JPA映射:"QuerySyntaxException:foobar未映射..."

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)
Run Code Online (Sandbox Code Playgroud)

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:

  1. 插入一行
  2. 返回所有行

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
}
Run Code Online (Sandbox Code Playgroud)

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();
    }
}
Run Code Online (Sandbox Code Playgroud)

{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>
Run Code Online (Sandbox Code Playgroud)

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)
Run Code Online (Sandbox Code Playgroud)

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"
Run Code Online (Sandbox Code Playgroud)

  • 是.虽然看起来像sql,但是你正在执行JPA查询语言; 所以你要从实体中选择,而不是从表中选择. (30认同)
  • 那么,到类名而不是表名? (6认同)
  • 你一定是在开玩笑吧!我只花了2个小时拆分配置文件,检查类路径并重新设计构建文件.我读了你的帖子并将"tableName"更改为"TableName"并且它有效! (6认同)

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

<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 {
Run Code Online (Sandbox Code Playgroud)

您需要为搜索编写类名.
FooBar