Oracle 1Z0-898 echte frage : Java EE 6 Java Persistence API Developer Certified Expert Exam

Wer die echte Prüfungssituation bereits erlebt hat, geht gelassener in die 1Z0-898 Prüfung. Die Test Engines von EchteFrage simulieren den Ablauf der Oracle Java EE 6 Java Persistence API Developer Certified Expert und bieten Ihnen 2026 mit 63 Praxisfragen die realistischste Vorbereitung.

Oracle 1Z0-898 Prüfungsübersicht:

Zertifizierungsanbieter:	Oracle
Prüfungsname:	Zertifizierungsprüfung zum Oracle Certified Expert für die Entwicklung von Datenbankanwendungen mit der Java Persistence API in der Java Platform, Enterprise Edition 6
Prüfungsnummer:	1Z0-898
Mindestpunktzahl:	61 %
Prüfungsdauer:	135–150
Gültigkeitsdauer des Zertifikats:	Keine Befristung (unbegrenzt gültig)
Verfügbare Sprachen:	Englisch
Prüfungsgebühr:	245 USD
Verwandte Zertifizierungen:	Oracle Certified Expert, Java EE 6 EJB Developer Oracle Certified Expert, Java EE 6 Web Services Developer Oracle Certified Professional, Java SE Programmer
Anzahl der Fragen:	58–64
Prüfungsformat:	Einfachauswahl, Mehrfachauswahl
Empfohlenes Training:	Java EE 6: Entwickeln von Datenbankanwendungen mit JPA
Prüfungsanmeldung:	Anmeldung über Pearson VUE Oracle-Zertifizierungsportal
Beispielfragen:
Prüfungsmethode:	Computergestützte Prüfung in Prüfungszentren von Pearson VUE; keine beaufsichtigte Online-Prüfung verfügbar
Voraussetzungen:	Empfohlen: Zertifizierung als Oracle Certified Professional, Java SE 5/6/7 Programmer; keine zwingende Voraussetzung
Offizielle Syllabus-URL:	https://education.oracle.com/pls/web_prod-plq-dad/db_pages.getpage?page_id=5001&get_params=p_exam_id:1Z0-898

Oracle 1Z0-898 Prüfungsthemen:

Alles Wichtige zur Oracle Java EE 6 Java Persistence API Developer Certified Expert im Überblick

Oracle Java EE 6 Java Persistence API Developer Certified Expert 1Z0-898 Prüfungsfragen mit Lösungen

Frage #1

A developer wrote an entity class with the following method:
Private static Logger logger = Logger.getLogger ("myLogger");
@PrePersist @PreUpdate Public void doA () { Logger.info ("A"); } @PostPersist @PostUpdate Public void doB () { logger.info ("B"); }
What will the log message contain when an application does the following?
1.Begins a transaction
2.Creates the entity
3.Persists the entity
4.Commits the transaction
5.Begins the entity data
6.Modifies the entity data
7.Merges the entity
8.Commits the second transaction

A. A B A B
B. A A B B
C. A B B A B
D. The application will throw an exception because multiple lifecycle callback annotations applied to a single method.

Frage #2

An application uses optimistic locking by defining version attributes in its entity classes. The application performs a bulk update of the entities using a JPQL query.
Which of the following is correct?

A. The value of the Version attributes of the updated entitles should be also be explicitly updated by the query.
B. The persistence provider will ensure that the version value in each table is updated.
C. The persistence provider will create a new transaction for the bulk update.
D. An OptimisticLockException will be thrown by the persistence provider.

Frage #3

The developer wants to override the default mappings for an embeddable class Address used by the customer entity.
The Address class is defined as follows:
@Embeddable public class Address (
private String street;
private String city;
private String country;
. . .
)
Assume that NO mapping descriptor is present. Which code segment below shows the correct way to override the default mapping for address?

A. @AttributeOverrides ({
@AttributeOverride (name = "addr.street", column = @Column (name = ADDR_STREET)),
@AttributeOverride (name = "addr.city", column = @Column (name = ADDR_CITY)),
@AttributeOverride (name = "addr.country", column = @Column (name = ADDR_COUNTRY)),
}}
@Embedded Address addr;
B. @ AttributeOverrides ({
@AttributeOverride (name = "street", column (name = "name_STREET")),
@AttributeOverride (name = "city, column (name = "name_CITY")),
@AttributeOverride (name = "country, column (name = "name_COUNTRY")),
}}
@Embedded Address addr;
C. @ AttributeOverrides ({
@AttributeOverride (name = "street", column = @Column (name = "name_STREET")),
@AttributeOverride (name = "city, column = @Column (name = "name_CITY")),
@AttributeOverride (name = "country, column = @Column (name = "name_COUNTRY")),
}}
@Embedded Address addr;
D. @AttributeOverrides ({
@AttributeOverride (name = "street", column = @Column (name = ADDR_STREET)),
@AttributeOverride (name = "city, column = @Column (name = ADDR_CITY)),
@AttributeOverride (name = "country, column = @Column (name = ADDR_COUNTRY)),
}}
@Embedded Address addr;

Frage #4

Persistence application locks entity x with a LockModeType. PESSIMISTIC_READ lock type, which statement is true?

A. This operation will result in a TransactionRolledbackException if the lock cannot be obtained.
B. LockModeType. PESSIMISTIC_READ is the synonym of the LockModeType.REA
C. This operation will force serialization among transactions attempting to read the entity data.
D. If the application later updates the entity, and the changes are flushed to the database, the lock will be converted to an exclusive lock.

Frage #5

The developer wants to define a unidirectional relationship from the customer entity to the order entity and map this relationship using a foreign key mapping strategy.
Which one of the pairs of classes below correctly achieves this task?

A. @Entity public class Customer {
@Id int customerId;
@OneToMany @JoinColumn (name = "CUST_ID") Set <Order> orders;
. . .
}
@Entity public class order {
@Id int orderId;
. . .
}
B. @ Entity public class Customer {
@Id int customerId;
@OneToMany (JoinColumn = @JoinColumn (name = "CUST_ID"), table = ""ORDER) Set
<Order> orders;
. . .
}
@Entity public class order {
@Id int orderId;
. . .
}
C. @Entity public class Customer {
@Id int customerId;
@OneToMany Set <Order> orders;
. . .
}
@Entity
@JoinColumn (names = "CUST-ID", referencedColumnName = "customerId")
public class order {
@Id int order Id;
. . .
}
D. @Entity public class Customer {
@Id int customerId;
@OneToMany (JoinColumn = @joinColumn (name = "CUST_ID") Set <Order> orders;
. . .
}
@Entity public class order {
@Id int orderId;
. . .
}

Fragen und Antworten: