FK Position

Does it make any difference where we define the FKs in a table? I mean, do I speed up the query if I define it as the second field or the last one? What about the other fields, the ones that are not FKs, but are used as filters in a query?

Raul:

In general, none of this makes much practical difference to speed of execution. The two things that do matter are (1) do you have indexes on the foreign table in place that correspond to your foreign key and (2) does use of that index with a specific query also require a bookmark lookup for that specific query. If the foreign key has a correspondence to the clustered index of the other table -- and that is often the case -- then no bookmark lookup is necessary. If the foreign key has correspondence to a non-clustered index of the other table but there are fields in the other table that are referenced and are not part of the non-clustered index then a bookmark lookup will be necessary.

If a query references only a short list of records then the non-clustered index will often get used to optimize the query. However, if the query references a very large number of records in the foreign table the optimizer may "decide" that the cost of performing the random reads necessary to support bookmark lookups is too high. In these circumstances the optimizer will often opt to perform a table scan of the foreign table instead of of an index seek.

Bleah. Somebody please say this in a better way.

|||

Column position is virtually meaningless in defining PK's.

Column position is virtually menaingless except:

1. In UNIQUE and PRIMARY key constraints, and all indexes, order of columns has meaning.

2. In relatively rare cases, order of column conditions in a WHERE clause is involved (and only when the criteria is so large as to make it impossible for SQL Server to check all possible uses in a timely manner)

The position of a column in a table has no little if any bearing on performance, as it is just a representation of what is physically implemented in bits and bytes down in the physical table. If it were advantageous to reorganize the data on the page, the data could be reorganized by the storage engine without you knowing. So rest easy, it should make no difference at all.

Fix the position of graphs while scrolling

I have a report with a graph on top and a table at the bottom.
Sometimes the table becomes large enough to go to the second page.
The problem is - the graph doesn't show up on the second page.
Is there a way to fix the position of the graphs - so that it stays in
one place as we scroll down...
Without - spoiling the export to pdf... ?You could make the graph repeat on the second page or you could increase the
page length to avoid the break. We are considering adding the ability to
"float" headers in the SQL 2005 release.
--
Brian Welcker
Group Program Manager
Microsoft SQL Server
This posting is provided "AS IS" with no warranties, and confers no rights.
"Harry" <harshwardhan@.mailcity.com> wrote in message
news:e48fb32a.0411081427.2da88d72@.posting.google.com...
>I have a report with a graph on top and a table at the bottom.
> Sometimes the table becomes large enough to go to the second page.
> The problem is - the graph doesn't show up on the second page.
> Is there a way to fix the position of the graphs - so that it stays in
> one place as we scroll down...
> Without - spoiling the export to pdf... ?

First, Last, Middle ??

I have a Matrix that always displays two rows of data.
One row show values from Jan 1 of the current year.
The second row shows values for today.
In the foot I show the % the values have changed so YTD with the following
formula:
=(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
First(Fields!Core.Value) * 100
This works great and looks like this:
Date | Core
--
Jan 1 2005 | $400
Aug 31 2005 | $500
--
Footer +25%
I have been asked to add in the value from a year ago today but still
display the change in value from just Jan 1.
Date | Core
--
Aug 31 2004 | $300
Jan 1 2005 | $400
Aug 31 2005 | $500
--
Footer +25% (Diff between Jan 1 and Aug 31 2005)
How would I calc the % change in the footer. My formula will not work as the
"First" value is a year ago today not Jan 1. Is there a "Middle" function ?
:) :)
Thoughts ?
Thanks in Advance
Pete MitchellTry something like this:
=(Last(Fields!Core.Value) - CDate("1/1/"&CStr(Year(First(Fields!Core.Value)))))
/
CDate("1/1/"&CStr(Year(First(Fields!Core.Value)))) * 100
GeoSynch
"PeteMitchell" <PeteMitchell@.discussions.microsoft.com> wrote in message
news:EE7A94F6-8F3B-46EC-87E3-7DA323C927D0@.microsoft.com...
>I have a Matrix that always displays two rows of data.
> One row show values from Jan 1 of the current year.
> The second row shows values for today.
> In the foot I show the % the values have changed so YTD with the following
> formula:
> =(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
> First(Fields!Core.Value) * 100
> This works great and looks like this:
> Date | Core
> --
> Jan 1 2005 | $400
> Aug 31 2005 | $500
> --
> Footer +25%
> I have been asked to add in the value from a year ago today but still
> display the change in value from just Jan 1.
> Date | Core
> --
> Aug 31 2004 | $300
> Jan 1 2005 | $400
> Aug 31 2005 | $500
> --
> Footer +25% (Diff between Jan 1 and Aug 31 2005)
> How would I calc the % change in the footer. My formula will not work as the
> "First" value is a year ago today not Jan 1. Is there a "Middle" function ?
> :) :)
> Thoughts ?
> Thanks in Advance
> Pete Mitchell|||Actually, it probably shoud be:
=(Last(Fields!Core.Value) - CDate("1/1/"&CStr(Year(Last(Fields!Core.Value)))))
/ CDate("1/1/"&CStr(Year(Last(Fields!Core.Value)))) * 100
GeoSynch
"GeoSynch" <SpamSlayed@.Casablanca.com> wrote in message
news:ecLK3yprFHA.3884@.TK2MSFTNGP11.phx.gbl...
> Try something like this:
> =(Last(Fields!Core.Value) -
> CDate("1/1/"&CStr(Year(First(Fields!Core.Value))))) /
> CDate("1/1/"&CStr(Year(First(Fields!Core.Value)))) * 100
>
> GeoSynch
>
> "PeteMitchell" <PeteMitchell@.discussions.microsoft.com> wrote in message
> news:EE7A94F6-8F3B-46EC-87E3-7DA323C927D0@.microsoft.com...
>>I have a Matrix that always displays two rows of data.
>> One row show values from Jan 1 of the current year.
>> The second row shows values for today.
>> In the foot I show the % the values have changed so YTD with the following
>> formula:
>> =(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
>> First(Fields!Core.Value) * 100
>> This works great and looks like this:
>> Date | Core
>> --
>> Jan 1 2005 | $400
>> Aug 31 2005 | $500
>> --
>> Footer +25%
>> I have been asked to add in the value from a year ago today but still
>> display the change in value from just Jan 1.
>> Date | Core
>> --
>> Aug 31 2004 | $300
>> Jan 1 2005 | $400
>> Aug 31 2005 | $500
>> --
>> Footer +25% (Diff between Jan 1 and Aug 31 2005)
>> How would I calc the % change in the footer. My formula will not work as the
>> "First" value is a year ago today not Jan 1. Is there a "Middle" function ?
>> :) :)
>> Thoughts ?
>> Thanks in Advance
>> Pete Mitchell
>|||Thanks a bunch.
How does that work ?
There are two fields in play here : Date and Core
How is that get the Core.value when the Date.value = Jan 1 2005 ?
Pete
"GeoSynch" wrote:
> Actually, it probably shoud be:
> =(Last(Fields!Core.Value) - CDate("1/1/"&CStr(Year(Last(Fields!Core.Value)))))
> / CDate("1/1/"&CStr(Year(Last(Fields!Core.Value)))) * 100
>
> GeoSynch
>
> "GeoSynch" <SpamSlayed@.Casablanca.com> wrote in message
> news:ecLK3yprFHA.3884@.TK2MSFTNGP11.phx.gbl...
> > Try something like this:
> > =(Last(Fields!Core.Value) -
> > CDate("1/1/"&CStr(Year(First(Fields!Core.Value))))) /
> > CDate("1/1/"&CStr(Year(First(Fields!Core.Value)))) * 100
> >
> >
> > GeoSynch
> >
> >
> > "PeteMitchell" <PeteMitchell@.discussions.microsoft.com> wrote in message
> > news:EE7A94F6-8F3B-46EC-87E3-7DA323C927D0@.microsoft.com...
> >>I have a Matrix that always displays two rows of data.
> >> One row show values from Jan 1 of the current year.
> >> The second row shows values for today.
> >>
> >> In the foot I show the % the values have changed so YTD with the following
> >> formula:
> >>
> >> =(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
> >> First(Fields!Core.Value) * 100
> >>
> >> This works great and looks like this:
> >> Date | Core
> >> --
> >> Jan 1 2005 | $400
> >> Aug 31 2005 | $500
> >> --
> >> Footer +25%
> >>
> >> I have been asked to add in the value from a year ago today but still
> >> display the change in value from just Jan 1.
> >>
> >> Date | Core
> >> --
> >> Aug 31 2004 | $300
> >> Jan 1 2005 | $400
> >> Aug 31 2005 | $500
> >> --
> >> Footer +25% (Diff between Jan 1 and Aug 31 2005)
> >>
> >> How would I calc the % change in the footer. My formula will not work as the
> >> "First" value is a year ago today not Jan 1. Is there a "Middle" function ?
> >> :) :)
> >>
> >> Thoughts ?
> >>
> >> Thanks in Advance
> >>
> >> Pete Mitchell
> >
> >
>
>|||CDate("1/1/"&CStr(Year(Last(Fields!Core.Value)))) evaluates thusly:
Last(Fields!Core.Value) = '08/31/2005' data type Date
Year(Last(Fields!Core.Value)) = '2005' data type Integer
CStr(Year(Last(Fields!Core.Value))) converts it to a string value
so that when concatenated with "1/1/" it will evaluate to string value
"1/1/2005"
CDate converts it back to an actual date value of '01/01/2005'
GeoSynch
"PeteMitchell" <PeteMitchell@.discussions.microsoft.com> wrote in message
news:940A01C6-D8A5-4C44-8BA4-3AF232AD790F@.microsoft.com...
> Thanks a bunch.
> How does that work ?
> There are two fields in play here : Date and Core
> How is that get the Core.value when the Date.value = Jan 1 2005 ?
> Pete
>
> "GeoSynch" wrote:
>> Actually, it probably shoud be:
>> =(Last(Fields!Core.Value) -
>> CDate("1/1/"&CStr(Year(Last(Fields!Core.Value)))))
>> / CDate("1/1/"&CStr(Year(Last(Fields!Core.Value)))) * 100
>>
>> GeoSynch
>>
>> "GeoSynch" <SpamSlayed@.Casablanca.com> wrote in message
>> news:ecLK3yprFHA.3884@.TK2MSFTNGP11.phx.gbl...
>> > Try something like this:
>> > =(Last(Fields!Core.Value) -
>> > CDate("1/1/"&CStr(Year(First(Fields!Core.Value))))) /
>> > CDate("1/1/"&CStr(Year(First(Fields!Core.Value)))) * 100
>> >
>> >
>> > GeoSynch
>> >
>> >
>> > "PeteMitchell" <PeteMitchell@.discussions.microsoft.com> wrote in message
>> > news:EE7A94F6-8F3B-46EC-87E3-7DA323C927D0@.microsoft.com...
>> >>I have a Matrix that always displays two rows of data.
>> >> One row show values from Jan 1 of the current year.
>> >> The second row shows values for today.
>> >>
>> >> In the foot I show the % the values have changed so YTD with the following
>> >> formula:
>> >>
>> >> =(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
>> >> First(Fields!Core.Value) * 100
>> >>
>> >> This works great and looks like this:
>> >> Date | Core
>> >> --
>> >> Jan 1 2005 | $400
>> >> Aug 31 2005 | $500
>> >> --
>> >> Footer +25%
>> >>
>> >> I have been asked to add in the value from a year ago today but still
>> >> display the change in value from just Jan 1.
>> >>
>> >> Date | Core
>> >> --
>> >> Aug 31 2004 | $300
>> >> Jan 1 2005 | $400
>> >> Aug 31 2005 | $500
>> >> --
>> >> Footer +25% (Diff between Jan 1 and Aug 31 2005)
>> >>
>> >> How would I calc the % change in the footer. My formula will not work as
>> >> the
>> >> "First" value is a year ago today not Jan 1. Is there a "Middle" function
>> >> ?
>> >> :) :)
>> >>
>> >> Thoughts ?
>> >>
>> >> Thanks in Advance
>> >>
>> >> Pete Mitchell
>> >
>> >
>>

First, Last, Again

I have a Matrix that always displays two rows of data.
One row show values from Jan 1 of the current year.
The second row shows values for today.
In the foot I show the % the values have changed so YTD with the following
formula:
=(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
First(Fields!Core.Value) * 100
This works great and looks like this:
Date | Core
--
Jan 1 2005 | $400
Aug 31 2005 | $500
--
Footer +25%
I have been asked to add in the value from a year ago today but still
display the change in value from just Jan 1.
Date | Core
--
Aug 31 2004 | $300
Jan 1 2005 | $400
Aug 31 2005 | $500
--
Footer +25% (Diff between Jan 1 and Aug 31 2005)
How would I calc the % change in the footer. My formula will not work as the
"First" value is a year ago today not Jan 1. Is there a "Middle" function ?
:) :)
Thoughts ?
Thanks in Advance
Pete Mitchellmaybe you could use groups to accomplish the division of the first row from
the other two. Then you could just hide the group header and footer of
group1 and hide the header of group 2. I am not sure how the first and last
will work with groups ... just a thought
> Date | Core
> --
> Aug 31 2004 | $300 ======== > group 1
> Jan 1 2005 | $400 ========> group 2
> Aug 31 2005 | $500 ========> group 2
> --
> Footer +25% (Diff between Jan 1 and Aug 31 2005)
"PeteMitchell" wrote:
> I have a Matrix that always displays two rows of data.
> One row show values from Jan 1 of the current year.
> The second row shows values for today.
> In the foot I show the % the values have changed so YTD with the following
> formula:
> =(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
> First(Fields!Core.Value) * 100
> This works great and looks like this:
> Date | Core
> --
> Jan 1 2005 | $400
> Aug 31 2005 | $500
> --
> Footer +25%
> I have been asked to add in the value from a year ago today but still
> display the change in value from just Jan 1.
> Date | Core
> --
> Aug 31 2004 | $300
> Jan 1 2005 | $400
> Aug 31 2005 | $500
> --
> Footer +25% (Diff between Jan 1 and Aug 31 2005)
> How would I calc the % change in the footer. My formula will not work as the
> "First" value is a year ago today not Jan 1. Is there a "Middle" function ?
> :) :)
> Thoughts ?
> Thanks in Advance
> Pete Mitchell
>|||Thanks, good suggestion.
"MJ Taft" wrote:
> maybe you could use groups to accomplish the division of the first row from
> the other two. Then you could just hide the group header and footer of
> group1 and hide the header of group 2. I am not sure how the first and last
> will work with groups ... just a thought
> > Date | Core
> > --
> > Aug 31 2004 | $300 ======== > group 1
> > Jan 1 2005 | $400 ========> group 2
> > Aug 31 2005 | $500 ========> group 2
> > --
> > Footer +25% (Diff between Jan 1 and Aug 31 2005)
>
> "PeteMitchell" wrote:
> > I have a Matrix that always displays two rows of data.
> > One row show values from Jan 1 of the current year.
> > The second row shows values for today.
> >
> > In the foot I show the % the values have changed so YTD with the following
> > formula:
> >
> > =(Last(Fields!Core.Value) - First(Fields!Core.Value)) /
> > First(Fields!Core.Value) * 100
> >
> > This works great and looks like this:
> > Date | Core
> > --
> > Jan 1 2005 | $400
> > Aug 31 2005 | $500
> > --
> > Footer +25%
> >
> > I have been asked to add in the value from a year ago today but still
> > display the change in value from just Jan 1.
> >
> > Date | Core
> > --
> > Aug 31 2004 | $300
> > Jan 1 2005 | $400
> > Aug 31 2005 | $500
> > --
> > Footer +25% (Diff between Jan 1 and Aug 31 2005)
> >
> > How would I calc the % change in the footer. My formula will not work as the
> > "First" value is a year ago today not Jan 1. Is there a "Middle" function ?
> > :) :)
> >
> > Thoughts ?
> >
> > Thanks in Advance
> >
> > Pete Mitchell
> >

First(, Last(,

There's no Second(?
Or Third(Nope.
We've considered adding Nth(), but it's rather expensive so it's fairly far
down on our future features wishlist.
--
This post is provided 'AS IS' with no warranties, and confers no rights. All
rights reserved. Some assembly required. Batteries not included. Your
mileage may vary. Objects in mirror may be closer than they appear. No user
serviceable parts inside. Opening cover voids warranty. Keep out of reach of
children under 3.
"Cindy Lee" <cindylee@.hotmail.com> wrote in message
news:eLDWIBBfEHA.3964@.TK2MSFTNGP12.phx.gbl...
> There's no Second(?
> Or Third(
>
>
>|||Actually I don't need it now.
I just used SUM(iff(
to get the exact value I wanted.
"Chris Hays [MSFT]" <chays@.online.microsoft.com> wrote in message
news:e976FNBfEHA.3412@.TK2MSFTNGP11.phx.gbl...
> Nope.
> We've considered adding Nth(), but it's rather expensive so it's fairly
far
> down on our future features wishlist.
> --
> This post is provided 'AS IS' with no warranties, and confers no rights.
All
> rights reserved. Some assembly required. Batteries not included. Your
> mileage may vary. Objects in mirror may be closer than they appear. No
user
> serviceable parts inside. Opening cover voids warranty. Keep out of reach
of
> children under 3.
> "Cindy Lee" <cindylee@.hotmail.com> wrote in message
> news:eLDWIBBfEHA.3964@.TK2MSFTNGP12.phx.gbl...
> > There's no Second(?
> > Or Third(
> >
> >
> >
> >
> >
> >
>

first form normalizations and second form normalization

Hi everyone,
What is the main difference between first form normalizations and second form normalization ?
In my opinion, they are both generated for the same operation which is to prevent redundancy(in other words; duplication of data in several records).
So would you please explain it to me ?

Thanks

Taken from: http://databases.about.com/od/specificproducts/a/normalization.htm

First normal form (1NF) sets the very basic rules for an organized database:

Eliminate duplicative columns from the same table.

Create separate tables for each group of related data and identify each row with a unique column or set of columns (the primary key).