Exam Name: Querying Microsoft SQL Server 2012/2014 (70-461日本語版)
Certification Provider: Microsoft
Corresponding Certification: Microsoft SQL Server 2012
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Microsoft 70-461日本語 Exam Overview:
| Certification Vendor: | Microsoft |
|---|---|
| Exam Name: | Querying Microsoft SQL Server 2012/2014 |
| Exam Number: | 70-461 |
| Real Exam Qty: | 40–60 (varies) |
| Related Certifications: | MCSA: SQL Server 2012/2014 MCSE Data Platform (legacy) |
| Certificate Validity Period: | Retired (legacy certification, no longer active) |
| Exam Duration: | 120 minutes |
| Exam Price: | USD 165 (varies by region) |
| Available Languages: | Japanese, Korean, German, English, Spanish, French, Chinese (Simplified) |
| Passing Score: | 700 (on a 1000-point scale) |
| Exam Format: | Multiple choice, Multiple response, Case studies, T-SQL query tasks, Drag and drop |
| Recommended Training: | SQL Server Querying Courses (general training platforms) Microsoft Learn SQL Server training paths |
| Exam Registration: | Microsoft Credentials (Learn) Pearson VUE Microsoft Exams |
| Sample Questions: | DOWNLOAD DEMO |
| Exam Way: | Available via Pearson VUE test centers and online proctored exams (where offered) |
| Pre Condition: | No formal prerequisites. Recommended to have basic knowledge of SQL Server and T-SQL. |
Microsoft 70-461日本語 Exam Syllabus Topics:
| Section | Objectives |
|---|---|
| Topic 1: Troubleshoot and optimize queries | - Performance tuning basics
|
| Topic 2: Implement T-SQL queries | - Query data
|
| Topic 3: Create database objects | - Design and implement tables
|
| Topic 4: Modify data | - Insert, update, and delete data
|
Microsoft 70-461日本語 Exam: Spare-Time Answers
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The Microsoft Querying Microsoft SQL Server 2012/2014 (70-461日本語版) is delivered Available via Pearson VUE test centers and online proctored exams (where offered) — pick the arrangement that suits you when booking.
The Microsoft Querying Microsoft SQL Server 2012/2014 (70-461日本語版) is Microsoft's certification exam for MCSA: SQL Server 2012/2014, at the Associate level. Office workers without time for classes prepare best in spare time with focused material. Related credentials include MCSA: SQL Server 2012/2014, MCSE Data Platform (legacy).
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The Microsoft Querying Microsoft SQL Server 2012/2014 (70-461日本語版) blueprint spans 4 domains — including Troubleshoot and optimize queries, Implement T-SQL queries, Modify data. Daily checks keep our material aligned; the complete outline above lists every subtopic.
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Microsoft Querying Microsoft SQL Server 2012/2014 (70-461日本語版) Sample Questions:
あなたのデータベースは、プロダクトとProductsPriceLogという名前のテーブルを含みます。プロダクト・テーブルは、ProductCodeとプライスという名のカラムを含んでいます。 ProductsPriceLogテーブルは、ProductCode、OldPriceとNewPriceという名前のコラムを含みます。ProductsPriceLogテーブルはOldPriceカラムに前の価格を格納し、NewPriceカラムに新しい価格を格納します。
あなたは5％のProductsテーブル内のすべての製品の価格の列の値を増やす必要があります。またProductsPriceLogテーブルへの変更を記録する必要があります。どのTransact-SQLクエリを使うべきでしょうか。
- A. UPDATE Products SET Price = Price * 1.05 OUTPUT inserted.ProductCode, deleted.Price, inserted.Price * 1.05 INTO ProductsPriceLog(ProductCode, OldPrice, NewPrice)
- B. UPDATE Products SET Price = Price * 1.05 OUTPUT inserted.ProductCode, deleted.Price, inserted.Price INTO ProductsPriceLog(ProductCode, OldPrice, NewPrice)
- C. UPDATE Products SET Price = Price * 1.05 INSERT INTO ProductsPriceLog(ProductCode, OldPrice, NewPrice) SELECT ProductCode, Price, Price * 1.05 FROM Products
- D. UPDATE Products SET Price = Price * 1.05 OUTPUT inserted.ProductCode, inserted.Price, deleted.Price INTO ProductsPriceLog(ProductCode, OldPrice, NewPrice)
Correct Answer: B 🗳️
ContosoDbという名前のMicrosoft Azure SQLデータベースインスタンスを使用します。 ContosoDbには、既存のレコードを持つProductsという名前のテーブルが含まれています。
Productsテーブルには、CodeおよびQuantityOnHandという名前の列があります。
Productsテーブルに、Categoryという名前の新しい列を作成して、null値を許可し、既存のすべてのレコードのCategory列の値をGeneralに設定する必要があります。
ソリューションの開発に使用する必要があるTransact-SQLセグメントはどれですか？回答するには、適切なTransact-SQLセグメントをTransact-SQLセグメントのリストから回答エリアに移動し、正しい順序で並べます。
Correct Answer:
Explanation:
NULL or NOT NULL specifies whether the column can accept null values. Columns that do not allow null values can be added with ALTER TABLE only if they have a default specified or if the table is empty. NOT NULL can be specified for computed columns only if PERSISTED is also specified. If the new column allows null values and no default is specified, the new column contains a null value for each row in the table. If the new column allows null values and a default definition is added with the new column, WITH VALUES can be used to store the default value in the new column for each existing row in the table.
Reference:
https://docs.microsoft.com/en-us/sql/t-sql/statements/alter-table-transact-sql
あなたは、Microsoft SQL Server 2012のデータベースを管理します。データベースには、Employeeという名前のテーブルが含まれています。Employeeテーブルの一部は、展示に示されています。（Exhibitボタンをクリックしてください。）
従業員に関する機密情報はEmployeeDataという別のテーブルに格納されます。1つのレコードは従業員テーブルの各レコードのEmployeeDataに存在します。あなたは、データの整合性と視認性を確保するために適切な制約やテーブルのプロパティを割り当てる必要があります。 On which column in the Employee table should you create a unique constraint?
- A. EmployeeID
- B. JobTitle
- C. EmployeeNum
- D. MiddleName
- E. DateHired
- F. DepartmentID
- G. LastName
- H. ReportsToID
- I. FirstName
Correct Answer: C 🗳️
あなたは、Microsoft SQL Server 2012データベースを開発します。あなたは、次の要件を満たしているバッチ処理を作成する必要があります。
指定されたパラメータに基づいた結果セットを返します。
返された結果セットがテーブルで接点を実行するのを可能にします。
どのオブジェクトを使うべきでしょうか。
- A. インラインuser-definedの機能
- B. ストアドプロシージャ
- C. スカラーuser-defined機能
- D. Table-valued user-defined機能
Correct Answer: D 🗳️
あなたは、EmployeeとPersonという名前のテーブルを含むMicrosoft SQL Server 2012のデータベースを開発すしまする。テーブルには、以下の定義があります。
ユーザーは、単一のINSERTステートメントまたはINSERT...SELECTステートメント使用してこのビューに挿入することができます。あなたはそのユーザーがVwEmployeeビューを使用して、両方の従業員とPersonテーブルにレコードを挿入するために単一のステートメントを使用することができることを確認する必要があります。どのTransact-SQLステートメントを使用する必要がありますか。
- A. CREATE TRIGGER TrgVwEmployee
ON VwEmployee
INSTEAD OF INSERT
AS
BEGIN
INSERT INTO Person(Id, FirstName, LastName)
SELECT Id, FirstName, LastName FROM VwEmployee
INSERT INTO Employee(PersonID, EmployeeNumber)
SELECT Id, EmployeeNumber FROM VwEmployee
End - B. CREATE TRIGGER TrgVwEmployee
ON VwEmployee
FOR INSERT
AS
BEGIN
INSERT INTO Person(Id, FirstName, LastName)
SELECT Id, FirstName, LastName, FROM inserted
INSERT INTO Employee(PersonId, EmployeeNumber)
SELECT Id, EmployeeNumber FROM inserted
END - C. CREATE TRIGGER TrgVwEmployee
ON VwEmployee
INSTEAD OF INSERT
AS
BEGIN
INSERT INTO Person(Id, FirstName, LastName)
SELECT Id, FirstName, LastName, FROM inserted
INSERT INTO Employee(PersonId, EmployeeNumber)
SELECT Id, EmployeeNumber FROM inserted
END - D. CREATE TRIGGER TrgVwEmployee
ON VwEmployee
INSTEAD OF INSERT
AS
BEGIN
DECLARE @ID INT, @FirstName NVARCHAR(25), @LastName NVARCHAR(25),
@PersonID
INT, @EmployeeNumber NVARCHAR(15)
SELECT @ID = ID, @FirstName = FirstName, @LastName = LastName,
@EmployeeNumber
= EmployeeNumber
FROM inserted
INSERT INTO Person(Id, FirstName, LastName)
VALUES(@ID, @FirstName, @LastName)
INSERT INTO Employee(PersonID, EmployeeNumber)
VALUES(@PersonID, @EmployeeNumber
End
Correct Answer: C 🗳️
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