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- Name: Enterprise Routing and Switching, Specialist (JNCIS-ENT)
- Exam Code: JN0-351
- Certification: JN0-351
- Vendor: Juniper
- Total Question: 630
- Price: 49$
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NEW QUESTION: 1
Sie müssen eine Abfrage erstellen, die die folgenden Anforderungen erfüllt:
* Die Abfrage muss eine Liste von Verkäufern zurückgeben, die nach Umsatzmenge geordnet und nach Postleitzahl geordnet sind.
* Der Verkäufer mit dem höchsten Umsatz muss an erster Stelle stehen.
Ein Teil des korrekten Transact-SQL wurde im Antwortbereich unten bereitgestellt. Geben Sie den Code in den Antwortbereich ein, der das Problem löst und die angegebenen Ziele oder Anforderungen erfüllt. Sie können Code innerhalb des bereitgestellten Codes sowie darunter hinzufügen.
Verwenden Sie die Schaltfläche 'Syntax prüfen', um Ihre Arbeit zu überprüfen. Alle Syntax- oder Rechtschreibfehler werden nach Zeilen- und Zeichenposition gemeldet.
A. 1 SELECT RowNumber () OVER (PARTITION BY PostalCode ORDER BY SalesYTd DESC) AS "Ranking",
2 p.LastName, s.SalesYTD, a.PostalCode
3 FROM Sales.SalesPerson AS a
etc
Fügen Sie in Zeile 1 Folgendes hinzu: RowNumber
Eine Zeile 1 hinzufügen: PARTITION BY
ROW_NUMBER () nummeriert die Ausgabe einer Ergebnismenge. Gibt insbesondere die fortlaufende Nummer einer Zeile innerhalb einer Partition einer Ergebnismenge zurück, beginnend bei 1 für die erste Zeile in jeder Partition.
SYNTAX für ÜBER:
ÜBER (
[<PARTITION BY-Klausel>]
[<ORDER BY-Klausel>]
[<ROW- oder RANGE-Klausel>]
)
Beispiel: Verwenden der OVER-Klausel mit der Funktion ROW_NUMBER
Im folgenden Beispiel wird die ROW_NUMBER für Vertriebsmitarbeiter basierend auf dem zugewiesenen Vertriebskontingent zurückgegeben.
SELECT ROW_NUMBER () OVER (ORDER BY SUM (SalesAmountQuota) DESC) AS RowNumber, Vorname, Nachname, CONVERT (varchar (13), SUMME (SalesAmountQuota), 1) AS SalesQuota FROM dbo.DimEmployee AS e INNER JOIN dbo.Fact ON e.EmployeeKey = sq.EmployeeKey WHERE e.SalesPersonFlag = 1 GROUP BY LastName, FirstName; Hier ist eine Teilergebnismenge.
Zeilennummer Vorname Nachname SalesQuota
--------- --------- ------------------ -------------
1 Jillian Carson 12.198.000,00
2 Linda Mitchell 11.786.000,00
3 Michael Blythe 11.162.000,00
4 Jae Pak 10.514.000,00
B. 1 SELECT RowNumber () OVER (PARTITION BY PostalCode ORDER BY SalesYTd DESC) AS "Ranking",
2 p.LastName, s.SalesYTD, a.PostalCode
3 FROM Sales.SalesPerson AS a
etc
Gibt insbesondere die fortlaufende Nummer einer Zeile innerhalb einer Partition einer Ergebnismenge zurück, beginnend bei 1 für die erste Zeile in jeder Partition.
SYNTAX für ÜBER:
ÜBER (
[<PARTITION BY-Klausel>]
[<ORDER BY-Klausel>]
[<ROW- oder RANGE-Klausel>]
)
Beispiel: Verwenden der OVER-Klausel mit der Funktion ROW_NUMBER
Im folgenden Beispiel wird die ROW_NUMBER für Vertriebsmitarbeiter basierend auf dem zugewiesenen Vertriebskontingent zurückgegeben.
SELECT ROW_NUMBER () OVER (ORDER BY SUM (SalesAmountQuota) DESC) AS RowNumber, Vorname, Nachname, CONVERT (varchar (13), SUMME (SalesAmountQuota), 1) AS SalesQuota FROM dbo.DimEmployee AS e INNER JOIN dbo.Fact ON e.EmployeeKey = sq.EmployeeKey WHERE e.SalesPersonFlag = 1 GROUP BY LastName, FirstName; Hier ist eine Teilergebnismenge.
Zeilennummer Vorname Nachname SalesQuota
--------- --------- ------------------ -------------
1 Jillian Carson 12.198.000,00
2 Linda Mitchell 11.786.000,00
3 Michael Blythe 11.162.000,00
4 Jae Pak 10.514.000,00
Answer: A
Explanation:
References:
https://docs.microsoft.com/en-us/sql/t-sql/functions/row-number-transact-sql
https://docs.microsoft.com/en-us/sql/t-sql/queries/select-over-clause-transact-sql
NEW QUESTION: 2
ストーリーが小さすぎることもあります。エクストリームプログラミングチームに共通の小さなストーリーに対する良いアプローチは、それらを組み合わせて、半日から数日の作業を表す大きなストーリーにまとめることです。結合されたストーリーには名前が付けられ、他のストーリーと同様にスケジュールされて作業されます。次のうちどれが小さな物語の例ではありませんか？
A. バグレポート
B. ユーザーインターフェイスの変更
C. バグ
D. AとC
Answer: C
NEW QUESTION: 3
在一台运行 RIP 的 MSR 路由器上看到如下信息:
<MSR>display rip
Public VPN-instance name:
RIP process: 1
RIP version: 2
Preference: 100
Checkzero: Enabled
Default-cost: 0
Summary: Enabled
Hostroutes: Enabled
Maximum number of balanced paths: 8
那么从显示信息可以分析出 ______ 。
A. 本路由器发送或者接收的 RIP 路由的 Cost 都是 0
B. 该路由器运行的是 RIPv2
C. RIP 的自动聚合功能是开启的
D. 支持 8 条路由实现负载分担
Answer: B,C,D
NEW QUESTION: 4
Which of the following is NOT a valid selection for tracking and controlling packets in R71?
A. Reject
B. Hold
C. Accept
D. Session Auth
Answer: B
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