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Index defaults 2


Index defaults 2

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Hugo Kornelis
Hugo Kornelis
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Comments posted to this topic are about the item Index defaults 2


Hugo Kornelis, SQL Server MVP
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Toby Harman
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You have two answers that are potentially correct here.

None, and None unless the rest of the command specifies one.

Shame I picked the wrong one!

Edit: I stand corrected. The CREATE INDEX has to be a separate command. Missed that.
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Nice question .. learned something

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

Ron

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Thanks, Hugo -- a nice one to end my day with!
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I can create an index by means of a unique constraint:

CREATE TABLE dbo.QotD
(KeyColumn int NOT NULL PRIMARY KEY,
RefColumn int NOT NULL,
-- More column definitions
CONSTRAINT FK_RefTab FOREIGN KEY (RefColumn)
REFERENCES dbo.RefTab(RefTabKey),
-- More constraints
CONSTRAINT UQ_RefColumn UNIQUE(RefColumn)
);


Doesn't it count as answer #2? :-)
Koen Verbeeck
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vk-kirov (6/26/2012)
I can create an index by means of a unique constraint:

CREATE TABLE dbo.QotD
(KeyColumn int NOT NULL PRIMARY KEY,
RefColumn int NOT NULL,
-- More column definitions
CONSTRAINT FK_RefTab FOREIGN KEY (RefColumn)
REFERENCES dbo.RefTab(RefTabKey),
-- More constraints
CONSTRAINT UQ_RefColumn UNIQUE(RefColumn)
);


Doesn't it count as answer #2? :-)


It's possible, but I doubt a 1-1 foreign key relationship is very useful :-D

I also took answer number 2. Forgot that you cannot specify an index directly in a create table statement. D'oh!


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I also took answer number 2. Forgot that you cannot specify an index directly in a create table statement. D'oh!


Thanks - I was wondering why my answer was incorrect too. I thought it strange that Hugo might have made a mistake!

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Hugo - nice question.
Surprised only 15% have this correct so far...
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vk-kirov (6/26/2012)
I can create an index by means of a unique constraint:

CREATE TABLE dbo.QotD
(KeyColumn int NOT NULL PRIMARY KEY,
RefColumn int NOT NULL,
-- More column definitions
CONSTRAINT FK_RefTab FOREIGN KEY (RefColumn)
REFERENCES dbo.RefTab(RefTabKey),
-- More constraints
CONSTRAINT UQ_RefColumn UNIQUE(RefColumn)
);


Doesn't it count as answer #2? :-)


That's exactly what I thought - but then decided that given this was a question from Hugo Kornelis it wouldn't be that sneaky, so went for option 1.

Thanks for the question. ;-)
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vk-kirov (6/26/2012)
I can create an index by means of a unique constraint


That's exactly the basis on which I answered "None unless the rest of the statement creates one", which I therefore believe should be the correct answer. Particularly given that the previous "index defaults" question was about the possibility of a Unique constraint appearing later in the statement!
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