Asking Email Address Up David Portas and Erland Sommarskog

Dear Mr Portas and Mr Sommarskog,

Cause I am a newbie in sql, can I ask for your email address?

My email Address is mike@.ptmb.com

Thanx

Best Regards

Michael

--
Message posted via http://www.sqlmonster.comI would prefer to answer requests for help in the groups rather than by
email. Please do post your questions to the newsgroups so that everyone can
share the discussion and learn from it. I can't guarantee to respond
promptly to email but if you post here others may be able to give you a
quicker or better answer.

My disguised email address is in my message header. Just take out the
REMOVE_BEFORE_REPLYING_ prefix. Most of us don't post a plain email address
because spammers collect addresses from newgroup posts.

--
David Portas
SQL Server MVP
--|||Michael Teja via SQLMonster.com (forum@.SQLMonster.com) writes:
> Dear Mr Portas and Mr Sommarskog,
> Cause I am a newbie in sql, can I ask for your email address?

It happens that people mail me SQL questions, and I try to answer.
However, it's always better to ask question in public forums, rather
than mailing an individual. If you post to a newsgroup, there are more
people who see the question, and who may know the answer. If you mail me,
say, a DTS question, you are only wasting your time, because that's area
that I have no knowledge about. Furthermore, the person you mail, may be
away, so you will have to wait for the answer.

And one thing you definitely not should do is to mail more than one
person, because that means that several of us might spend our time
answering the same question.

Finally, another advantage with posting to a public forum, is that
other people who have the same problem might see the question and
answer and benefit from that.

--
Erland Sommarskog, SQL Server MVP, esquel@.sommarskog.se

Books Online for SQL Server SP3 at
http://www.microsoft.com/sql/techin.../2000/books.asp|||As great as David Portas and Erland Sommarskog are( and they have helped me
greatly).

I'd like to think that I can offer something as well.

The whole is greater than the sum of the parts.

"Michael Teja via SQLMonster.com" <forum@.SQLMonster.com> wrote in message
news:37a558fbb2c645dba637d06eb0419173@.SQLMonster.c om...
> Dear Mr Portas and Mr Sommarskog,
> Cause I am a newbie in sql, can I ask for your email address?
> My email Address is mike@.ptmb.com
> Thanx
> Best Regards
> Michael
> --
> Message posted via http://www.sqlmonster.comsql

Arithmetic overflow error converting numeric to data type nume

My results can never be > 9.99 as it is being divided by itelsef plus
something else.
I.e. possible is 1.00 to 0......
"Erland Sommarskog" wrote:

> dpc (dpc@.discussions.microsoft.com) writes:
> I guess the problem is that the count does not account for results that
> are > 9.99.
> But without access to the data it is very difficult to say exactly what
> happens.
> --
> Erland Sommarskog, SQL Server MVP, esquel@.sommarskog.se
> Books Online for SQL Server SP3 at
> http://www.microsoft.com/sql/techin.../2000/books.asp
>examnotes (dpc@.discussions.microsoft.com) writes:
> My results can never be > 9.99 as it is being divided by itelsef plus
> something else.
That "something else" can be negative. At least from my corner of
ignorance.
Anyway, the best you can do is this:
SELECT *
FROM #tblOutput
WHERE cast(cnSR / isnull(cn210,0) + isnull( nSR, 0) as decimal(13,2)) > 9.99
and (isnull(cn210,0) + isnull(cnSR,0) ) > 0
and cnsr is not null
and cnsr <> 0
and cn210 is not null ;
That will give you some clue of the data that is causing you problems.
Erland Sommarskog, SQL Server MVP, esquel@.sommarskog.se
Books Online for SQL Server SP3 at
http://www.microsoft.com/sql/techin.../2000/books.asp|||On Tue, 23 Aug 2005 23:10:04 -0700, dpc wrote:

>My results can never be > 9.99 as it is being divided by itelsef plus
>something else.
Hi dpc,
It is not. As I already mentioned in my previous reply, you missed some
parentheses. The formula
X = Y / Z + Y
will divide Y by Z, then add Y to that. It's the same as Y * (1 + 1/Z)
What you need is this:
X = Y / (Z + Y)
The extra parentheses ensure that Z + Y is calculated first, and that Y
is then divided by this value.
Best, Hugo
--
(Remove _NO_ and _SPAM_ to get my e-mail address)|||Oh, I see what your problem is, you are not using parenthesis correctly.
Update #tblOutput
SET pctProcSR = cast (cnSR / isnull(cn210,0) + isnull( cnSR, 0) as
decimal(3,2) )
Where ( isnull(cn210,0) + isnull(cnSR,0) ) > 0 and cnsr is not null and
cnsr <> 0 and cn210 is not null ;
Where pctProcSR is Decimal(3,2)
cast (cnSR / isnull(cn210,0) + isnull( cnSR, 0) as decimal(3,2) )
You are doing:
Division comes before addition in precedence, so you are doing the divsion
first. Change to:
cnSR / (isNull(cn210,0) + isnull(cnSR,0))
And you should have your problem fixed.
----
Louis Davidson - http://spaces.msn.com/members/drsql/
SQL Server MVP
"Arguments are to be avoided: they are always vulgar and often convincing."
(Oscar Wilde)
"dpc" <dpc@.discussions.microsoft.com> wrote in message
news:C5CC542D-3253-4772-B90B-2FE83173043D@.microsoft.com...
> My results can never be > 9.99 as it is being divided by itelsef plus
> something else.
> I.e. possible is 1.00 to 0......
> "Erland Sommarskog" wrote:
>|||Sorry Hugo, didn't see you had already said that. Messy messy thread :)
----
Louis Davidson - http://spaces.msn.com/members/drsql/
SQL Server MVP
"Arguments are to be avoided: they are always vulgar and often convincing."
(Oscar Wilde)
"Hugo Kornelis" <hugo@.pe_NO_rFact.in_SPAM_fo> wrote in message
news:jpbpg11e74g15mm3gjmmli02mfni6ha@.
4ax.com...
> On Tue, 23 Aug 2005 23:10:04 -0700, dpc wrote:
>
> Hi dpc,
> It is not. As I already mentioned in my previous reply, you missed some
> parentheses. The formula
> X = Y / Z + Y
> will divide Y by Z, then add Y to that. It's the same as Y * (1 + 1/Z)
> What you need is this:
> X = Y / (Z + Y)
> The extra parentheses ensure that Z + Y is calculated first, and that Y
> is then divided by this value.
> Best, Hugo
> --
> (Remove _NO_ and _SPAM_ to get my e-mail address)