A merry christmas to everyone!

Hmmm, I have never used the Import/Export columns before. I really should look into...

paul.goldstraw (12/22/2010)

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I got it right based on the fact that an update statement shouldn't ever be able to change the number of records in a table - since 8 rows...

Toreador (12/22/2010)

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I get a pile of errors when I run it - Cannot insert the value NULL into column 'Seq'.

I eventually worked out that this is because I have ANSI_NULL_DFLT_ON...

dfine (12/22/2010)

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Yes, the answer is "Ann". But when you say most of the time we would get “Ann” since the clustered index is created on “first name”.

Why can’t we...

Allright, I'm installing Denali now.

I created a whole new virtual machine with Windows Server 2008 R2 x64 and there it works perfectly.

Still don't know what the original problem was though.

You can use a for each loop to loop over the table. In the for each loop, place a data flow task that will write the corresponding row to the...

Do you receive any errors or warnings?

Is the flat file destination green at the end of the package?

Are you looking at the right file?

Do you use any variables or expressions?

The question itself was OK, as a test to see what output the T-SQL would give.

But not OK for the purpose mentioned in the explanation, as there are other statements...

sqlapprentice (12/21/2010)

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I agree that this is a region setting issue, because I've tested if I use my own account to run the sql server agent and execute this the SSIS...

I'm confused: the correct answer is the row that is returned "most of the time"?

If yes, then it isn't a good question, as other answers might be possible. If you...

Maybe the regional settings on the server are different, so that a . is interpreted as a 1000-separator instead of a decimal separator.

May I ask...

ian.grace (12/20/2010)

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I don't understand how the first transaction could succeed given the attempt to insert the null value into the not-null field "a". Also, why is the last insert...

philip.cullingworth (12/20/2010)

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I have to disagree with the answer. I ran the code on SQL server 2005 and got the following output:

abcd

4join14join2

4join14join2

2join32one

2join32one

1oneNULLNULL

NULLNULLNULLtwo

NULLNULLNULLthree

As I had worked out the same output before...