JPA 映射:“QuerySyntaxException: foobar 未映射...”

2022-08-31 12:04:47

我一直在玩一个非常简单的JPA示例,并试图将其调整为现有的数据库。但是我无法克服这个错误。(下图。它只是一些我没有看到的简单事情。

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)

在下面的 DocumentManager 类(一个简单的 servlet,因为这是我的目标)中,执行了两件事:

  1. 插入一行
  2. 返回所有行

插入工作完美 - 一切都很好。问题出在检索上。我已经尝试了各种参数值,但没有运气,并且我尝试了各种更解释类的注释(如列类型),但没有成功。Query q = entityManager.createQuery

请把我从我自己身上救出来。我敢肯定这是一件小事。我对JPA的缺乏经验使我无法走得更远。

My ./src/ch/geekomatic/jpa/FooBar.java file:

@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
}

My ./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();
    }
}

The {tomcat-home}/webapps/ROOT/WEB-INF/classes/METE-INF/persistance.xml file

<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>

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)

答案 1

JPQL大多不区分大小写。区分大小写的一件事是 Java 实体名称。将您的查询更改为:

"SELECT r FROM FooBar r"

答案 2

此错误还有另一个可能的来源。在一些 J2EE / Web 容器中(根据我在 Jboss 7.x 和 Tomcat 7.x 下的经验),您必须将要用作休眠实体的每个类添加到文件持久性中.xml

<class>com.yourCompanyName.WhateverEntityClass</class>

在jboss的情况下,这涉及每个实体类(本地 - 即在您正在开发的项目中或在库中)。在Tomcat 7.x的情况下,这只涉及库中的实体类。


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