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NEW QUESTION: 1
Sie müssen eine Transact-SQL-Anweisung entwickeln, die die folgenden Anforderungen erfüllt:
- Die Anweisung muss einen benutzerdefinierten Fehler zurückgeben, wenn beim Aktualisieren einer Tabelle Probleme auftreten.
- Die Fehlernummer muss den Wert 50555 haben.
- Der Schweregrad des Fehlers muss 14 sein.
- Eine Microsoft SQL Server-Warnung muss ausgelöst werden, wenn die Fehlerbedingung auftritt.
Welches Transact-SQL-Segment sollten Sie für jede Anforderung verwenden? Wählen Sie zum Beantworten die entsprechenden Transact-SQL-Segmente im Antwortbereich aus.

Answer:
Explanation:

Erläuterung

Kasten 1: VERSUCHEN ... FANGEN
Das TRY ... CATCH Transact-SQL-Konstrukt implementiert eine Fehlerbehandlung für Transact-SQL, die der Ausnahmebehandlung in den Sprachen Microsoft Visual C # und Microsoft Visual C ++ ähnelt. Eine Gruppe von Transact-SQL-Anweisungen kann in einen TRY-Block eingeschlossen werden. Wenn im TRY-Block ein Fehler auftritt, wird die Steuerung an eine andere Gruppe von Anweisungen übergeben, die in einem CATCH-Block enthalten sind.
Feld 2: RAISERROR (50555, 14, 1 'Das Update ist fehlgeschlagen. ") WITH LOG
Wir müssen RAISERROR verwenden, um den erforderlichen Schweregrad von 14 angeben zu können, und wir sollten auch die Option LOG verwenden, die den Fehler im Fehlerprotokoll und im Anwendungsprotokoll für die Instanz des Microsoft SQL Server-Datenbankmoduls protokolliert Aktivieren Sie eine MS MS SQL Server-Warnung ausgelöst werden.
Hinweis: RAISERROR generiert eine Fehlermeldung und leitet die Fehlerverarbeitung für die Sitzung ein. RAISERROR kann entweder auf eine benutzerdefinierte Nachricht verweisen, die in der sys.messages-Katalogansicht gespeichert ist, oder eine Nachricht dynamisch erstellen. Die Nachricht wird als Serverfehlermeldung an die aufrufende Anwendung oder an einen zugeordneten CATCH-Block eines TRY ... CATCH-Konstrukts zurückgegeben.

NEW QUESTION: 2
Where is the succession Talent Card configuration located in Admin Tools? Please choose the correct answer.
A. Presentation
B. Succession
C. Reporting
D. Employee Files
Answer: D

NEW QUESTION: 3
Refer to the exhibit.

BigBids Incorporated is a worldwide auction provider. The network uses EIGRP as its routing protocol throughout the corporation. The network administrator does not understand the convergence of EIGRP. Using the output of the show ip eigrp topology all-links command, answer the administrator's question.
Which two networks does the Core1 device have feasible successors for? (Choose two)
A. 172.17.0.0/30
B. 172.17.3.0/25
C. 172.17.1.0/24
D. 10.140.0.0/24
E. 172.17.3.128/25
F. 172.17.2.0/24
Answer: A,D
Explanation:
Explanation To understand the output of the "show ip eigrp topology all-links command" command, let's analyze an entry (we choose the second entry because it is better for demonstration than the first one)

The first line tells us there is only 1 successor for the path to 10.140.0.0/24 network but there are 2 lines below. So we can deduce that one line is used for successor and the other is used for another route to that network. Each of these two lines has 2 parameters: the first one ("156160 or "157720) is the Feasible Distance (FD) and the second ("128256 or "155160) is the Advertised Distance (AD) of that route. The next thing we want to know is: if the route via 172.17.10.2 (the last line) would become the feasible successor for the 10.140.0.0/24 network. To figure out, we have to compare the Advertised Distance of that route with the Feasible Distance of the successor's route, if AD < FD then it will become the feasible successor. In this case, because AD (155160) < FD (156160) so it will become the feasible successor. Therefore we can conclude the network 10.140.0.0/24 has 1 feasible successor. Because the question asks about feasible successor so we just need to focus on entries which have more paths than the number of successor. In this case, we find 3 entries that are in blue boxes because they have only 1 successor but has 2 paths, so the last path can be the feasible successor. By comparing the value of AD (of that route) with the FD (of successor's route) we figure out there are 2 entries will have the feasible successor: the first and the second entry. The third entry has AD = FD (30720) so we eliminate it.

NEW QUESTION: 4

A. Option E
B. Option A
C. Option C
D. Option B
E. Option F
F. Option D
Answer: B,D,E