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70-442GB2312 Exam

PRO:Design & Optimize Data Access by Using MS SQL Serv 2005

  • Exam Number/Code : 70-442GB2312
  • Exam Name : PRO:Design & Optimize Data Access by Using MS SQL Serv 2005
  • Questions and Answers : 91 Q&As
  • Update Time: 2011-02-02
  • Price: $ 119.00 $ 69.00
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70-442GB2312

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Microsoft Microsoft Licensing 70-442GB2312 exam braindumps questions and answers

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


1. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£¹«Ë¾ÕýÔÚ´Ó SQL Server 2000 Éý¼¶µ½ SQL Server 2005¡£¹«Ë¾ÐµıàÂë±ê×¼ÒªÇó·ûºÏ ANSI ±ê×¼¡£
ÄúµÄ¾­ÀíÒªÇóÄú½«ÒÔϲéѯת»»ÎªÊ¹Óà ANSI ±ê×¼´ë´Ê¡£
SELECT E.EmployeeID, E.Hiredate, J.Resume
FROM HumanResources.Employee AS E,
HumanResources.JobCandidate AS J
WHERE E.EmployeeID *= J.EmployeeID
вéѯ±ØÐëʼÖÕÉú³ÉÓëÏÖÓвéѯÏàͬµÄ½á¹û¼¯¡£
ÄúÐèÒª±àд²éѯÒÔÂú×ãÕâЩҪÇó¡£
ÄúÓ¦¸ÃʹÓÃÄĸö Transact-SQL ²éѯ£¿
A. SELECT E.EmployeeID, E.Hiredate, J.Resume
FROM HumanResources.Employee AS E RIGHT OUTER JOIN
HumanResources.JobCandidate AS J
ON E.EmployeeID = J.EmployeeID
B. SELECT E.EmployeeID, E.Hiredate, J.Resume
FROM HumanResources.Employee AS E LEFT OUTER JOIN
HumanResources.JobCandidate AS J
ON E.EmployeeID = J.EmployeeID
C. SELECT E.EmployeeID, E.Hiredate, J.Resume
FROM HumanResources.Employee AS E INNER JOIN
HumanResources.JobCandidate AS J
ON E.EmployeeID = J.EmployeeID
D. SELECT E.EmployeeID, E.Hiredate, J.Resume
FROM HumanResources.Employee AS E FULL OUTER JOIN
HumanResources.JobCandidate AS J
ON E.EmployeeID = J.EmployeeID
Answer: B

2. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£Äú¸ºÔðÔÚ¹«Ë¾Ð嵀 SQL Server 2005 ¼ÆËã»úÉϽ«ÃûΪ DB1 ºÍ DB2 µÄÁ½¸öÏÖÓÐ SQL Server 2000 Êý¾Ý¿âºÏ²¢ÎªÒ»¸öÃûΪ NewDB µÄÊý¾Ý¿â¡£
ÒÔϲéѯÔÚÓÃÓÚ·ÃÎÊ DB1 Êý¾Ý¿âµÄ¾ÉÓ¦ÓóÌÐòÖпÉÒÔÕý³£¹¤×÷¡£
SELECT * FROM Customer WHERE FaxNumber = NULL
µ«ÊÇ£¬¸Ã²éѯÔÚÐ嵀 SQL Server 2005 ¼ÆËã»úÉÏÔËÐÐʱ½«²»»á·µ»ØÈκÎÐУ¬¼´Ê¹ FaxNumber ÁÐÖдæÔÚ NULL Öµ¡£
ÄúÐèÒª½â¾ö¸ÃÎÊÌ⣬ÒÔʹ¸Ã²éѯÔÚ SQL Server 2005 ¼ÆËã»úÉÏÄÜÕý³£¹¤×÷¡£ÄúÐèҪȷ±£ÄúµÄ½â¾ö·½°¸²»»áÓ°Ïì DB2 ÖбíºÍ²éѯµÄÐÐΪ¡£
ÄúÓ¦¸ÃÔõô°ì£¿
A. ʹÓà ISNULL º¯Êý£¬¶ø²»ÊÇ = NULL¡£
B. ½« = NULL ¸ü¸ÄΪ IS NULL¡£
C. ½« = NULL ¸ü¸ÄΪ = 'NULL'¡£
D. ÔÚ²éѯǰÃæÌí¼ÓÒÔÏÂÃüÁSET ANSI_NULLS ON
E. ʹÓà ALTER DATABASE ÃüÁ ANSI_NULLS Êý¾Ý¿âÑ¡Ïî¸ü¸ÄΪ ON¡£
Answer: B

3. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£Óû§ÐèÒªÓйØÔÚ TotalDue ÁÐÖоßÓÐ×î¸ßÖµµÄ¶©µ¥µÄÐÅÏ¢¡£Óû§Ï£ÍûÄܹ»ÔÚÿ´ÎÇëÇóÐÅϢʱ·µ»Ø²»Í¬µÄ¼Ç¼°Ù·Ö±È¡£
ÄúÐèÒªÔÚ SQL Server 2005 ÉÏÉú³É´úÂ룬¸Ã´úÂ뽫ÔÊÐíÇ°¶ËÓ¦ÓóÌÐò´«µÝ×îÖÕÓû§½«ÎªÒª·µ»ØµÄ¼Ç¼°Ù·Ö±ÈÌṩµÄ±äÁ¿¡£
ÄúÓ¦¸ÃÈçºÎ±àд´úÂ룿
A. CREATE PROCEDURE TopPercentTotalDue
@P int
AS
SELECT TOP(@P) PERCENT
TotalDue, OrderDate
FROM Purchasing.PurchaseOrderHeader
ORDER BY TotalDue DESC
B. CREATE PROCEDURE TopPercentTotalDue
@P int
AS
SELECT TOP(@P) PERCENT
TotalDue, OrderDate
FROM Purchasing.PurchaseOrderHeader
ORDER BY TotalDue ASC
C. CREATE PROCEDURE TopPercentTotalDue
@R int
AS
SET rowcount @R
SELECT TotalDue, OrderDate
FROM Purchasing.PurchaseOrderHeader
ORDER BY TotalDue DESC
D. CREATE PROCEDURE TopPercentTotalDue
@R int
AS
SET rowcount @R
SELECT TotalDue, OrderDate
FROM Purchasing.PurchaseOrderHeader
ORDER BY TotalDue ASC
Answer: A

4. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£ÏÖ±ØÐëÓÅ»¯ Stockholm Êж©µ¥µÄ´¦ÀíÀý³Ì£¬ÒÔ»ñµÃ¸üºÃµÄÐÔÄÜ¡£
µ±Ç°Àý³ÌÈçÒÔÏ´úÂë¶ÎËùʾ¡£
...
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
...
ÄúÐèÒªÓÅ»¯¸ÃÀý³ÌÒÔ»ñµÃ×î¼ÑÐÔÄÜ¡£
ÄúÓ¦¸ÃÔõô°ì£¿
A. ½«Óαê¸ü¸ÄΪÒÔ DYNAMIC ÐÎʽÉùÃ÷¡£
B. ½«Óαê¸ü¸ÄΪÒÔ KEYSET ÐÎʽÉùÃ÷¡£
C. ½«Óαê¸ü¸ÄΪÒÔ STATIC ÐÎʽÉùÃ÷¡£
D. ¸ü¸ÄÓαêÖÐµÄ SELECT Óï¾äÒÔ°üÀ¨ WHERE City = 'Stockholm'¡£
E. ½« IF Óï¾ä¸ü¸ÄΪ¼ì²é @City = 'Stockholm'£¬´Ó¶ø±ÜÃâʹÓà GOTO Óï¾ä¡£
Answer: D

5. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£¹«Ë¾×î½üÉý¼¶µ½ÁË SQL Server 2005¡£
ÄúÕýÔÚ´´½¨Ò»¸ö²éѯ£¬ÒԱ㷵»ØÏÔʾָ¶¨Ô±¹¤×éËùÏ´ï²É¹º¶©µ¥×ÜÊýµÄÐÅÏ¢£¬°´ÇëÇ󶩵¥µÄ¹©Ó¦ÉÌ·Ö×é¡£
½á¹û¼¯Ó¦ÔÚÆäµ¥¶ÀÒ»ÁÐÖÐÏÔʾÿλԱ¹¤µÄ²É¹º¶©µ¥×ÜÊý¡£Ã¿¸ö¹©Ó¦ÉÌÔÚ½á¹û¼¯Öж¼Ó¦ÓÐ×Ô¼ºµÄÐС£
ÄúÐèÒª´´½¨²éѯÀ´Âú×ãÕâЩҪÇó¡£
ÄúÓ¦¸ÃʹÓÃÄĸö Transact-SQL ÃüÁ
A. CROSS APPLY
B. UNION
C. PIVOT
D. INTERSECT
Answer: C

6. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£Ã¿Ì콫ÒÔ XML ÎĵµµÄÐÎʽ·¢ËÍ Parts Êý¾Ý¿âµÄÐÅÏ¢£¬²¢½«¸ÃÐÅÏ¢·ÅÔÚ ProductXML ±íµÄ ProductInfo ÁÐÖеĵ¥Ò»ÐÐÖС£¸Ã XML ÎĵµÖеÄÐÅÏ¢µÄʾÀýÈçÏÂËùʾ¡£
<Products>
<Product ID="..." Name="..." Number="..."/>
<Product ID="..." Name="..." Number="..."/>
...
</Products>
ÿÌì½áÊøʱ½«ÔËÐÐÒ»¸öÅú´¦Àí×÷Òµ¡£Õâ¸öÅú´¦Àí×÷Òµ½« XML Îĵµ´Ó ProductXML ±íµ¼Èë Product ±í¡£½«ÎªÃ¿¸ö Product ÔªËØ´Ó XML ÎĵµÖÐÌáÈ¡²úÆ·µÄ ID¡¢Ãû³ÆºÍ±àºÅ£¬²¢½«ÕâЩÐÅÏ¢²åÈë Product ±íµÄ ProductID¡¢Name¡¢ºÍ Number ÁÐÖС£ÔÚÅú´¦Àí½áÊøʱ»á½Ø¶Ï ProductXML ±í¡£
ÄúÐèÒªÉè¼Æ´Ó ProductXML ±íÖмìË÷ÐÐÒÔ²åÈë Product ±íµÄ²éѯ¡£
ÄúÓ¦¸ÃʹÓÃÄĸö Transact-SQL ²éѯ´Ó ProductXML ±íÖвéѯÊý¾Ý£¿
A. SELECT
ProductInfo.value('(/Products/Product)[1]/@ID', 'integer')
AS ProductID,
ProductInfo.value('(/Products/Product)[1]/@Name', 'varchar(50)')
AS Name,
ProductInfo.value('(/Products/Product)[1]/@Number', 'varchar(25)')
AS ProductNumber
FROM ProductXML
B. SELECT
col.value('@ID', 'integer') AS ProductID,
col.value('@Name', 'varchar(50)') AS Name,
col.value('@Number', 'varchar(25)') AS ProductNumber
FROM ProductXML
CROSS APPLY ProductInfo.nodes('/Products/Product') AS x (col)
C. SELECT
col.query('data(@ID), data(@Name), data(@Number)')
FROM ProductXML
CROSS APPLY ProductInfo.nodes('/Products/Product') AS x (col)
D. SELECT
ProductInfo.query('
for $p in /Products/Product
return fn:concat($p/@ID, $p/@Name, $p/@Number)
') AS Result
FROM ProductXML
Answer: B

7. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£ÄúÕýÔÚ´´½¨Ò»¸ö²ÎÊý»¯´æ´¢¹ý³Ì£¬¸Ã´æ´¢¹ý³Ì½«²éѯÊý¾Ý²¢·µ»ØһЩÐУ¬ÆäÖаüº¬×÷Ϊ¸Ã´æ´¢¹ý³ÌÊäÈëÌṩµÄ¶¯´ÊµÄÈκζ¯´Êʱ̬¡£²éѯ»¹Ó¦·µ»Øͬʱ°üº¬Ãû´ÊµÄµ¥ÊýºÍ¸´ÊýÐÎʽµÄÐС£Êý¾Ý°üÀ¨ÔÚÈô¸É varchar(max) ºÍ varbinary(max) ÁÐÖС£
ÄúÐèÒª´´½¨´æ´¢¹ý³ÌÀ´Âú×ãÕâЩҪÇó¡£
ÄúÓ¦¸ÃÖ´ÐÐÄÄÁ½Ïî²Ù×÷£¿£¨Ã¿¸öÕýÈ·´ð°¸¶¼½ö¸ø³öÁ˲¿·Ö½â¾ö·½°¸¡£ÇëÑ¡ÔñÁ½¸ö´ð°¸¡££©
A. ʹÓà LIKE ÔËËã·û¡£
B. ʹÓà CONTAINS ν´Ê¡£
C. Õë¶ÔÊʵ±µÄÒ»¸ö»ò¶à¸öÁд´½¨È«ÎÄË÷Òý¡£
D. Õë¶ÔÊʵ±µÄÒ»¸ö»ò¶à¸öÁд´½¨·Ç¾Û¼¯Ë÷Òý¡£
Answer: BC

8. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£Ò»¸öÃûΪ Articles µÄ±í°üº¬±¨Ö½ÎÄÕ£¬ÆäÉè¼ÆÈçÏÂËùʾ¡£
ArticleID (PK)
Title
ArticleXml
CategoryID (FK)
int
nvarchar(50)
xml
int
ArticleXml ÁÐÖÐµÄ XML ÎĵµÈçÏÂËùʾ¡£
<article>
<paragraph>...</paragraph>
<image>...</image>
...
<image>...</image>
...
</article>ÄúÐèÒªÉè¼ÆÒ»¸ö²éѯ£¬¸Ã²éѯ½«°´ÕÕ CategoryID ÖµÏÔʾͼÏñµÄ×ÜÊý¡£
ÄúÓ¦¸ÃʹÓÃÄĸö²éѯ£¿
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

9. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£ÔÚÄúµÄÊý¾Ý¿âÖУ¬Employees ±í°üº¬´óÔ¼ 1,000 ÐУ¬Orders ±í°üº¬Êý°ÙÍòÐС£
Employees ±íÉè¼ÆΪÈçÏÂËùʾ¡£
EmployeeID (PK)
Firstname
Lastname
...
int
nvarchar(30)
nvarchar(30)
...
Orders ±íÉè¼ÆΪÈçÏÂËùʾ¡£
OrderID (PK)
SoldByEmployeeID (FK)
OrderDate
...
int
int
datetime
...
ÄúÐèÒª±àдһ¸ö²éѯ£¬¸Ã²éѯ·µ»ØÿλԱ¹¤ËùϵĶ©µ¥Êý¡£
ÄúÓ¦¸ÃʹÓÃÄĸö²éѯ£¿
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

10. ÄúÊǹ«Ë¾µÄÊý¾Ý¿â¿ª·¢ÈËÔ±¡£ÓÐÒ»¸ö½«¶©µ¥Ïî±£´æµ½ OrderItems ±íµÄ¹ý³Ì¡£Èç¹û¸ÃÏî²»´æÔÚ£¬ÔòÓ¦Ö´ÐвåÈë¡£Èç¹û¸ÃÏî´æÔÚ£¬ÔòÓ¦Ö´ÐиüС£OrderItems ±íÉè¼ÆΪÈçÏÂËùʾ¡£
OrderID (PK, FK)
ProductID (PK, FK)
Quantity
int
int
int
ÄúÐèÒª¿ª·¢Ò»¸öʹÓÃ×îÉÙÁ¿×ÊÔ´µÄÀý³Ì¡£
ÄúÓ¦¸ÃʹÓÃÄĸöÀý³Ì£¿
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


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