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NEW QUESTION: 1
Your database contains a table named Purchases. The table includes a DATETIME column named PurchaseTime that stores the date and time each purchase is made. There is a non-clustered index on the PurchaseTime column. The business team wants a report that displays the total number of purchases made on the current day. You need to write a query that will return the correct results in the most efficient manner.
Which Transact-SQL query should you use?
A：
B：
C：
D：
A. Option C
B. Option D
C. Option B
D. Option A
Answer: B
Explanation:
Explanation/Reference:
Reference: http://technet.microsoft.com/en-us/library/ms181034.aspx
NEW QUESTION: 2
Which one of the following queries selects the customer whose order has the highest total price?
A. CriteriaBuilder cb = ...
Criteria Query <Customer> cq = cb.create Query (Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
B. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.createquery(customer.class)
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
C. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.cteateQuery(Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
D. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.createQuery(Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = sq.from(Order.class);
sq.select (cb.max ( subo.get (Order_ . Total Price) ) ) ;
cq.where(sq.all(o.get(Order_.totalPrice)));
Answer: D
Explanation:
Explanation/Reference:
Incorrect: Not a, not b, not C: use .from not .correlate.
Example:
CriteriaBuilder qb = em.getCriteriaBuilder();
CriteriaQuery<Number> cq = qb.createQuery(Number.class);
Root<Event> root = cq.from(Event.class);
cq.select(qb.max(root.get("dateProcessed")));
cq.where(qb.equal(Event.get("org"), qb.parameter(MyOrgType.class, "myOrg"))); em.createQuery(cq).setParameter("myOrg", myOrg).getSingleResult();
Note:
max(Expression<N> x)
Create an aggregate expression applying the numerical max operation.
Reference: javax.persistence.criteria, Interface CriteriaBuilder
NEW QUESTION: 3
What is a consideration for using a VPLEX logging volume?
A. VIAS can be used to create a logging volume at each cluster
B. A distributed device must be created before a logging volume can be created
C. Automatically provisioned at each VPLEX cluster when the distributed device is created
D. Must be created on each VPLEX cluster before creating a distributed device
Answer: D
NEW QUESTION: 4
会社は、タイ語が異なるソースからのさまざまな種類のデータを保存するアプリケーションを開発しています。
ストレージソリューションを推奨する必要があります。
どのストレージソリューションをお勧めしますか？ 答えるには、適切なストレージソリューションを
正しいデータ型。 各ストレージソリューションは、1回、複数回使用することも、まったく使用しないこともできます。 ドラッグする必要があるかもしれません
ペイン間の分割バーまたはスクロールしてコンテンツを表示します。
注：それぞれの正しい選択には1ポイントの価値があります。
Answer:
Explanation:
Explanation