My entity class :
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
my 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>
Now persist account and it was successful
ACCOUNT ac=new ACCOUNT("admin","123");
em.persist(ac);
I checked http://localhost:8888/_ah/admin and yes ,there is ACOUNT entity
with 1 record admin and 123
but when I use JPQL to get all entity , it returned null
Query q=new Query("SELECT ac FROM ACCOUNT ac");
List<ACCOUNT> list=q.getResultList();
even with
em.find(ACCOUNT.class,"admin"); return null too
Did i miss something ? Please help me , very grateful for your help
--
You received this message because you are subscribed to the Google Groups
"Google App Engine" group.
To view this discussion on the web visit
https://groups.google.com/d/msg/google-appengine/-/_EC0SHCHncYJ.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/google-appengine?hl=en.
