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Variant order 1


Variant order 1

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TomThomson
TomThomson
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Comments posted to this topic are about the item Variant order 1

Tom

bitbucket-25253
bitbucket-25253
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Rather simple question .....

If everything seems to be going well, you have obviously overlooked something.

Ron

Please help us, help you -before posting a question please read

Before posting a performance problem please read
Henrico Bekker
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even though x=y, and y=x, your case statement catered for the following result:
'v3 = v2'

and no where specifies v2=v3.

"None of the Above" should be correct as well.

----------------------------------------------
Msg 8134, Level 16, State 1, Line 1
Divide by zero error encountered.
wloong
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I agreed!

From the programming logic, "v2 = v3" could not be displayed! In fact, the right answer should be "None of the above". If the editor admit that it is a typo mistake, then both "v2 = v3" and "None of the above" should be correct in order to be fair.
Revenant
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Tom

Thanks for a REALLY interesting QotD. I wish I came up with this one.
Tom Brown
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Nice question Tom.
CaptainStu72
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Must say I agree with Henrico Bekker, the case statements at no point say v2 = v3, but rather v3 = v2. Whilst the mean the same thing, I agree that None of the Above should also be an applicable answer for those who are being pedantic.
Cadavre
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Got it wrong because I believe it should say "v3 = v2" not "v2 = v3"


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vinucool35
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Answer is v1 > v2   v1 > v3   v3 = v2   v4 > v1
and not "v1 > v2   v1 > v3   v2 = v3   v4 > v1"
rfr.ferrari
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good question and explanation!!!!

thanks Tom!!!


rfr.ferrari
DBA - SQL Server 2008
MCITP | MCTS

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