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Foreign key


Foreign key

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Igor Micev
Igor Micev
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Comments posted to this topic are about the item Foreign key


Igor Micev,
SQL Server developer at Seavus
www.seavus.com
Britt Cluff
Britt Cluff
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Good question. Thanks for submitting.

http://brittcluff.blogspot.com/
bitbucket-25253
bitbucket-25253
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Well my answer was declared to be incorrect, so went to read the explanation, following the link given ..... nada / nothing / not a word to support what is deemed the correct answer.

So went and entered the code in SQL 2008R2, triple checked what I had entered and it matched what was given in the question. When pressing the F5 key for SSMS (2008R2) .. and again using SSMS checked after the error message and what do you know ... what was shown did NOT match what was give as the correct answer.

Hmmmmmmmmmmm

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

Ron

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baabhu
baabhu
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Good Question.
Hardy21
Hardy21
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Good One.
If Table2 declared as following than all 3 tables created without any issue.
create table Table2
( id_tb2 int primary key references table1(id_tb1),
value2 varchar(100))



Thanks
Andre Guerreiro
Andre Guerreiro
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I wasn't paying enough attention to detail here.
Table2's int column was just an int and nothing else. Hehe
Time to go to bed for me.
Thank you for the question.

I still see documentation around that never mentions unique constraints only PKs that can be used in a FK reference, even though we know that both can be used.

Best regards,

Andre Guerreiro Neto

Database Analyst
http://www.softplan.com.br
MCITPx1/MCTSx2/MCSE/MCSA
kapfundestanley
kapfundestanley
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Nice and trick question,thank you.

“When I hear somebody sigh, ‘Life is hard,’ I am always tempted to ask, ‘Compared to what?’” - Sydney Harris
gemisigo
gemisigo
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Yeah, tricky question that bled me to death :-) I knew that creation of the third table would have some issues but I still voted for three being created and dropping some warning. Obviously, creating the table and the constraint at the same time does not equal to creating the table and adding the constraint after. Though the error message is slightly misleading here. It warns about being unable to create constraint but does not mention anything for the table itself. The devil is in the details.

Good question ;-)
Deepty Singh
Deepty Singh
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i actually tried this on my database..
My second table was not created..
still it is showing incorrect answer Sad
Raunak Jhawar
Raunak Jhawar
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Good one.

Regards/Raunak
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