1Z0-898最も有効な質問と解答で勉強

本番と同じ緊張感の中で時間配分を試せるかどうかが、Oracle Java EE 6 Java Persistence API Developer Certified Expertの仕上がりを左右します。Tech4Examのテストエンジンは1Z0-898本試験の環境を再現し、63問の練習問題を模擬試験形式で繰り返し解けます。

Oracle 1Z0-898 試験概要:

Oracle 1Z0-898 試験シラバストピック:

Oracle Java EE 6 Java Persistence API Developer Certified Expertに関するよくあるご質問

Oracle Java EE 6 Java Persistence API Developer Certified Expert 認定 1Z0-898 試験問題：

問題 #1

Which cascade option can be specified in a mapping descriptor so that it applies to all relationships in a persistent e unit?

A. cascade all
B. cascade detach
C. cascade-persist
D. cascade remove

問題 #2

A user entity is retrieved in a query and stored in an instance variable user. The user entity has a single valued name property that uses the mapping defaults, and a photo property, which is lazily loaded. The application then calls the following method:
PersistenceUtil.isLoaded (user);
Which two of the following statements are correct?

A. The photo property was NOT loaded from the database.
B. The photo property may or may not have been loaded from the database.
C. The name property may or may not have been loaded from the database.
D. The name property was NOTbeloaded from the database.
E. The photo property was loaded from the database.
F. The name property was loaded from the database.

問題 #3

Given two entities with many-to-many bidirectional association between them: What set of annotations correctly defines the association?

A. @manyToMany ()targetEntity = project.class) on the projects field,
@manyToMany (mappedBy = "projects") on the emps field
B. @manyToMany on the projects field,
@manyToMany (mappedBy= "projects") on the emps field
C. @manyToMany (mappedBy = emps) on the projects field,
@manyToMany on the emps field
D. @manyToMany (targetEntity = Project.class) on the projects field,
@manyToMany on the emps field

問題 #4

Entity lifecycle callback methods may be defined in which three classes? (Choose three)

A. Concrete non-entity superclasses
B. Embedded classes
C. Abstract classes
D. Mapped superclasses
E. Entity classes
F. Entity listener classes

問題 #5

Which of the following Criteria query snippets demonstrates the correct way to create and execute strongly typed queries? Assume that cb references an instance of the CriteriaBuilder interface and em references an EntityManager instance.

A. CriteriaQuery <office> cq = cb.createQuery (Office.class);
. . .
TypedQuery<Office> tq = em.ceateQuery (cq) ; L
1st <office> offices = tq.getResultList ();
B. CriteriaQuery cq = cb.createQuery (Office.class)
. . .
TypedQuery<office> tq = em.createQuery (cq, office.class);
List <office> offices = tq.getresult ();
C. CriteriaQuery<office> cq = em.createQuery (cq, office.class);
. . .
TypedQuery<Office> tq = em.createQuery (cq);
List <office> offices = tq.getresult ();
D. CriteriaQuery <office> cq = cb.createQuery (Office.class);
. . .
TypedQuery<Office> tq = em.ceateQuery (cq);
List<office> Offices = tq.getResultList ();

解説：