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70-461 PDF DEMO:

QUESTION NO: 1
You administer a Microsoft SQL Server database that supports a banking transaction management application.
You need to retrieve a list of account holders who live in cities that do not have a branch location.
Which Transact-SQL query or queries should you use? (Each correct answer presents a complete solution. Choose all that apply.)
A. SELECT AccountHolderID
FROM AccountHolder
WHERE CityID <> SOME (SELECT CityID FROM BranchMaster)
B. SELECT AccountHolderID
FROM AccountHolder
WHERE CityID <> ALL (SELECT CityID FROM BranchMaster)
C. SELECT AccountHolderID
FROM AccountHolder
WHERE CityID <> ANY (SELECT CityID FROM BranchMaster)
D. SELECT AccountHolderID
FROM AccountHolder
WHERE CityID NOT IN (SELECT CityID FROM BranchMaster)
Answer: B,D
Explanation:
Verified the answers as correct.

QUESTION NO: 2
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. SELECT COUNT(*)
FROM Purchases
WHERE PurchaseTime = GETDATE()
B. SELECT COUNT(*)
FROM Purchases
WHERE PurchaseTime >= CONVERT(DATE, GETDATE())
AND PurchaseTime < DATEADD(DAY, 1, CONVERT(DATE, GETDATE()))
C. SELECT COUNT(*)
FROM Purchases
WHERE CONVERT(VARCHAR, PurchaseTime, 112) = CONVERT(VARCHAR, GETDATE(), 112)
D. SELECT COUNT(*)
FROM Purchases
WHERE PurchaseTime = CONVERT(DATE, GETDATE())
Answer: B
Explanation:
Two answers will return the correct results (the "WHERE CONVERT..." and "WHERE ... AND ... " answers).
The correct answer for Microsoft would be the answer that is most "efficient". Anybody have a clue as to which is most efficient? In the execution plan, the one that I've selected as the correct answer is the query with the shortest duration. Also, the query answer with "WHERE CONVERT..." threw warnings in the execution plan...something about affecting CardinalityEstimate and SeekPlan.
I also found this article, which leads me to believe that I have the correct answer:
http://technet.microsoft.com/en-us/library/ms181034.aspx

QUESTION NO: 3
You use Microsoft SQL Server 2012 database to develop a shopping cart application.
You need to invoke a table-valued function for each row returned by a query.
Which Transact-SQL operator should you use?
A. PIVOT
B. UNPIVOT
C. CROSS APPLY
D. CROSS JOIN
Answer: C
Reference:
http://msdn.microsoft.com/en-us/library/ms175156.aspx

QUESTION NO: 4
You administer a Microsoft SQL Server 2012 database named ContosoDb. Tables are defined as shown in the exhibit. (Click the Exhibit button.)
You need to display rows from the Orders table for the Customers row having the CustomerId value set to 1 in the following XML format:
<row OrderId="1" OrderDate="2000-01-01T00:00:00" Amount="3400.00" Name="Customer A"
Country="Australia" />
<row OrderId="2" OrderDate="2001-01-01T00:00:00" Amount="4300.00" Name="Customer A"
Country="Australia" /> Which Transact-SQL query should you use?
A. SELECT OrderId, OrderDate, Amount, Name, Country
FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId WHERE
Customers.CustomerId = 1 FOR XML RAW
B. SELECT Name AS 'Customers/Name', Country AS 'Customers/Country', OrderId, OrderDate,
Amount FROM Orders INNER JOIN Customers ON Orders.CustomerId= Customers.CustomerId
WHERE Customers.CustomerId = 1 FOR XML PATH ('Customers')
C. SELECT OrderId, OrderDate, Amount, Name, Country
FROM Orders INNER JOIN Customers ON Orders.CustomerId - Customers.CustomerId WHERE
Customers.CustomerId = 1 FOR XML AUTO, ELEMENTS
D. SELECT Name, Country, OrderId, OrderDate, Amount
FROM Orders INNER JOIN Customers ON Orders.CustomerId= Customers.CustomerId WHERE
Customers.CustomerId = 1 FOR XML AUTO, ELEMENTS
E. SELECT Name, Country, OrderId, OrderDate, Amount
FROM Orders INNER JOIN Customers ON Orders.CustomerId= Customers.CustomerId WHERE
Customers.CustomerId = 1 FOR XML AUTO
F. SELECT OrderId, OrderDate, Amount, Name, Country
FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId WHERE
Customers.CustomerId = 1 FOR XML AUTO
G. SELECT Name AS '@Name', Country AS '@Country', OrderId, OrderDate, Amount FROM Orders
INNER JOIN Customers ON Orders.CustomerId= Customers.CustomerId WHERE
Customers.CustomerId = 1 FOR XML PATH ('Customers')
H. SELECT OrderId, OrderDate, Amount, Name, Country
FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId WHERE
Customers.CustomerId = 1 FOR XML RAW, ELEMENTS
Answer: A

QUESTION NO: 5
You need to create a query that meets the following requirements:
* The query must return a list of salespeople ranked by amount of sales and organized by postal code.
* The salesperson who has the highest amount of sales must be ranked first.
Part of the correct Transact-SQL has been provided in the answer area below. Enter the code in the answer area that resolves the problem and meets the stated goals or requirements. You can add code within code that has been provided as well as below it.
Use the 'Check Syntax' button to verify your work. Any syntax or spelling errors will be reported by line and character position.
A. 1 SELECT RowNumber() OVER(PARTITION BY PostalCode ORDER BY SalesYTd DESC) AS "Ranking",
2 p.LastName, s.SalesYTD, a.PostalCode
3 FROM Sales.SalesPerson AS a
etc
On line 1 add: RowNumber
One line 1 add: PARTITION BY
ROW_NUMBER() numbers the output of a result set. More specifically, returns the sequential number of a row within a partition of a result set, starting at 1 for the first row in each partition.
SYNTAX for OVER:
OVER (
[ <PARTITION BY clause> ]
[ <ORDER BY clause> ]
[ <ROW or RANGE clause> ]
)
Example: Using the OVER clause with the ROW_NUMBER function
The following example returns the ROW_NUMBER for sales representatives based on their assigned sales quota.
SELECT ROW_NUMBER() OVER(ORDER BY SUM(SalesAmountQuota) DESC) AS RowNumber,
FirstName, LastName, CONVERT(varchar(13), SUM(SalesAmountQuota),1) AS SalesQuota FROM dbo.DimEmployee AS e INNER JOIN dbo.FactSalesQuota AS sq ON e.EmployeeKey = sq.EmployeeKey
WHERE e.SalesPersonFlag = 1 GROUP BY LastName, FirstName; Here is a partial result set.
RowNumber FirstName LastName SalesQuota
--------- --------- ------------------ -------------
1 Jillian Carson 12,198,000.00
2 Linda Mitchell 11,786,000.00
3 Michael Blythe 11,162,000.00
4 Jae Pak 10,514,000.00
B. 1 SELECT RowNumber() OVER(PARTITION BY PostalCode ORDER BY SalesYTd DESC) AS "Ranking",
2 p.LastName, s.SalesYTD, a.PostalCode
3 FROM Sales.SalesPerson AS a
etc
More specifically, returns the sequential number of a row within a partition of a result set, starting at
1 for the first row in each partition.
SYNTAX for OVER:
OVER (
[ <PARTITION BY clause> ]
[ <ORDER BY clause> ]
[ <ROW or RANGE clause> ]
)
Example: Using the OVER clause with the ROW_NUMBER function
The following example returns the ROW_NUMBER for sales representatives based on their assigned sales quota.
SELECT ROW_NUMBER() OVER(ORDER BY SUM(SalesAmountQuota) DESC) AS RowNumber,
FirstName, LastName, CONVERT(varchar(13), SUM(SalesAmountQuota),1) AS SalesQuota FROM dbo.DimEmployee AS e INNER JOIN dbo.FactSalesQuota AS sq ON e.EmployeeKey = sq.EmployeeKey
WHERE e.SalesPersonFlag = 1 GROUP BY LastName, FirstName; Here is a partial result set.
RowNumber FirstName LastName SalesQuota
--------- --------- ------------------ -------------
1 Jillian Carson 12,198,000.00
2 Linda Mitchell 11,786,000.00
3 Michael Blythe 11,162,000.00
4 Jae Pak 10,514,000.00
Answer: A
Explanation:
References:
https://docs.microsoft.com/en-us/sql/t-sql/functions/row-number-transact-sql
https://docs.microsoft.com/en-us/sql/t-sql/queries/select-over-clause-transact-sql

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Updated: May 28, 2022

70-461合格体験談 - 70-461試験合格攻略 & Querying Microsoft SQL Server 2012/2014

PDF問題と解答

試験コード:70-461
試験名称:Querying Microsoft SQL Server 2012/2014
最近更新時間:2024-05-09
問題と解答:全 252
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模擬試験

試験コード:70-461
試験名称:Querying Microsoft SQL Server 2012/2014
最近更新時間:2024-05-09
問題と解答:全 252
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オンライン版

試験コード:70-461
試験名称:Querying Microsoft SQL Server 2012/2014
最近更新時間:2024-05-09
問題と解答:全 252
Microsoft 70-461 資格トレーニング

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70-461 テスト資料