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NEW QUESTION: 1
会社の組織情報を保存するデータストアを設計しています。 データは、ユーザー間の関係を識別するために使用されます。 データはAzure Cosmos DBデータベースに格納され、数百万のオブジェクトが含まれます。
データベースに使用するAPIを推奨する必要があります。 APIは、ユーザー関係を照会するための複雑さを最小限に抑える必要があります。 ソリューションは高速トラバーサルをサポートする必要があります。
どのAPIをお勧めしますか?
A. MongoDB
B. Gremlin
C. Cassandra
D. Table
Answer: B
Explanation:
Gremlin features fast queries and traversals with the most widely adopted graph query standard.
References:
https://docs.microsoft.com/th-th/azure/cosmos-db/graph-introduction?view=azurermps-5.7.0
NEW QUESTION: 2
Your development team is using access keys to develop an application that has access to S3 and DynamoDB. A new security policy has outlined that the credentials should not be older than 2 months, and should be rotated. How can you achieve this?
Please select:
A. Use the application to rotate the keys in every 2 months via the SDK
B. Delete the user associated with the keys after every 2 months. Then recreate the user again.
C. Delete the 1AM Role associated with the keys after every 2 months. Then recreate the 1AM Role again.
D. Use a script to query the creation date of the keys. If older than 2 months, create new access key and update all applications to use it inactivate the old key and delete it.
Answer: D
Explanation:
One can use the CLI command list-access-keys to get the access keys. This command also returns the "CreateDate" of the keys. If the CreateDate is older than 2 months, then the keys can be deleted.
The Returns list-access-keys CLI command returns information about the access key IDs associated with the specified 1AM user. If there are none, the action returns an empty list Option A is incorrect because you might as use a script for such maintenance activities Option C is incorrect because you would not rotate the users themselves Option D is incorrect because you don't use 1AM roles for such a purpose For more information on the CLI command, please refer to the below Link:
http://docs.aws.amazon.com/cli/latest/reference/iam/list-access-keys.htmll The correct answer is: Use a script to query the creation date of the keys. If older than 2 months, create new access key and update all applications to use it inactivate the old key and delete it.
Submit your Feedback/Queries to our Experts
NEW QUESTION: 3
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 für diese Frage.
Sie haben eine Datenbank für ein Bankensystem. Die Datenbank enthält zwei Tabellen mit den Namen tblDepositAcct und tblLoanAcct, in denen Einlagen- und Darlehenskonten gespeichert sind. Beide Tabellen enthalten folgende Spalten:
Sie müssen die Gesamtzahl der Kunden ermitteln, die entweder Einlagenkonten oder Darlehenskonten haben, jedoch nicht beide Kontotypen.
Welche Transact-SQL-Anweisung sollten Sie ausführen?
A. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT (*) FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
C. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
D. SELECT COUNT (DISTINCT L.CustNo) VON tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
E. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT (DISTINCT D.CustNo) FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
G. SELECT COUNT (*) FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
H. SELECT COUNT (DISTINCT COALESCE (D.CustNo, L.CustNo)) FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNoWhere
Answer: H
Explanation:
Erläuterung
SQL Server stellt den vollständigen Outer-Join-Operator FULL OUTER JOIN bereit, der alle Zeilen aus beiden Tabellen enthält, unabhängig davon, ob die andere Tabelle einen übereinstimmenden Wert hat oder nicht.
Betrachten Sie einen Join der Product-Tabelle und der SalesOrderDetail-Tabelle in ihren ProductID-Spalten. Die Ergebnisse zeigen nur die Produkte mit Kundenaufträgen. Der Operator ISO FULL OUTER JOIN gibt an, dass alle Zeilen aus beiden Tabellen in die Ergebnisse einbezogen werden sollen, unabhängig davon, ob die Tabellen übereinstimmende Daten enthalten.
Sie können eine WHERE-Klausel mit einem vollständigen Outer-Join einfügen, um nur die Zeilen zurückzugeben, bei denen keine übereinstimmenden Daten zwischen den Tabellen vorhanden sind. Die folgende Abfrage gibt nur die Produkte zurück, die keine übereinstimmenden Kundenaufträge haben, sowie die Kundenaufträge, die nicht mit einem Produkt übereinstimmen.
USE AdventureWorks2008R2;
GEHEN
- Das Schlüsselwort OUTER nach dem Schlüsselwort FULL ist optional.
SELECT p.Name, sod.SalesOrderID
AB Produktion.Produkt p
VOLLSTÄNDIGE AUSSENVERBINDUNG Sales.SalesOrderDetail sod
ON p.ProductID = sod.ProductID
WHERE p.ProductID IST NULL
OR sod.ProductID IS NULL
ORDER BY p.Name;
Referenzen: https://technet.microsoft.com/en-us/library/ms187518(v=sql.105).aspx