Stu*_*son 66 java mysql hibernate jpa
我一直在玩一个非常简单的JPA示例,并试图将其调整到现有数据库.但我无法克服这个错误.(下面.)它只是一些我没有看到的简单的事情.
org.hibernate.hql.internal.ast.QuerySyntaxException: FooBar is not mapped [SELECT r FROM FooBar r]
org.hibernate.hql.internal.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:180)
org.hibernate.hql.internal.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:110)
org.hibernate.hql.internal.ast.tree.FromClause.addFromElement(FromClause.java:93)
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在下面的DocumentManager类(一个简单的servlet,因为这是我的目标目标)做两件事:
插入效果很好 - 一切都很好.问题在于检索.我已经为Query q = entityManager.createQuery参数尝试了各种各样的值,但没有运气,我尝试了各种更复杂的类注释(如列类型),但都没有成功.
请救我自己.我确定它很小.我对JPA的经验不足使我无法继续前进.
我的./src/ch/geekomatic/jpa/FooBar.java文件:
@Entity
@Table( name = "foobar" )
public class FooBar {
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="id")
private int id;
@Column(name="rcpt_who")
private String rcpt_who;
@Column(name="rcpt_what")
private String rcpt_what;
@Column(name="rcpt_where")
private String rcpt_where;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getRcpt_who() {
return rcpt_who;
}
public void setRcpt_who(String rcpt_who) {
this.rcpt_who = rcpt_who;
}
//snip...the other getters/setters are here
}
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我的./src/ch/geekomatic/jpa/DocumentManager.java类
public class DocumentManager extends HttpServlet {
private EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory( "ch.geekomatic.jpa" );
protected void tearDown() throws Exception {
entityManagerFactory.close();
}
@Override
public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException {
FooBar document = new FooBar();
document.setRcpt_what("my what");
document.setRcpt_who("my who");
persist(document);
retrieveAll(response);
}
public void persist(FooBar document) {
EntityManager entityManager = entityManagerFactory.createEntityManager();
entityManager.getTransaction().begin();
entityManager.persist( document );
entityManager.getTransaction().commit();
entityManager.close();
}
public void retrieveAll(HttpServletResponse response) throws IOException {
EntityManager entityManager = entityManagerFactory.createEntityManager();
entityManager.getTransaction().begin();
// *** PROBLEM LINE ***
Query q = entityManager.createQuery( "SELECT r FROM foobar r", FooBar.class );
List<FooBar> result = q.getResultList();
for ( FooBar doc : result ) {
response.getOutputStream().write(event.toString().getBytes());
System.out.println( "Document " + doc.getId() );
}
entityManager.getTransaction().commit();
entityManager.close();
}
}
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{tomcat-home} /webapps/ROOT/WEB-INF/classes/METE-INF/persistance.xml文件
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0">
<persistence-unit name="ch.geekomatic.jpa">
<description>test stuff for dc</description>
<class>ch.geekomatic.jpa.FooBar</class>
<properties>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://svr:3306/test" />
<property name="javax.persistence.jdbc.user" value="wafflesAreYummie" />
<property name="javax.persistence.jdbc.password" value="poniesRock" />
<property name="hibernate.show_sql" value="true" />
<property name="hibernate.hbm2ddl.auto" value="create" />
</properties>
</persistence-unit>
</persistence>
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MySQL表描述:
mysql> describe foobar;
+------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| rcpt_what | varchar(255) | YES | | NULL | |
| rcpt_where | varchar(255) | YES | | NULL | |
| rcpt_who | varchar(255) | YES | | NULL | |
+------------+--------------+------+-----+---------+----------------+
4 rows in set (0.00 sec)
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Chr*_*ris 162
JPQL 主要是不区分大小写的.区分大小写的一个是Java实体名称.将您的查询更改为:
"SELECT r FROM FooBar r"
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小智 14
还有另一种可能的错误来源.在一些J2EE/Web容器中(根据我在Jboss 7.x和Tomcat 7.x下的经验)您必须将要用作hibernate Entity的每个类添加到文件persistence.xml中
<class>com.yourCompanyName.WhateverEntityClass</class>
在jboss的情况下,这涉及每个实体类(本地 - 即您正在开发的项目或库中).对于Tomcat 7.x,这只涉及库中的实体类.
小智 5
您已将您的班级声明为:
@Table( name = "foobar" )
public class FooBar {
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您需要为搜索编写类名.
从FooBar
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