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60

@Column( name = "columnName") doesn't work

{ @Id @GeneratedValue @Column(name = "KEY_ID") private long id; private Long timestamp; @ManyToOne @Column(name = "KEY_DESCRIPTOR") private AfwKeyDescriptor keyDescriptor; @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL) @Column( name = "DATA_ITEMS
59

Getting a list of all the entity class names

I was wondering if there is any way to get a list of the entity names in the database? dmoshal ... . support Support Actually, there is a simpler way to get the class names (in one line of code ... because it is portable and provide better functionality (for example you can use it to get entity names or class names - which is not the same). support Support
58

How best to unbind all names?

portable way, by adding a 'name' field, which may be set as a primary key or as a unique index ... for this was I presumed that it would provide fast lookup, and the bookmark name I use is in fact simply
57

javax.persistence.PersistenceException: No Persistence provider for EntityManager named in Karaf 4.0.7 and OSGi DS test

.persistence.PersistenceException: No Persistence provider for EntityManager named objectdb://localhost:6136
11

Issues with JDO Casting & Full Qualified Names

Hi, As I've feared.. Fully Qualified Names in casting not working, throws an identifier expected ... base = "TEST"; } @Embeddable public static class EmbeddedA extends EmbeddedBase {   String name ... class EmbeddedA extends EmbeddedBase {     String name = "Alex"; } @Embeddable class
11

Literals in JPQL and Criteria Queries

values, but the fully qualified name of the enum type should always be specified. For example, assuming ... retrieval by type. In JPQL an entity type literal is written simply as the name of the entity class (e.g. Country). That is equivalent to Country.class in Java code. Notice that the name of the entity class
11

JPA Persistence Unit

://java.sun.com/xml/ns/persistence/persistence_1_0.xsd" version="1.0"> <persistence-unit name ... > <properties> <property name="javax.persistence.jdbc.url" value="objectdb://localhost/my.odb"/> <property name="javax.persistence.jdbc.user" value="admin
2

Step 4: Add a Controller Class

) and selecting New > Class. The package name should be guest. Enter GuestController as the class name - use exactly that case sensitive class name. Click Finish to create the new Spring Controller class. Now ... (if any): String name = request.getParameter("name"); if (name != null
2

Step 4: Add a Servlet Class

... Enter GuestServlet as the class name - use exactly that case sensitive class name. The Java package name should be guest. Click Finish to create the new servlet class. Now replace the content ... .servlet.http.HttpServletResponse; @WebServlet(name="GuestServlet", urlPatterns={"/guest
2

Step 4: Add a Servlet Class

> Other... > Web > Servlet and clicking Next. The Java package name should be guest. Enter GuestServlet as the class name - use exactly that case sensitive class name. Click Finish to create the new ... .http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; @WebServlet(name

Getting Started

ObjectDB is very easy to use. Follow the Getting Started Tutorial and the Quick Tour manual chapter and in minutes you may be able to write and run first Java programs against ObjectDB.

Prior knowledge or experience in database programming (SQL, JDBC, ORM, JPA, etc.) is not required, but some background in using the Java language is essential.

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