我試圖在我的DerbyDB的模式gomobile
中查詢我的user
表的所有數據。'gomobile.user u'不能成爲FROM子句的第一個聲明
我已成功建立與我的數據庫的連接並創建了一個JPA實體,其所有列都對應於數據庫表。
@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();
}
}
這是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
在persistence.xml我的持久性聲明如下所示:
<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>
編輯
如果我使用:
String queryString = "SELECT * FROM gomobile.user u";
我得到這個錯誤:
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
您是否嘗試過對您的Derby數據庫交互的運行只是1條SELECT語句?另外,什麼是完整的Derby異常:http://wiki.apache.org/db-derby/UnwindExceptionChain –
@BryanPendleton我沒有得到'SQL異常'。我得到一個'IllegalArgumentException'。如果我嘗試通過JDBC運行上面的查詢,我得到一個錯誤。我必須使用'SELECT * FROM gomobile.user'。我會更新我的線程。 – Niklas