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試験番号：70-461試験科目：「Querying Microsoft SQL Server 2012/2014」
一年間無料で問題集をアップデートするサービスを提供いたします
最近更新時間：2017-10-26
問題と解答：全165問 70-461 勉強の資料
NO.1 DRAG DROP
You have a SQL Server database named CUSTOMERS.
You need to sign a stored procedure named SelectCustomers in the CUSTOMERS database.
Which four statements should you execute in sequence? To answer, move the appropriate
statements from the list of statements to the answer area and arrange them in the correct order.
Answer:
Explanation:
There are four steps involved in signing a module:
References:https://msdn.microsoft.com/en-us/library/bb669102(v=vs.110).aspx
NO.2 You administer several Microsoft SQL Server 2012 database servers. Merge replication has been
configured for an application that is distributed across offices throughout a wide area network
(WAN). Many of the tables involved in replication use the XML and varchar (max) data types.
Occasionally, merge replication fails due to timeout errors. You need to reduce the occurrence of
these timeout errors. What should you do?
A. Change the Merge agent on the problem subscribers to run continuously.
B. Set the Merge agent on the problem subscribers to use the slow link agent profile.
C. Create a snapshot publication, and reconfigure the problem subscribers to use the snapshot
publication.
D. Set the Remote Connection Timeout on the Publisher to 0.
Answer: B
70-461 参考書
NO.3 You are a database developer for an application hosted on a Microsoft SQL Server 2012
server.
The database contains two tables that have the following definitions:
Global customers place orders from several countries.
You need to view the country from which each customer has placed the most orders.
Which Transact-SQL query do you use?
A. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM Customer c
INNER JOIN
(SELECT CustomerID, ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY COUNT(OrderAmount) DESC) AS Rnk
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WHERE o.Rnk = 1
B. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM
(SELECT c.CustomerID, c.CustomerName, o.ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY COUNT(o.OrderAmount) ASC) AS Rnk
FROM Customer c
INNER JOIN Orders o
ON c.CustomerID = o.CustomerID
GROUP BY c.CustomerID, c.CustomerName, o.ShippingCountry) cs
WHERE Rnk = 1
C. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM Customer c
INNER JOIN
(SELECT CustomerID, ShippingCountry,
COUNT(OrderAmount) DESC) AS OrderAmount
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
ORDER BY OrderAmount DESC
D. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM Customer c
INNER JOIN
(SELECT CustomerID, ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY OrderAmount DESC) AS Rnk
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WHERE o.Rnk = 1
Answer: A
70-461 対応 70-461 答案
NO.4 You are a database developer of a Microsoft SQL Server 2012 database.
You are designing a table that will store Customer data from different sources. The table will include
a column that contains the CustomerID from the source system and a column that contains the
SourceID.
A sample of this data is as shown in the following table.
You need to ensure that the table has no duplicate CustomerID within a SourceID. You also need to
ensure that the data in the table is in the order of SourceID and then
CustomerID.
Which Transact- SQL statement should you use?
A. CREATE TABLE Customer
(SourceID int NOT NULL PRIMARY KEY CLUSTERED,
CustomerID int NOT NULL UNIQUE,
CustomerName varchar(255) NOT NULL);
B. CREATE TABLE Customer
(SourceID int NOT NULL,
CustomerID int NOT NULL,
CustomerName varchar(255) NOT NULL,
CONSTRAINT PK_Customer PRIMARY KEY CLUSTERED
(SourceID, CustomerID));
C. CREATE TABLE Customer
(SourceID int NOT NULL IDENTITY,
CustomerID int NOT NULL IDENTITY,
CustomerName varchar(255) NOT NULL);
D. CREATE TABLE Customer
(SourceID int NOT NULL,
CustomerID int NOT NULL PRIMARY KEY CLUSTERED,
CustomerName varchar(255) NOT NULL);
Answer: B
70-461 テスト
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