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NEW QUESTION: 1
Why would you use the Mediator component in a Service Component Architecture (SCA) application Instead of using Oracle Service Bus?
A. The Mediator is much faster than Oracle Service Bus.
B. Oracle Service Bus runs out side of a SCA application. Therefore, the Mediator is the only choice for translation and transformation tasks in an SCA application.
C. It is always better to use Oracle service Bus Instead of the Mediator.
D. The Mediator supports transactions, but Oracle Service Bus does not.
Answer: B
Explanation:
Whereas Oracle Service Bus provides service virtualization and protocol transformations for an Oracle SOA Suite application containing multiple composites, Oracle Mediator is an intra-composite mediation component that is deployed within an application.
Note:
* Analogous to a load balancer routing HTTP traffic, the Oracle Mediator routes data from service providers to external partners. In addition, it can subscribe to and publish business events.
Using Oracle Mediator, you create routing services and rules for them. A routing service is the key
component for moving a message across the enterprise service bus - from its entry point to its exit
point. The rules determine how a message instance processed by the routing service gets to its
next destination. Using the rules, Oracle Mediator can perform the following actions:
Route: Determines the service component (BPEL process, business rule, human task, and
mediator) to which to send the messages.
Validate: Provides support for validating the incoming message payload by using a schematron or
an XSD file.
Filter: If specified in the rules, applies a filter expression that specifies the contents (payload) of a
message be analyzed before any service is invoked.
Transformation: If specified in the rules, transforms document data from one XML schema to
another, thus enabling data interchange among applications using different schemas.
NEW QUESTION: 2
HOTSPOT
Answer:
Explanation:
Explanation:
Box 1: No
The LastWriteTime of DC01.vhd was on June 21, 2016, and the current date is also June
21, 2016, but the MinimumFileAgeDays is 3.
MinimumFileAgeDays specifies a number of days. The deduplication engine optimizes files that users have not accessed in the number of days that you specify. If the last access time is not available, then the deduplication engine uses the last modified time.
Box 2: No
The size of Readme.txt, 12400bytes, is less than the Minimum File size, 32768 bytes.
MinimumFileSize specifies the minimum size threshold, in bytes, for files that are optimized. The deduplication engine does not optimize files that do not meet the minimum threshold.
Box 3: Yes
The Software ISO file is both large and old enough for deduplication.
References: https://technet.microsoft.com/en-us/library/hh848438.aspx
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 die folgenden Spalten:
Sie müssen eine Abfrage ausführen, um die Gesamtzahl der Kunden zu ermitteln, die sowohl Einzahlungs- als auch Darlehenskonten haben.
Welche Transact-SQL-Anweisung sollten Sie ausführen?
A. SELECT COUNT (DISTINCT L.CustNo) VON tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
B. SELECT COUNT (*) FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
C. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
D. SELECT COUNT (*) FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
E. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT (DISTINCT COALESCE (D.CustNo, L.CustNo)) FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNoWhere
G. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT (DISTINCT D.CustNo) FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
Answer: B
Explanation:
Erläuterung
Der SQL INTERSECT-Operator wird verwendet, um die Ergebnisse von 2 oder mehr SELECT-Anweisungen zurückzugeben. Es werden jedoch nur die Zeilen zurückgegeben, die von allen Abfragen oder Datensätzen ausgewählt wurden. Wenn ein Datensatz in einer Abfrage und nicht in der anderen vorhanden ist, wird er in den INTERSECT-Ergebnissen nicht berücksichtigt.
Referenzen: https://www.techonthenet.com/sql/intersect.php