Question

我正在尝试使用JPA选择查询，但我收到了一些错误：

这是我的CountryDto.java文件

@Entity    
@Table(name = "COUNTRY")
public class CountryDto {

   @Id
   @GeneratedValue(strategy = GenerationType.AUTO)  
   @Column(name="COUNTRY_ID")
   private int Country_Id;
   @Column(name="COUNTRY")
   private String Country;  

   // Getters & Setters go here

   public CountryDto() {
    // TODO Auto-generated constructor stub
   }
   public CountryDto(int country_Id, String country) {
   //   super();
       Country_Id = country_Id;
       Country = country;
   }
}

My Persistence.xml文件如下：

<persistence-unit name="texttiledb" transaction-type="RESOURCE_LOCAL">
    <class>com.textileworld.mill.Dto.CountryDto</class>
    <properties>
        <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
        <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/texttiledb" />
        <property name="javax.persistence.jdbc.user" value="root" />
        <property name="javax.persistence.jdbc.password" value="root" />
        <property name="eclipselink.logging.level" value="FINE" />
        <property name="eclipselink.ddl-generation" value="create-tables" />
    </properties>
</persistence-unit>

这是我试图运行的代码

    EntityManagerFactory emfactory = Persistence.createEntityManagerFactory("texttiledb");
    EntityManager entitymanager = emfactory.createEntityManager();
    entitymanager.getTransaction( ).begin( );
    Query query = entitymanager.createQuery("SELECT c FROM COUNTRY c",CountryDto.class);
         List<CountryDto> rol= (List<CountryDto> ) query.getResultList();
         for (CountryDto con : rol){
             System.out.println(con);
         }

    entitymanager.getTransaction( ).commit( );
    entitymanager.close( );
    emfactory.close( );

但是我不知道为什么它显示错误，他抽象模式类型'COUNTRY'是未知的。

这是我得到的错误：

Exception Description: Problem compiling [SELECT c FROM COUNTRY c]
[14, 21] The abstract schema type 'COUNTRY' is unknown.
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1625)
    at com.textileworld.mill.Dao.CountryDao.main(CountryDao.java:24)
Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.5.2.v20140319-9ad6abd): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Problem compiling [SELECT c FROM COUNTRY c]. 
[14, 21] The abstract schema type 'COUNTRY' is unknown.
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildException(HermesParser.java:155)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.validate(HermesParser.java:347)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.populateQueryImp(HermesParser.java:278)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildQuery(HermesParser.java:163)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:142)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:116)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:102)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:86)
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1603)

Answer 1

试试这个......首先你应该使用NamedQuery来提高性能。

    EntityManagerFactory emfactory = Persistence.createEntityManagerFactory("texttiledb");
    EntityManager entitymanager = emfactory.createEntityManager();
    entitymanager.getTransaction( ).begin( );
    Query query = entitymanager.createNamedQuery("selectAllCountry");
         List<CountryDto> rol= (List<CountryDto> ) query.getResultList();
         for (CountryDto con : rol){
             System.out.println(con);
         }

    entitymanager.getTransaction( ).commit( );
      entitymanager.close( );
      emfactory.close( );

在CountryDto中定义您的命名查询

@Entity
@NamedQueries({
        @NamedQuery(name = "selectAllCountry", query = "SELECT C FROM CountryDto C")
})
@Table(name = "COUNTRY")
public class CountryDto {

@Id
@GeneratedValue(strategy = GenerationType.AUTO)     
@Column(name="COUNTRY_ID")
private int Country_Id;
@Column(name="COUNTRY")
private String Country;

public String getCountry() {
    return Country;
}
public int getCountry_Id() {
    return Country_Id;
}
public void setCountry(String country) {
    Country = country;
}
public void setCountry_Id(int country_Id) {
    Country_Id = country_Id;
}
public CountryDto() {
    // TODO Auto-generated constructor stub
}
public CountryDto(int country_Id, String country) {
//  super();
    Country_Id = country_Id;
    Country = country;
}

Answer 2

尝试使用此查询：

"SELECT c FROM CountryDto c"

而不是：

"SELECT c FROM COUNTRY c"

@Table(name = "COUNTRY")注释用于DB。

Answer 3

按如下方式更改查询。

Query query = entitymanager.createQuery("SELECT c FROM CountryDao c ",CountryDto.class);

请记住以下内容。

@Table是可选的。将POJO类注释为实体

需要@Entity

如果实体创建如下，

 @Entity(name="MyEntityName")
 @Table(name="MyEntityTableName")
 class MyEntity {

然后创建名为 MyEntityTableName 的表格，实体名称为 MyEntityName 。

您的JPQL查询将是：

select * from MyEntityName

或者，如果实体创建如下，

@Entity
 class MyEntity {

将创建名称为 MyEntity 的表格，实体名称将为 MyEntity 。

您的JPQL查询将是：

select * from MyEntity

希望这有帮助。

使用JPA执行Select语句时出错

3 个答案: