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java - Hibernate 多对多映射的问题

问题描述

我有一个hibernate + oracle程序。在那，我通过“ApplicationIdentityProviders”表在“Applications”和“IdentityProviders”表之间建立了多对多的关系。运行程序时出现以下错误：

Exception in thread "main" java.lang.RuntimeException: org.hibernate.MappingException: 
Could not determine type for: String, at table: identity_providers, for columns: [org.hibernate.mapping.Column(issuer)]
at com.oracle.hibernate.OCIAuthManager.setupTenant1Pdb1(OCIAuthManager.java:246)
at com.oracle.hibernate.OCIAuthManager.main(OCIAuthManager.java:283)

我无法弄清楚为什么。Applications.java 在这里。IdentityProviders.java 在这里。ApplicationIdentityProviders.java 在这里。主课在这里。Applications.hbm.xml 如下：

<?xml version = "1.0" encoding = "utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC 
"-//Hibernate/Hibernate Mapping DTD//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"> 

<hibernate-mapping>
 <class name = "Applications" table = "applications">

  <meta attribute = "class-description">
     This class contains the applications details. 
  </meta>

  <id name = "application_id" type = "int" column = "application_id">
     <generator class="native"/>
  </id>

  <set name = "application_id_providers" cascade="save-update" table="application_identity_providers">
     <key column = "application_id"/>
     <many-to-many column = "identity_provider_id" class="IdentityProviders"/>
  </set>

  <property name = "application_name" column = "application_name" type = "string"/>
  <property name = "attr" column = "attr" type = "short"/>
  <property name = "salary" column = "salary" type = "int"/>

 </class>

 <class name = "IdentityProviders" table = "identity_providers">

  <meta attribute = "class-description">
     This class contains the identity providers' details. 
  </meta>

  <id name = "identity_provider_id" type = "int" column = "identity_provider_id">
     <generator class="native"/>
  </id>

  <property name = "attr" column = "attr" type = "short"/>
  <property name = "protocols" column = "protocols" type = "short"/>
  <property name = "issuer" column = "issuer" type = "String"/>
  <property name = "cert1" column = "cert1" type = "String"/>
  <property name = "cert2" column = "cert2" type = "String"/>
  <property name = "is_tenant_default" column = "is_tenant_default" type = "short"/>

 </class>

</hibernate-mapping>

hibernate-1.cfg.xml如下：

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
    "-//Hibernate/Hibernate Configuration DTD 3.0//EN"
    "http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
  <session-factory>

    <property name="connection.url">jdbc:oracle:thin:@//localhost:1521/t1p1.oradev.oraclecorp.com</property>
    <property name="connection.username">username</property>
    <property name="connection.password">password</property>
    <property name="connection.driver_class">oracle.jdbc.OracleDriver</property>
    <property name="dialect">org.hibernate.dialect.Oracle12cDialect</property>

    <property name="show_sql">true</property>

    <property name="format_sql">true</property>
    <property name="hbm2ddl.auto">create</property>

    <!-- JDBC connection pool (use the built-in) -->
    <property name="connection.pool_size">1</property>
    <property name="current_session_context_class">thread</property>

    <mapping resource = "Applications.hbm.xml"/>

  </session-factory>
</hibernate-configuration>

我哪里错了？请帮忙。

更新 1：@Guillaume 的回答有效。但我现在收到以下错误：

Exception in thread "main" java.lang.RuntimeException:
org.hibernate.MappingException: An association from the table 
application_identity_providers refers to an unmapped class: 
IdentityProviders at
com.oracle.hibernate.OCIAuthManager.setupTenant1Pdb1(OCIAuthManager.java:246) 
at com.oracle.hibernate.OCIAuthManager.main(OCIAuthManager.java:283)

更新 2：我解决了更新 1 中的错误。现在我收到以下错误：错误：ORA-01400：无法将 NULL 插入 ("SYS"."APPLICATION_IDENTITY_PROVIDERS"."APPLICATION_IDENTITY_PROVIDER_ID")

此错误发生在包含其他两个表的主键的中间表中，这些表具有多对多映射。在 ApplicationIdentityProviders.java 中，我有：

@Id
@SequenceGenerator(name="seq",sequenceName="oracle_seq",allocationSize=1)        
@GeneratedValue(strategy=GenerationType.SEQUENCE, generator="seq")
private int application_identity_provider_id;

当我运行程序时，相关的输出片段是： Hibernate:

create table application_identity_providers (
application_identity_provider_id number(10,0) not null,
application_id number(10,0) not null,
identity_provider_id number(10,0) not null,
is_application_default number(5,0) not null,
primary key (application_id, identity_provider_id)
)

我哪里错了？

更新 3：给定“applications”表中的“application_name”，获取“identity_providers”和“saml_identity_providers”表中的全部内容。怎么做？

我尝试了以下方法：

Query query = entitymanager.createQuery("select a from Applications a join 
fetch a.ApplicationIdentityProviders ap " +
"join fetch ap.IdentityProviders ip where a.application_name = 
'newApplicationName'");

它给了我这个错误：

org.springframework.beans.factory.BeanCreationException: Error creating bean 
with name 'applicationsRepository' defined in file 
[D:\filepath\ApplicationsRepository.class]: Bean instantiation via 
constructor failed; nested exception is 
org.springframework.beans.BeanInstantiationException: Failed to instantiate 
[ApplicationsRepository]: Constructor threw exception; nested exception is 
java.lang.IllegalArgumentException: org.hibernate.QueryException: could not 
resolve property: ApplicationIdentityProviders of: Applications [select a 
from com.package.entities.Applications a join fetch 
a.ApplicationIdentityProviders ap join fetch 
ap.IdentityProviders ip where a.application_name = 'newApplicationName']

标签： javaxmlhibernatemany-to-many

解决方案

更快的方法是使用 Hibernate 工具自动生成映射文件。请参考以下文章。

如何在 Eclipse 中安装 Hibernate 工具？
http://www.mkyong.com/hibernate/how-to-generate-code-with-hibernate-tools/

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