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NEW QUESTION: 1
どのITプロセスがクラウドコンピューティングでより多くの努力を必要とする可能性がありますか？
A. オペレーティングシステムのパッチのインストール
B. サーバーのメンテナンス
C. パフォーマンス監視
D. ハードウェア保守
Answer: C

NEW QUESTION: 2
The developer wants to write a criteria query that will return the number of orders made by customer of each county.
Assume that customer is an entity with a unidirectional one-to-many relationship to the Order entity and that Address is an embeddable class, with an attribute country of type String.
Which one of the queries below correctly achieves this?
A. CriteriaBuilder cb = ...
CriteriaQuery cq = cb.createQueryO;
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
Join<Address, String> country = c.join(Customer,.address) .join(Address cq.multiselect(cq.count(o), country );
cq.groupBy(c.get(Customer.address) - get (Address_ . country) ) ;
B. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Customer> c = cq.from(Customer.class); cq.select (cb.count(c.join
(customer_. Orders)) , c.get(customers(0), c.get(customer_.address) . get (Address_'country)); (c.get(Customer_.address). get(address_.country));
C. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.multiselect(cb.count(0), c,get(customer_.address.get(address_.country) cq.groupBy (c.get(customer_.address) .get(address_.country))
D. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Custower> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.select(cb.count(o));
cq.groupBy(c.qet(Customer__.address) - get(Address_.country)) ;
Answer: C
Explanation:
Explanation/Reference:
Incorrect: Not B, Not C: Use multiselect, not select. Not D: Use one Join, not two.
Use multiselect, count and Group By. multiselect() method is used because we are going to get compound result and not one entity type.
Example:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Object[]> query = cb.createQuery(Object[].class);
Root<Department> d = query.from(Department.class);
Join<Department,Teacher> teachers = d.join("teachers");
query.multiselect(d.get("name"),cb.count(teachers)).groupBy(d.get("name")); Reference: Criteria group by clause

NEW QUESTION: 3
Linuxユーザーは、SMTPポートを介してmail.foo.comという名前のリモート電子メールサーバーに接続する必要があります。
次のコマンドのどれがこのタスクを実行しますか？
A. traceroute mail.foo.com 25
B. traceroute mail.foo.com 110
C. netcat mail.foo.com 110
D. netcat mail.foo.com 25
Answer: D

NEW QUESTION: 4
A solutions architect is designing a publicly accessible web application that is on an Amazon CloudFront distribution with an Amazon S3 website endpoint as the origin.
When the solution is deployed, the website returns an Error 403: Access Denied message.
Which steps should the solutions architect take to correct the issue? (Select TWO.)
A. Disable S3 object versioning
B. Change the storage class from S3 Standard to S3 One Zone-Infrequent Access (S3 One Zone- IA).
C. Remove the origin access identity (OAI) from the CloudFront distribution.
D. Remove the requester pays option from the S3 bucket.
E. Remove the S3 block public access option from the S3 bucket.
Answer: D,E