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NEW QUESTION: 1
Hinweis: Diese Frage ist Teil einer Reihe von Fragen, bei denen die gleichen oder ähnliche Antwortmöglichkeiten verwendet werden. Eine Antwortauswahl kann für mehr als eine Frage in der Reihe richtig sein. Jede Frage ist unabhängig von den anderen Fragen in dieser Reihe. In einer Frage angegebene Informationen und Details gelten nur für diese Frage.
Sie erstellen eine Tabelle, indem Sie die folgende Transact-SQL-Anweisung ausführen:
Sie müssen alle Kundendaten prüfen.
Welche Transact-SQL-Anweisung sollten Sie ausführen?
A. Option B
B. Option A
C. Option F
D. Option G
E. Option G
F. Option C
G. Option E
H. Option D
Answer: A
Explanation:
Erläuterung
Die FOR SYSTEM_TIME ALL-Klausel gibt alle Zeilenversionen aus der Temporal- und der History-Tabelle zurück.
Hinweis: Eine vom System versionierte temporale Tabelle, die durch definiert wird, ist ein neuer Benutzertabellentyp in SQL Server 2016, der in der letzten Zeile WITH (SYSTEM_VERSIONING = ON ... definiert wird, um einen vollständigen Verlauf der Datenänderungen beizubehalten und zu vereinfachen Zeitpunktanalyse.
Um temporäre Daten abzufragen, enthält die SELECT-Anweisung FROM <table> eine neue Klausel FOR SYSTEM_TIME mit fünf temporalspezifischen Unterklauseln zum Abfragen von Daten in der aktuellen Tabelle und in der Verlaufstabelle.
Referenzen: https://msdn.microsoft.com/en-us/library/dn935015.aspx
NEW QUESTION: 2
View the exhibit and examine the structure of the EMPLOYEEStable.
You want to display all employees and their managers having 100 as the MANAGER_ID. You want the output in two columns: the first column would have the LAST_NAMEof the managers and the second column would have LAST_NAMEof the employees.
Which SQL statement would you execute?
A. SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON e.employee_id = m.manager_id
WHERE m.manager_id = 100;
B. SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE m.manager_id = 100;
C. SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
WHERE m.employee_id = e.manager_id and AND e.manager_id = 100
D. SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE e.manager_id = 100;
Answer: D
NEW QUESTION: 3
企業は、ミッションクリティカルなアプリケーションをサポートするためにAzure SQL Databaseを使用する予定です。
アプリケーションは、メンテナンス期間中にパフォーマンスが低下することなく、可用性が高くなければなりません。
ソリューションを実装する必要があります。
どの3つのテクノロジーを実装する必要がありますか?それぞれの正解はソリューションの一部を示しています。
注:それぞれの正しい選択には1ポイントの価値があります。
A. Premium service tier
B. Virtual machine Scale Sets
C. Always On availability groups
D. Zone-redundant configuration
E. SQL Data Sync
F. Basic service tier
Answer: A,C,D
Explanation:
Premium/business critical service tier model that is based on a cluster of database engine processes. This architectural model relies on a fact that there is always a quorum of available database engine nodes and has minimal performance impact on your workload even during maintenance activities.
In the premium model, Azure SQL database integrates compute and storage on the single node. High availability in this architectural model is achieved by replication of compute (SQL Server Database Engine process) and storage (locally attached SSD) deployed in 4-node cluster, using technology similar to SQL Server Always On Availability Groups.
Zone redundant configuration
By default, the quorum-set replicas for the local storage configurations are created in the same datacenter. With the introduction of Azure Availability Zones, you have the ability to place the different replicas in the quorum-sets to different availability zones in the same region. To eliminate a single point of failure, the control ring is also duplicated across multiple zones as three gateway rings (GW).
References:
https://docs.microsoft.com/en-us/azure/sql-database/sql-database-high-availability
NEW QUESTION: 4
2つのテーブルからデータを選択し、結果を構造体として保存します。
テーブルPARTNERには、フィールドPARTJDおよびKINDが含まれています。
テーブルCONTRACTには、項目CONTJD、CONT_TYPE、およびDIVISIONが含まれています。
構造は次のように定義されます
データ:wa_resultの始まり、
Part_idタイプpartner-partjd、cont_idタイプcontract-cont_id、
ConMype TYPEのcontract-cont_type、
wa_resultの終わり、
wa_resultのLt_resultタイプテーブル。
次のSELECTステートメントを外部結合に置き換えるにはどうすればよいですか?
パートナーからpartjdを選択しますINTO wa_result WHERE kind = 'Residential'。
contjdをCONTRACTからwa_result-cont_id WHEREパートEQに選択します
wa_partner-part_idそしてDIVISION eq 'Water'。
wa_resultをlt_resultに追加します。
ENDSELECT。
sy-subrc <> 0の場合。クリアwa_result-cont_id
wa_resultをlt_resultに追加します。 ENDIF。
ENDSELECT。
正しい答えを選んでください。
A. パートナーからのpartjdcontjdを選択します。partner-partjdの左結合コントラクト= contract-partjd AND契約区分EQ '水' INTOテーブルこれにより、WHERE種類EQ '住宅'になります。
B. a-partjd = b-partjdのパートナーAS AS LEFT JOINコントラクトからpartjdcontjdを選択し、テーブルlt_result WHERE kind = 'Residential'およびAND区分EQ 'Water'の対応するフィールドに入力します。
C. a-partjd = b-partjdAND b-division EQ 'Water' INTO TABLEIt_result WHERE kind = 'Residential'のパートナーAS A LEFT JOINコントラクトASからpartjdcontjdを選択します。
D. partner-partjdのパートナーLEFT JOINコントラクトからpartjdcontjdを選択します= contract-partjdおよびpartner-kind EQ 'Residential' INTO CORRESPONDING FIELDS OF TABLE lt_result WHERE Division eq 'Water'。
Answer: B