltTranDef")
@Cascade({ org.hibernate.annotations.CascadeType.ALL,
org.hibernate.annotations.CascadeType.DELETE_ORPHAN })
private Set<SaltTranUser> saltTranUsers;
SaltTranUser实体类是
And the SaltTranUser entity class is
@Id
@Column(name="salt_id")
private Integer saltId;
@Id
@Column(name="salt_property")
private String saltProp;
@Column(name="salt_value")
private String saltValue;
@ManyToOne
@JoinColumn(name="salt_id")
private SaltTranDef saltTranDef;
以上两个实体类都扩展了一个appeddSuperclass
Both the above entity classes extend a mappedSuperclass
@Column(name="cre_suid")
private String creatorId;
@Column(name="mod_suid")
private String modifiedId;
@Column(name="cre_date")
private Timestamp creationDate;
@Column(name="mod_date")
private Timestamp modificationDate;
从JUnit插入时:
@Test
public void testInsert(){
SaltTranDef std = new SaltTranDef();
SaltTranUser stu1 = new SaltTranUser();
SaltTranUser stu2 = new SaltTranUser();
SaltTranUser stu3 = new SaltTranUser();
Set<SaltTranUser> set1 = new HashSet<SaltTranUser>();
Transaction tx = session.beginTransaction();
std.setParentSystem("A");
std.setParentSystemReference("123");
std.setStatus(10);
std.setTransactionType("A");
std.setUserId("1234");
std.setCreationDate(new Timestamp(new Date().getTime()));
std.setCreatorId("1234");
session.persist(std);
// session.flush();
stu1.setCreationDate(new Timestamp(new Date().getTime()));
stu1.setCreatorId("1234");
stu1.setSaltProp("Fname");
stu1.setSaltValue("Swateek");
stu1.setSaltId(std.getSaltId());
stu2.setCreationDate(new Timestamp(new Date().getTime()));
stu2.setCreatorId("1234");
stu2.setSaltProp("Lname");
stu2.setSaltValue("Jena");
stu2.setSaltId(std.getSaltId());
stu3.setCreationDate(new Timestamp(new Date().getTime()));
stu3.setCreatorId("1234");
stu3.setSaltProp("Phone");
stu3.setSaltValue("9900056668");
stu3.setSaltId(std.getSaltId());
set1.add(stu1);
set1.add(stu2);
set1.add(stu3);
std.setSaltTranUsers(set1);
session.save(std);
tx.commit();
}
我收到一条错误消息:
严重:参数索引超出范围(8>参数数,即7). 2015年3月25日,上午8:06:35 org.hibernate.event.def.AbstractFlushingEventListener performExecutions 严重:无法将数据库状态与会话同步 org.hibernate.exception.GenericJDBCException:无法插入:[com.salt.entity.SaltTranUser] 在org.hibernate.exception.SQLStateConverter.handledNonSpecificException(SQLStateConverter.java:103) 由以下原因引起:java.sql.SQLException:参数索引超出范围(8>参数数量,即7). 在com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1055)
SEVERE: Parameter index out of range (8 > number of parameters, which is 7). Mar 25, 2015 8:06:35 AM org.hibernate.event.def.AbstractFlushingEventListener performExecutions SEVERE: Could not synchronize database state with session org.hibernate.exception.GenericJDBCException: could not insert: [com.salt.entity.SaltTranUser] at org.hibernate.exception.SQLStateConverter.handledNonSpecificException(SQLStateConverter.java:103) Caused by: java.sql.SQLException: Parameter index out of range (8 > number of parameters, which is 7). at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1055)
推荐答案
这种问题几乎总是与双列映射有关.的确如此.我们可以看到,此映射两次使用了一次列 "salt_id"
:
This kind of problem is almost always related to double column mapping. And indeed. We can see, that this mapping uses one column twice "salt_id"
:
SaltTranUser
实体类:
@Id
@Column(name="salt_id")
private Integer saltId;
...
@ManyToOne
@JoinColumn(name="salt_id")
private SaltTranDef saltTranDef;
那是错误的.最后,Hibernate会在一列中插入两次,即在INSERT,UPDATE中添加更多的参数,然后再插入列
And that is wrong. Hibernate is at the end inserting into one column twice, i.e. more arguments then columns in INSERT, UPDATE
这里的解决方案很可能非常简单-因为@ManyToOne
似乎是错误的.我希望有一些特殊的列可供参考,例如: SaltTranDef_id
Solution here would be mostlikely very simple - because the @ManyToOne
seems to be wrong. I would expect some special column for reference like: SaltTranDef_id
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