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我的实体类：

     public class ACCOUNT implements Serializable{

private static final long serialVersionUID = 1L;

@Id
@Column(name = "USERNAME")
private String Username;

@Column(name = "PASSWORD")
private String Password;

public ACCOUNT(String user,String pass)
{
    this.Username=user;
    this.Password=pass;
}

// geter and setter

我的persistence.xml

<?xml version="1.0" encoding="UTF-8" ?>
<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_1_0.xsd" version="1.0">

<persistence-unit name="transactions-optional">
    <provider>org.datanucleus.api.jpa.PersistenceProviderImpl</provider>
    <class>com.materialshop.server.Datastore.ACCOUNT</class>
    <exclude-unlisted-classes/>
    <properties>
        <property name="datanucleus.NontransactionalRead" value="true"/>
        <property name="datanucleus.NontransactionalWrite" value="true"/>
        <property name="datanucleus.nontx.atomic" value="true"/>
        <property name="datanucleus.ConnectionURL" value="appengine"/>
    </properties>
</persistence-unit>
</persistence>

现在持久化帐户，它是成功的

ACCOUNT ac=new ACCOUNT("admin","123");
em.persist(ac);

我检查了

     http://localhost:8888/_ah/admin

是的，有 1 个记录管理员和 123 个 ACOUNT 实体

但是当我使用 JPQL 获取所有实体时，它返回 null

Query q=new Query("SELECT ac FROM ACCOUNT ac");
List<ACCOUNT> list=q.getResultList();

即使

 em.find(ACCOUNT.class,"admin");

也返回 null

我错过了什么 ？请帮助我，非常感谢您的帮助

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0 回答 0