Exam 70-761 topic 1 question 89 discussion

B correct, again sample code DROP TABLE #ERRORS CREATE TABLE #ERRORS ( ERRORID INT PRIMARY KEY, SERVERID INT, OCCURENCES INT, LOGMESSAGE NVARCHAR(MAX) ) INSERT INTO #ERRORS (ERRORID, SERVERID, OCCURENCES, LOGMESSAGE) VALUES (1, 1,3,'ERROR1'), (2,1,2,'ERROR2'), (3,1,10,'ERROR2'), (4,2,4,'ERROR2'), (5,2,14,'ERROR2'), (6,2,24,'ERROR1'), (7,3,10,'ERROR3'), (8,3,24,'ERROR3') SELECT DISTINCT SERVERID, LOGMESSAGE FROM #ERRORS AS E1 WHERE OCCURENCES > ALL ( SELECT E2.OCCURENCES FROM #ERRORS AS E2 WHERE E2.LOGMESSAGE = E1.LOGMESSAGE AND E2.SERVERID <> E1.SERVERID) SELECT DISTINCT SERVERID, LOGMESSAGE FROM #ERRORS AS E1 WHERE LOGMESSAGE IN ( SELECT TOP 1 E2.LOGMESSAGE FROM #ERRORS AS E2 WHERE E1.LOGMESSAGE = E2.LOGMESSAGE AND E1.SERVERID <> E2.SERVERID ORDER BY OCCURENCES )

I think В is correct. In A statement "ORDER BY" without DESC, hence TOP (1) returns min value of occurences.

B is closest

In A ORDER BY sorts Occurrencies in ascending order so we'll have a log message with the MIN number of occurrencies. In my opinion the closest one is B

The answer does not satisfy the requirement if there is more than one server with the same error message and the same number of occurrences.

There is A

for answer B, just one thing I want to know, why dont't use '>=' all(...), when using '>', how could 100>100?

A repeated question the closest answer should be B.