January 18, 2018 at 10:55 pm
Comments posted to this topic are about the item Basic COUNTs 1
January 18, 2018 at 11:32 pm
Nice simple one to end the week on, thanks Steve.
...
January 19, 2018 at 12:24 am
nice simple. added one condition in my mind while using count() if there is null value will not be calculated...😎😎
Manik
You cannot get to the top by sitting on your bottom.
January 19, 2018 at 1:52 am
'For the DISTINCT keyword, null values are considered to be duplicates of each other. When DISTINCT is included in a SELECT statement, only one NULL is returned in the results, regardless of how many null values are encountered.'
So I guessed 5.
January 19, 2018 at 2:10 am
Nice, easy one to end the week on, thanks Steve
____________________________________________
Space, the final frontier? not any more...
All limits henceforth are self-imposed.
“libera tute vulgaris ex”
January 19, 2018 at 5:16 am
paul s-306273 - Friday, January 19, 2018 1:52 AM'For the DISTINCT keyword, null values are considered to be duplicates of each other. When DISTINCT is included in a SELECT statement, only one NULL is returned in the results, regardless of how many null values are encountered.'So I guessed 5.
That's when selecting the values. If you run this code
SELECT DISTINCT charcol
FROM dbo.TheCounts AS tc
you will get 5 lines.
But NULLS are not included in COUNT (nor in other aggregates, like MIN and MAX)
Running the query in the question, you even get a message telling you:
Warning: Null value is eliminated by an aggregate or other SET operation.
January 19, 2018 at 9:20 am
A question I could actually answer! However, after all of the other more difficult questions, I doubted myself. I admit I had to run the code to make sure I wasn't missing anything! Nice discussion about Nulls and how they act with Distinct. I also did not look at the messages tab to see the warning, so that is a good reminder to check that tab!
Thanks!
January 19, 2018 at 5:52 pm
A coffee break question - thank you,. Steve!
Viewing 8 posts - 1 through 8 (of 8 total)
You must be logged in to reply to this topic. Login to reply