You administer a Microsoft SQL Server 2012 databasenamed ContosoDb. Tables are defined as shown in the
exhibit. (Refer to the Exhibit.)
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.
<CUSTOMERS Name=”Customer A” Country=”Australia”>
<ORDERS OrderID=”1″ OrderDate=”2001-01-01″ Amount=”3400.00″ />
<ORDERS OrderID=”2″ OrderDate=”2002-01-01″ Amount=”4300.00″ />
</CUSTOMERS>
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 OrderId, OrderDate, Amount, Name, Country
FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML RAW, ELEMENTS
C.
SELECT OrderId, OrderDate, Amount, Name, Country
FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML AUTO
D.
SELECT OrderId, OrderDate, Amount, Name, Country
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 Name, Country, Orderld, OrderDate, Amount
FROM Orders INNER JOIN Customers ON Orders.CustomerId= Customers.CustomerId
WHERE Customers.CustomerId= 1
FOR XML AUTO, ELEMENTS
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 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’)
Explanation:
Verified answer as correct.
G
E is correct. I tested.
E is correct..
E is giving following answer when i tested
and G is giving following answer
1
2012-01-01T00:00:00
3400
2
2002-01-01T00:00:00
4300
E is correct, here when i pasted xml formated it did not print
no, its E
Wrong, wrong, wrong… Its G:
For anyone that wants to test on a test environment:
use xmldb
go
create table Customers (
customerid int NOT NULL,
name varchar(255) NULL,
country varchar(255) null,
PRIMARY KEY (customerid)
)
go
use xmldb
go
create table Orders (
orderid int NOT NULL,
customerid int NOT NULL,
orderdate date NULL,
amount int null,
PRIMARY KEY (orderid)
)
go
use xmldb
go
Insert Orders values (100,1,’2002-01-01′,2000)
go
use xmldb
go
Insert Orders values (101,1,’2002-01-02′,4000)
go
use xmldb
go
Insert Customers values (1,’Jose Silva’,’Brazil’)
go
use xmldb
go
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’)
actually the answer is E…i tested you code,but it’s wrong.
OPTION E —–CORRECT ANSWER
OPTION G
100
2002-01-01
2000
101
2002-01-02
4000
New 70-461 Exam Questions and Answers Updated Recently (5/May/2017):
NEW QUESTION 169
You have a database that contains a table named Customer. The customer table contains a column named LastName that has a column definition of varchar(50). An application named App1 reads from the table frequently. You need to change the column definition to nvarchar(100). The solution must minimize the amount of time it takes for App1 to read the data. Which statement should you execute?
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1691.png
Answer: C
Explanation:
To change the data type of a column in a SQL Server (or Microsoft access) table, use the following syntax:
ALTER TABLE table_name
ALTER COLUMN column_name datatype
NEW QUESTION 170
You are maintaining a SQL Server database that uses the default settings. The database contains a table that is defined by the following Transact-SQL statement:
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1701.png
Answer: A
Explanation:
Char(13) is a carriage return. Use the IIF construct to return an empty string for NULL values of the Adressline2 column. IIF returns one of two values, depending on whether theBoolean expression evaluates to true or false in SQL Server.
NEW QUESTION 171
A table named Profits stores the total profit made each year within a territory. The Profits table has columns named Territory, Year, and Profit. You need to create a report that displays the profits made by each territory for each year and its previous year. Which Transact-SQL query should you use?
A. SELECT Territory, Year, Profit, LEAD(Profit, 1, 0) OVER (PARTITION BY Territory ORDER BY Year) AS PreviousYearProfit FROM Profits
B. SELECT Territory, Year, Profit, LAG(Profit, 1, 0) OVER (PARTITION BY Year ORDER BY Territory) AS PreviousYearProfit FROM Profits
C. SELECT Territory, Year, Profit, LAG(Profit, 1, 0) OVER (PARTITION BY Territory ORDER BY Year) AS PreviousYearProfit FROM Profits
D. SELECT Territory, Year, Profit, LEAD(Profit, 1, 0) OVER (PARTITION BY Year ORDER BY Territory) AS PreviousYearProfit FROM Profits
Answer: C
NEW QUESTION 172
Your database contains a table named Products that has columns named ProductID and Name. You want to write a query that retrieves data from the Products table sorted by Name listing 15 rows at a time. You need to view rows 31 through 45. Which Transact-SQL query should you create?
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1721.png
Answer: C
NEW QUESTION 173
A database named AdventureWorks contains two tables named Production.Product and Sales.SalesOrderDetail. The tables contain data on the available products and a detailed order history. The Production.Product table contains the following two columns:
– ProductID
– Name
The Sales.SalesOrderDetail table contains the following three columns:
– SalesOrderID
– ProductID
– OrderQty
You need to create a query listing all of the products that were never ordered. Which statements should you execute?
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1731.png
Answer: A
NEW QUESTION 174
You plan to write a query for a new business report that will contain several nested queries. You need to ensure that a nested query can call a table-valued function for each row in the main query. Which query operator should you use in the nested query?
A. CROSS APPLY
B. INNER JOIN
C. OUTER JOIN
D. PIVOT
Answer: A
Explanation:
The APPLY operator allows you to invoke a table-valued function for each row returned by an outer table expression of a query. The table-valued function acts as the right input and the outer table expression acts as the left input. The right input is evaluated for each row from the left input and the rows produced are combined for the final output. The list of columns produced by the APPLY operator is the set of columns in the left input followed by the list of columns returned by the right input. There are two forms of APPLY: CROSS APPLY and OUTER APPLY. CROSSAPPLY returns only rows from the outer table that produce a result set from the table-valued function. OUTER APPLY returns both rows that produce a result set, and rows that do not, with NULL values in the columns produced by the table-valued function.
NEW QUESTION 175
You are designing a table for a SQL Server database. The table uses horizontal partitioning. You have the following requirements:
– Each record in the table requires a unique key.
– You must minimize table fragmentation as the table grows.
You need to choose the appropriate data type for the key value. What should you do?
A. Use the NEWID function to create a unique identifier.
B. Use the NEWSEQUENTIALID function to create a unique identifier.
C. Generate a random value that uses the bigint datatype.
D. Generate a random value that uses the char(16) data type.
Answer: B
Explanation:
Horizontal partitioning divides a table into multiple tables. Each table then contains the same number of columns, but fewer rows. For example, a table that contains 1 billion rows could be partitioned horizontally into 12 tables, with each smaller table representing one month of data for a specific year. Any queries requiring data for a specific month only reference the appropriate table.
NEWSEQUENTIALID creates a GUID that is greater than any GUID previously generated by this function on a specified computer since Windows was started. After restarting Windows, the GUID can start again from a lower range, but is still globally unique. When a GUID column is used as a row identifier, using NEWSEQUENTIALID can be faster than using the NEWID function. This is because the NEWID function causes random activity and uses fewer cached data pages. Using NEWSEQUENTIALID also helps to completely fill the data and index pages.
NEW QUESTION 176
You are a database developer of a Microsoft SQL Server 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.
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1761.jpg
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,
CustomerID int NOT NULL,
CustomerName varchar(255) NOT NULL
CONSTRAINT UQ_Customer UNIQUE
(SourceID, CustomerID));
B. CREATE TABLE Customer
(SourceID int NOT NULL UNIQUE,
CustomerID int NOT NULL UNIQUE,
CustomerName varchar(255) NOT NULL);
C. CREATE TABLE Customer
(SourceID int NOT NULL PRIMARY KEY CLUSTERED,
CustomerID int NOT NULL UNIQUE,
CustomerName varchar(255) NOT NULL);
D. CREATE TABLE Customer
(SourceID int NOT NULL,
CustomerID int NOT NULL,
CustomerName varchar(255) NOT NULL,
CONSTRAINT PK_Customer PRIMARY KEY CLUSTERED
(SourceID, CustomerID));
Answer: D
Explanation:
A PRIMARY KEY is a constraint that enforces entity integrity for a specified column or columns by using a unique index. Only one PRIMARY KEY constraint can be created for each table. We need to use both SourceID and CustomerID, in that order, in the PRIMARY KEY constraint.
NEW QUESTION 177
Drag and Drop Question
You maintain a SQL Server database that is used by a retailer to track sales and salespeople. The database includes two tables and two triggers that is defined by the following Transact-SQL statements:
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1771.jpg
During days with a large sales volume, some new sales transaction fail and report the following error:
Arithmetic overflow error converting expression to data type int.
You need to ensure that the two triggers are applied once per sale, and that they do not interfere with each other. How should you complete the relevant Transact-SQL statement? To answer, drag the appropriate Transact-SQL segments to the correct location or locations. Each Transact-SQL segment may be used once, more than once, or not at all. You may need to drag the split bar between panes or scroll to view content.
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1772.jpg
Answer:
Image URL: passleader.org/wp-content/uploads/2017/05/passleader-70-461-dumps-1773.png
Explanation:
Box 1: OFF
Box 2: 0
Only direct recursion of AFTER triggers is prevented when the RECURSIVE_TRIGGERS database option is set to OFF. To disable indirect recursion of AFTER triggers, also set the nested triggers server option to 0.
Note:
Both DML and DDL triggers are nested whena trigger performs an action that initiates another trigger. These actions can initiate other triggers, and so on. DML and DDL triggers can be nested up to 32 levels. You can control whether AFTER triggers can be nested through the nested triggers server configuration option. If nested triggers are allowed and a trigger in the chain starts an infinite loop, the nesting level is exceeded and the trigger terminates.
NEW QUESTION 178
……
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