I Have a problem executing an hibernate query on Oracle:
my persistence.xml:
<description>ProfileEntityManager</description>
<class>com.entity.User</class>
<exclude-unlisted-classes>true</exclude-unlisted-classes>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.OracleDialect"/>
<property name="javax.persistence.jdbc.driver" value="oracle.jdbc.OracleDriver"/>
<property name="javax.persistence.jdbc.url" value="jdbc:oracle:thin:@XXX:9999/db"/>
<property name="javax.persistence.jdbc.user" value="user"/>
<property name="javax.persistence.jdbc.password" value="password"/>
<property name = "hibernate.show_sql" value = "true" />
</properties>
my entity class com.entity.User:
package com.entity;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;
@Entity
@Table(name ="USER", schema = "SCHEMA")
public class User{
@Id
@Column(name = "USER_ID")
private String userId;
@Column(name = "FIRST_NAME")
private String firstName;
@Column(name = "LAST_NAME")
private String lastName;
public String getUserId() {
return userId;
}
public void setUserId(String userId) {
this.userId = userId;
}
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
}
my java class:
//System.out.println(uid);
List<User> lp = null;
try {
lp = em.createNativeQuery("SELECT iu FROM User iu WHERE iu.userId = ?", User.class)
.setParameter(1, uid).getResultList();
} catch (Exception e) {
System.out.println(e.getMessage());
writeLog(fw, e.getMessage());
}
but when i execute my code, i obtain:
Hibernate: SELECT iu FROM User iu WHERE iu.userId = ?
gen 29, 2021 6:36:59 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 942, SQLState: 42000
gen 29, 2021 6:36:59 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: ORA-00942: tabella o vista inesistente
org.hibernate.exception.SQLGrammarException: could not extract ResultSet
Can you help me to find the error?
NOTE: if a use a PreparedStatement with the same connections and query i don’t have any issue, so it is not a problem of permissions
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Answer
You try to use HQL
query with EntityManager.createNativeQuery
method. This is a mistake.
Try to correct your query in this way:
em.createNativeQuery("SELECT * FROM SCHEMA.USER WHERE USER_ID = :usrid", User.class)
.setParameter("usrid", uid)
.getResultList();
If you use the SCHEMA
for all your entities, you can put it to the hibernate config:
<property name="hibernate.default_schema" value="SCHEMA"/>
then correct your entity definition in the following way:
@Entity
@Table(name ="USER")
public class User {
// ...
}
and then correct your query:
em.createNativeQuery("SELECT * FROM {h-schema}USER WHERE USER_ID = :usrid", User.class)
.setParameter("usrid", uid)
.getResultList();
P.S. And by the way, I would suggest you to correct your hibernate dialect definition. You should use as specific dialect as possible. For example if you use Oracle 10g, you should use org.hibernate.dialect.Oracle10gDialect
. You can still use org.hibernate.dialect.Oracle9iDialect
but some new features defined in Oracle10gDialect
will not be available.