70-442GB2312 Exam

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Exam : Microsoft 70-442GB2312
Title : PRO:Design & Optimize Data Access by Using MS SQL Serv 2005


1. 您是公司的数据库开发人员。一个名为 Articles 的表包含报纸文章，其设计如下所示。
ArticleID (PK)
Title
ArticleXml
CategoryID (FK)
int
nvarchar(50)
xml
int
ArticleXml 列中的 XML 文档如下所示。
<article>
<paragraph>...</paragraph>
<image>...</image>
...
<image>...</image>
...
</article>您需要设计一个查询，该查询将按照 CategoryID 值显示图像的总数。
您应该使用哪个查询？
A. SELECT
CategoryID
,COUNT(ArticleXml.value('count(/article/image)',
'INT')) AS NumberOfImages
FROM Articles
GROUP BY CategoryID;
B. SELECT
CategoryID
,COUNT(ArticleXml.exist('/article/image')
AS NumberOfImages
FROM Articles
GROUP BY CategoryID;
C. SELECT
CategoryID
,SUM(ArticleXml.value('count(/article/image)', 'INT'))
AS NumberOfImages
FROM Articles
GROUP BY CategoryID;
D. SELECT
CategoryID
,SUM(ArticleXml.exist('/article/image')
AS NumberOfImages
FROM Articles
GROUP BY CategoryID;
Answer: C

2. 您是公司的数据库开发人员。在您的数据库中，Employees 表包含大约 1,000 行，Orders 表包含数百万行。
Employees 表设计为如下所示。
EmployeeID (PK)
Firstname
Lastname
...
int
nvarchar(30)
nvarchar(30)
...
Orders 表设计为如下所示。
OrderID (PK)
SoldByEmployeeID (FK)
OrderDate
...
int
int
datetime
...
您需要编写一个查询，该查询返回每位员工所下的订单数。
您应该使用哪个查询？
A. SELECT e.Firstname, e.Lastname, COUNT(*)
FROM Employees AS e
LEFT OUTER JOIN Orders AS o
ON o.SoldByEmployeeID = e.EmployeeID
GROUP BY e.Firstname, e.Lastname
B. SELECT e.Firstname, e.Lastname, COUNT(*)
FROM Employees AS e
LEFT OUTER JOIN Orders AS o
ON o.SoldByEmployeeID = e.EmployeeID
GROUP BY e.EmployeeID, e.Firstname, e.Lastname
C. SELECT e.Firstname, e.Lastname, COUNT(o.OrderID)
FROM Employees AS e
LEFT OUTER JOIN Orders AS o
ON o.SoldByEmployeeID = e.EmployeeID
GROUP BY e.EmployeeID, e.Firstname, e.Lastname
D. SELECT e.Firstname, e.Lastname, COUNT(o.OrderID)
FROM Employees AS e
LEFT OUTER JOIN Orders AS o
ON o.SoldByEmployeeID = e.EmployeeID
GROUP BY e.Firstname, e.Lastname
Answer: C

3. 您是公司的数据库开发人员。现必须优化 Stockholm 市订单的处理例程，以获得更好的性能。
当前例程如以下代码段所示。
...
DECLARE OrderCursor CURSOR FOR
SELECT OrderID, City, CustomerID FROM Orders;
OPEN OrderCursor;
FETCH NEXT FROM OrderCursor INTO @OrderID, @CustomerID, @City;
WHILE(@@FETCH_STATUS = 0)
BEGIN
IF(@City <> 'Stockholm')
GOTO Next;
<Code for processing order>
Next:
FETCH NEXT FROM OrderCursor INTO
@OrderID, @CustomerID, @City;
END
...
您需要优化该例程以获得最佳性能。
您应该怎么办？
A. 将游标更改为以 DYNAMIC 形式声明。
B. 将游标更改为以 KEYSET 形式声明。
C. 将游标更改为以 STATIC 形式声明。
D. 更改游标中的 SELECT 语句以包括 WHERE City = 'Stockholm'。
E. 将 IF 语句更改为检查 @City = 'Stockholm'，从而避免使用 GOTO 语句。
Answer: D

4. 您是公司的数据库开发人员。您负责在公司新的 SQL Server 2005 计算机上将名为 DB1 和 DB2 的两个现有 SQL Server 2000 数据库合并为一个名为 NewDB 的数据库。
以下查询在用于访问 DB1 数据库的旧应用程序中可以正常工作。
SELECT * FROM Customer WHERE FaxNumber = NULL
但是，该查询在新的 SQL Server 2005 计算机上运行时将不会返回任何行，即使 FaxNumber 列中存在 NULL 值。
您需要解决该问题，以使该查询在 SQL Server 2005 计算机上能正常工作。您需要确保您的解决方案不会影响 DB2 中表和查询的行为。
您应该怎么办？
A. 使用 ISNULL 函数，而不是 = NULL。
B. 将 = NULL 更改为 IS NULL。
C. 将 = NULL 更改为 = 'NULL'。
D. 在查询前面添加以下命令：SET ANSI_NULLS ON
E. 使用 ALTER DATABASE 命令将 ANSI_NULLS 数据库选项更改为 ON。
Answer: B

5. 您是公司的数据库开发人员。有一个将订单项保存到 OrderItems 表的过程。如果该项不存在，则应执行插入。如果该项存在，则应执行更新。OrderItems 表设计为如下所示。
OrderID (PK, FK)
ProductID (PK, FK)
Quantity
int
int
int
您需要开发一个使用最少量资源的例程。
您应该使用哪个例程？
A. BEGIN TRY
UPDATE OrderItems SET Quantity = @Quantity
WHERE OrderID = @OrderID
AND ProductID = @ProductID;
END TRY
BEGIN CATCH
INSERT OrderItems (OrderID, ProductID, Quantity)
VALUES (@OrderID, @ProductID, @Quantity);
END CATCH
B. IF EXISTS (SELECT * FROM OrderItems
WHERE OrderID = @OrderID
AND ProductID = @ProductID
)
UPDATE OrderItems SET Quantity = @Quantity
WHERE OrderID = @OrderID
AND ProductID = @ProductID;
ELSE
INSERT OrderItems (OrderID, ProductID, Quantity)
VALUES (@OrderID, @ProductID, @Quantity);
C. IF NOT EXISTS (SELECT * FROM OrderItems
WHERE OrderID = @OrderID
AND ProductID = @ProductID
)
INSERT OrderItems (OrderID, ProductID, Quantity)
VALUES (@OrderID, @ProductID, @Quantity);
ELSE
UPDATE OrderItems SET Quantity = @Quantity
WHERE OrderID = @OrderID
AND ProductID = @ProductID;
D. UPDATE OrderItems SET Quantity = @Quantity
WHERE OrderID = @OrderID AND ProductID = @ProductID;
IF(@@ROWCOUNT = 0)
INSERT OrderItems (OrderID, ProductID, Quantity)
VALUES (@OrderID, @ProductID, @Quantity);
Answer: D