Question

我正在尝试查询DerbyDB架构user中gomobile表的所有数据。

我已成功建立与数据库的连接并创建了一个JPA实体，其所有列都对应于数据库表。

@Entity
@Table(name = "user", schema = "gomobile")
public class User implements Serializable {
    private static final long serialVersionUID = 1L;

    // all columns

    public static List<User> getAll() {
        String queryString = "SELECT u FROM gomobile.user u";
        EntityManager em = Persistence.createEntityManagerFactory("Eclipselink").createEntityManager();
        return em.createQuery(queryString, User.class).getResultList();
    }
}

这是stracktrace：

Exception in thread "main" java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Problem compiling [SELECT u FROM gomobile.user as u]. 
[14, 41] 'gomobile.user as u' cannot be the first declaration of the FROM clause.
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1625)
    at com.sap.sapchat.jpa.entities.User.getAll(User.java:45)
    at com.sap.sapchat.jpa.entities.InitDatabase.main(InitDatabase.java:50)
Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.5.1.v20130918-f2b9fc5): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Problem compiling [SELECT u FROM gomobile.user as u]. 
[14, 41] 'gomobile.user as u' cannot be the first declaration of the FROM clause.
    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)
    ... 3 more

我在persistence.xml中的持久性声明如下所示：

<persistence-unit name="Eclipselink" transaction-type="RESOURCE_LOCAL">
    <class>jpa.entities.User</class>
    <properties>
        <property name="javax.persistence.jdbc.url" value="jdbc:derby://localhost:1527/gomobile;create=true" />
        <property name="javax.persistence.jdbc.user" value="gomobile" />
        <property name="javax.persistence.jdbc.password" value="mypassword" />
        <property name="javax.persistence.jdbc.driver" value="org.apache.derby.jdbc.ClientDriver" />
    </properties>
</persistence-unit>

修改

如果我使用：

String queryString = "SELECT * FROM gomobile.user u";

我收到此错误：

Exception in thread "main" java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Syntax error parsing [SELECT * FROM gomobile.user u]. 
[38, 38] A select statement must have a FROM clause.
[7, 7] The left expression is missing from the arithmetic expression.
[9, 38] The right expression is not an arithmetic expression.
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1625)
    at com.sap.sapchat.jpa.entities.User.getAll(User.java:75)
    at com.sap.sapchat.jpa.entities.InitDatabase.main(InitDatabase.java:64)
Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.5.1.v20130918-f2b9fc5): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing [SELECT * FROM gomobile.user u]. 
[38, 38] A select statement must have a FROM clause.
[7, 7] The left expression is missing from the arithmetic expression.
[9, 38] The right expression is not an arithmetic expression.
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildException(HermesParser.java:155)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.validate(HermesParser.java:334)
    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)
    ... 3 more

Answer 1

没有名为gomobile.user的实体，因此您无法在JPQL查询中使用它。 JPQL是基于对象的，不像在SQL中那样直接使用表/模式和字段。

您应该只使用“SELECT u FROM User u”，因为您要查询的实体默认名为“用户”。

Answer 2

此问题是因为JPA希望不直接允许JPQL语法和像您这样的本机SQL。但是，有一种实体管理器支持的方法，这是一种可以直接使用普通SQL的简单方法。

关于如何在JPA中使用本机SQL的示例链接：http://www.thoughts-on-java.org/jpa-native-queries/

'gomobile.user u'不能是FROM子句的第一个声明

2 个答案: