Hi

Hi,

Somebody can speak (tips) about exam 70-431?

Tips?

Amount Questions?

Time ?

....

If you're looking for brain dumps this is not the site to get those answers. The 70-431 exam is a 200-level "how-to" exam on all the nuts and bolts of SQL Server 2005. There are free eLearning classes available on the Microsoft web site which will help you with the content information for the test, as well as a number of excellent books hitting the market about now.

I believe you'll get about 40 questions on the test, though it may be higher than that, and you'll get about 2 1/2 hours to answer them.

Go to www.microsoft.com/sql and click on Learning and you'll find out all you need to know to prepare for the test.

Good luck.

|||

What is the passing mark/percentage for MCTS?

|||

Hi Gunny,

is 700 scored.

|||

E X A M S P E C S
Exam Number:	70-431
Active / Retired:	Active
Prerequisites:
Exam Format:	Multiple-Choice; Multiple Answer
Num Questions:	55
Time Limit:	90
Cost (USD) $:	$125
Passing Score:	700

taken from cramsession's site: http://www.cramsession.com/certifications/exams/sql-server-2005.asp?exam_id=770

also use this code for 20% of the exam price

Call now to schedule your Microsoft exam call 1-800-247-8731 and use code MSUU5C7E0100 for your 20% discount.

|||

Many thanks WPH and rkohler.

That's 700 out of how many? What is the max possible score?

|||1000 max scored.|||

Thanks WPH.

|||

any one can respon

i have two server one is dpmain controller and second is exchange 2003 server i need to setup sql 2005

which one i choice ?

|||

a.m.m.e.

Your question really belongs in a separate/new thread. This thread is about MCTS.

But to answer your question, I would advise against putting it on the domain controller. The domain controller contains the Active Directory and security is extremely important. Adding any other role such as SQL Server would complicate security configuration for it.

Actually, if possible, I would set up a new dedicated SQL Server - not add it to the Exchange Server 2003. Mixing multiple roles on one server always complicates security. But if you cannot set up a new dedicated server I would add it to the Exchange Server 2003.

Other more experienced people may have different ideas.

|||

Hello everyone,

I would like to let you all know and be aware of the fact that microsoft has changed its exam pattern for

070-431 be very carefull as microsoft has changed the exam pattern and due to which i failed the exam today.

Its now based on simulations questions which comes after the multiple choice questions i got 15 simulations

separately apart from 35 multi choice questions along with exibits. I wasnt aware of this fact and it wasnt

mentioned ánywhere about simulations questions I think no one knows about it I am the first to face this type of

exam.. The exams comes up with 2 sections the first sections was questions that was 35 in 1hr and the next

was simulations that was 15 2hr 15mins so in total it was 195min exam and I got all correct in first section

and the second section was gred out 60% I got 560 out of 700 may not be difficult but very new.

If in case you have come across with any good site with simulations of this exam please forward me to

a_afroze@.yahoo.com

All the best to all of you

Abdul Afroze

|||

Hello everyone,

I would like to let you all know and be aware of the fact that microsoft has changed its exam pattern for

070-431 be very carefull as microsoft has changed the exam pattern and due to which i failed the exam today.

Its now based on simulations questions which comes after the multiple choice questions i got 15 simulations

separately apart from 35 multi choice questions along with exibits. I wasnt aware of this fact and it wasnt

mentioned ánywhere about simulations questions I think no one knows about it I am the first to face this type of

exam.. The exams comes up with 2 sections the first sections was questions that was 35 in 1hr and the next

was simulations that was 15 2hr 15mins so in total it was 195min exam and I got all correct in first section

and the second section was gred out 60% I got 560 out of 700 may not be difficult but very new.

If in case you have come across with any good site with simulations of this exam please forward me to

a_afroze@.yahoo.com

All the best to all of you

Abdul Afroze

|||

Thanks for the information. I'm going to be taking the test soon and was afraid they would do something like this. They did this to me on the 70-228... completely changed the content and format from what my rather expensive test prep materials were based on.

Could you give us a clue as to what the simulations focused on? (ie. TSQL, SSMS, etc.) I can do just about everything you need to do in SSMS and other GUI tools, but I don't have all the full syntaxes for the TSQL methods memorized (ie. create HTTP endpoint for SOAP vs. TCP endpoint for mirroring).

|||

Well, this really isn't a database documentation request, and using a questions-only site really isn't the best way to go. There are reputable firms, such as Transcender, that will help you with testing and so on. The point is that you should try and learn the trade, rather than just memorizing questions. I've posted an article that deals with certifications:

http://www.informit.com/guides/content.asp?g=sqlserver&seqNum=131

And another here dealing with being a DBA. I think you may find those more useful than looking up test questions on a site.

http://www.informit.com/guides/content.asp?g=sqlserver&seqNum=247

These are just my thoughts - I do wish you luck in your studies!

|||

Many thanks, Abdul, for that "heads up".

If you or anyone else here come to know of any test exams based on this new pattern I am sure we all would be eager to know about them as soon as possible.

Regards

Here's the Problem Again With Sample Data

I want the sum of the last payments (amount) for all customers. The last
payment is with one with most recent date. And if there are more than one
payment on the most recent date then the one with the higher paymentid is
the last payment. for example in the given data the insert statement that
starts with capital I is the last payment of that customer. The correct
answer should be 2100 as given below. both queries by Erland and Anith give
the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause from
both queries since right now I want current sum (not till some date). So
what should be the right query.
Thanks again for the help.
create table payments (
paymentid int,
customerid int,
amount int,
date datetime
)
insert payments values (1, 1, 100, '1/1/03')
insert payments values (2, 1, 200, '2/28/03')
Insert payments values (3, 1, 500, '5/15/03')
insert payments values (4, 2, 400, '1/16/03')
insert payments values (9, 2, 800, '4/30/03')
insert payments values (5, 2, 200, '6/15/03')
Insert payments values (6, 2, 900, '6/15/03')
insert payments values (7, 3, 700, '3/1/03')
insert payments values (10,3, 300, '7/10/03')
Insert payments values (8, 3, 600, '9/1/03')
insert payments values (11,4, 300, '8/1/03')
insert payments values (12,4, 400, '9/10/03')
Insert payments values (13,4, 100, '9/10/03')
customerid lastpayment amount
1 3 (on 5/15/03) 500
2 6 (on 6/15/03) 900
3 8 (on 9/1/03) 600
4 13 (on 9/10/03) 100
======== Result => 2100select customerid,paymentid,[date],amount
from payments p1
where paymentid=(select top 1 paymentid from payments p2 where
p2.customerid=p1.customerid order by [date] desc, paymentid desc)
-oj
RAC v2.2 & QALite!
http://www.rac4sql.net
"MAB" <dsfoalsdfsdfadouisdf@.yahoo.com> wrote in message
news:bjmcjb$klvrf$1@.ID-31123.news.uni-berlin.de...
> I want the sum of the last payments (amount) for all customers. The last
> payment is with one with most recent date. And if there are more than one
> payment on the most recent date then the one with the higher paymentid is
> the last payment. for example in the given data the insert statement that
> starts with capital I is the last payment of that customer. The correct
> answer should be 2100 as given below. both queries by Erland and Anith
give
> the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause from
> both queries since right now I want current sum (not till some date). So
> what should be the right query.
> Thanks again for the help.
> create table payments (
> paymentid int,
> customerid int,
> amount int,
> date datetime
> )
> insert payments values (1, 1, 100, '1/1/03')
> insert payments values (2, 1, 200, '2/28/03')
> Insert payments values (3, 1, 500, '5/15/03')
> insert payments values (4, 2, 400, '1/16/03')
> insert payments values (9, 2, 800, '4/30/03')
> insert payments values (5, 2, 200, '6/15/03')
> Insert payments values (6, 2, 900, '6/15/03')
> insert payments values (7, 3, 700, '3/1/03')
> insert payments values (10,3, 300, '7/10/03')
> Insert payments values (8, 3, 600, '9/1/03')
> insert payments values (11,4, 300, '8/1/03')
> insert payments values (12,4, 400, '9/10/03')
> Insert payments values (13,4, 100, '9/10/03')
>
> customerid lastpayment amount
> 1 3 (on 5/15/03) 500
> 2 6 (on 6/15/03) 900
> 3 8 (on 9/1/03) 600
> 4 13 (on 9/10/03) 100
> ========> Result => 2100
>
>|||Thanks! this works. However its too slow to run on the actual table with
thousands of rows but i've managed to eliminate most of the rows by creating
a temporary table and then I run this query on the temporary table.
"oj" <nospam_ojngo@.home.com> wrote in message
news:uqjta41dDHA.2320@.TK2MSFTNGP12.phx.gbl...
> select customerid,paymentid,[date],amount
> from payments p1
> where paymentid=(select top 1 paymentid from payments p2 where
> p2.customerid=p1.customerid order by [date] desc, paymentid desc)
>
> --
> -oj
> RAC v2.2 & QALite!
> http://www.rac4sql.net
>
> "MAB" <dsfoalsdfsdfadouisdf@.yahoo.com> wrote in message
> news:bjmcjb$klvrf$1@.ID-31123.news.uni-berlin.de...
> > I want the sum of the last payments (amount) for all customers. The last
> > payment is with one with most recent date. And if there are more than
one
> > payment on the most recent date then the one with the higher paymentid
is
> > the last payment. for example in the given data the insert statement
that
> > starts with capital I is the last payment of that customer. The correct
> > answer should be 2100 as given below. both queries by Erland and Anith
> give
> > the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause
from
> > both queries since right now I want current sum (not till some date). So
> > what should be the right query.
> >
> > Thanks again for the help.
> >
> > create table payments (
> > paymentid int,
> > customerid int,
> > amount int,
> > date datetime
> > )
> >
> > insert payments values (1, 1, 100, '1/1/03')
> > insert payments values (2, 1, 200, '2/28/03')
> > Insert payments values (3, 1, 500, '5/15/03')
> >
> > insert payments values (4, 2, 400, '1/16/03')
> > insert payments values (9, 2, 800, '4/30/03')
> > insert payments values (5, 2, 200, '6/15/03')
> > Insert payments values (6, 2, 900, '6/15/03')
> >
> > insert payments values (7, 3, 700, '3/1/03')
> > insert payments values (10,3, 300, '7/10/03')
> > Insert payments values (8, 3, 600, '9/1/03')
> >
> > insert payments values (11,4, 300, '8/1/03')
> > insert payments values (12,4, 400, '9/10/03')
> > Insert payments values (13,4, 100, '9/10/03')
> >
> >
> > customerid lastpayment amount
> >
> > 1 3 (on 5/15/03) 500
> > 2 6 (on 6/15/03) 900
> > 3 8 (on 9/1/03) 600
> > 4 13 (on 9/10/03) 100
> >
> > ========> > Result => 2100
> >
> >
> >
> >
>|||MAB (fkdfjdierkjflafdafa@.yahoo.com) writes:
> Thanks! this works. However its too slow to run on the actual table with
> thousands of rows but i've managed to eliminate most of the rows by
> creating a temporary table and then I run this query on the temporary
> table.
You might need to review your indexes. It is difficult to test
performance on the small sample, but I would try somthing like:
CREATE CLUSTERED INDEX payments_index ON
payments(customerid, date DESC, paymentid DESC)
--
Erland Sommarskog, SQL Server MVP, sommar@.algonet.se
Books Online for SQL Server SP3 at
http://www.microsoft.com/sql/techinfo/productdoc/2000/books.asp|||As Erland has suggested you should visit your indexing strategy. There is a
cost for creating the temp table and inserting data into it. With proper
index, this should be a breeze.
--
-oj
RAC v2.2 & QALite!
http://www.rac4sql.net
"MAB" <fkdfjdierkjflafdafa@.yahoo.com> wrote in message
news:bjoghs$l7pi0$1@.ID-31123.news.uni-berlin.de...
> Thanks! this works. However its too slow to run on the actual table with
> thousands of rows but i've managed to eliminate most of the rows by
creating
> a temporary table and then I run this query on the temporary table.
> "oj" <nospam_ojngo@.home.com> wrote in message
> news:uqjta41dDHA.2320@.TK2MSFTNGP12.phx.gbl...
> > select customerid,paymentid,[date],amount
> > from payments p1
> > where paymentid=(select top 1 paymentid from payments p2 where
> > p2.customerid=p1.customerid order by [date] desc, paymentid desc)
> >
> >
> > --
> > -oj
> > RAC v2.2 & QALite!
> > http://www.rac4sql.net
> >
> >
> >
> > "MAB" <dsfoalsdfsdfadouisdf@.yahoo.com> wrote in message
> > news:bjmcjb$klvrf$1@.ID-31123.news.uni-berlin.de...
> > > I want the sum of the last payments (amount) for all customers. The
last
> > > payment is with one with most recent date. And if there are more than
> one
> > > payment on the most recent date then the one with the higher paymentid
> is
> > > the last payment. for example in the given data the insert statement
> that
> > > starts with capital I is the last payment of that customer. The
correct
> > > answer should be 2100 as given below. both queries by Erland and Anith
> > give
> > > the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause
> from
> > > both queries since right now I want current sum (not till some date).
So
> > > what should be the right query.
> > >
> > > Thanks again for the help.
> > >
> > > create table payments (
> > > paymentid int,
> > > customerid int,
> > > amount int,
> > > date datetime
> > > )
> > >
> > > insert payments values (1, 1, 100, '1/1/03')
> > > insert payments values (2, 1, 200, '2/28/03')
> > > Insert payments values (3, 1, 500, '5/15/03')
> > >
> > > insert payments values (4, 2, 400, '1/16/03')
> > > insert payments values (9, 2, 800, '4/30/03')
> > > insert payments values (5, 2, 200, '6/15/03')
> > > Insert payments values (6, 2, 900, '6/15/03')
> > >
> > > insert payments values (7, 3, 700, '3/1/03')
> > > insert payments values (10,3, 300, '7/10/03')
> > > Insert payments values (8, 3, 600, '9/1/03')
> > >
> > > insert payments values (11,4, 300, '8/1/03')
> > > insert payments values (12,4, 400, '9/10/03')
> > > Insert payments values (13,4, 100, '9/10/03')
> > >
> > >
> > > customerid lastpayment amount
> > >
> > > 1 3 (on 5/15/03) 500
> > > 2 6 (on 6/15/03) 900
> > > 3 8 (on 9/1/03) 600
> > > 4 13 (on 9/10/03) 100
> > >
> > > ========> > > Result => 2100
> > >
> > >
> > >
> > >
> >
> >
>

Heres the Problem Again With Sample Data

I want the sum of the last payments (amount) for all customers. The last
payment is with one with most recent date. And if there are more than one
payment on the most recent date then the one with the higher paymentid is
the last payment. for example in the given data the insert statement that
starts with capital I is the last payment of that customer. The correct
answer should be 2100 as given below. both queries by Erland and Anith give
the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause from
both queries since right now I want current sum (not till some date). So
what should be the right query.

Thanks again for the help.

create table payments (
paymentid int,
customerid int,
amount int,
date datetime
)

insert payments values (1, 1, 100, '1/1/03')
insert payments values (2, 1, 200, '2/28/03')
Insert payments values (3, 1, 500, '5/15/03')

insert payments values (4, 2, 400, '1/16/03')
insert payments values (9, 2, 800, '4/30/03')
insert payments values (5, 2, 200, '6/15/03')
Insert payments values (6, 2, 900, '6/15/03')

insert payments values (7, 3, 700, '3/1/03')
insert payments values (10,3, 300, '7/10/03')
Insert payments values (8, 3, 600, '9/1/03')

insert payments values (11,4, 300, '8/1/03')
insert payments values (12,4, 400, '9/10/03')
Insert payments values (13,4, 100, '9/10/03')

customerid lastpayment amount

1 3 (on 5/15/03) 500
2 6 (on 6/15/03) 900
3 8 (on 9/1/03) 600
4 13 (on 9/10/03) 100

========
Result => 2100select customerid,paymentid,[date],amount
from payments p1
where paymentid=(select top 1 paymentid from payments p2 where
p2.customerid=p1.customerid order by [date] desc, paymentid desc)

--
-oj
RAC v2.2 & QALite!
http://www.rac4sql.net

"MAB" <dsfoalsdfsdfadouisdf@.yahoo.com> wrote in message
news:bjmcjb$klvrf$1@.ID-31123.news.uni-berlin.de...
> I want the sum of the last payments (amount) for all customers. The last
> payment is with one with most recent date. And if there are more than one
> payment on the most recent date then the one with the higher paymentid is
> the last payment. for example in the given data the insert statement that
> starts with capital I is the last payment of that customer. The correct
> answer should be 2100 as given below. both queries by Erland and Anith
give
> the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause from
> both queries since right now I want current sum (not till some date). So
> what should be the right query.
> Thanks again for the help.
> create table payments (
> paymentid int,
> customerid int,
> amount int,
> date datetime
> )
> insert payments values (1, 1, 100, '1/1/03')
> insert payments values (2, 1, 200, '2/28/03')
> Insert payments values (3, 1, 500, '5/15/03')
> insert payments values (4, 2, 400, '1/16/03')
> insert payments values (9, 2, 800, '4/30/03')
> insert payments values (5, 2, 200, '6/15/03')
> Insert payments values (6, 2, 900, '6/15/03')
> insert payments values (7, 3, 700, '3/1/03')
> insert payments values (10,3, 300, '7/10/03')
> Insert payments values (8, 3, 600, '9/1/03')
> insert payments values (11,4, 300, '8/1/03')
> insert payments values (12,4, 400, '9/10/03')
> Insert payments values (13,4, 100, '9/10/03')
>
> customerid lastpayment amount
> 1 3 (on 5/15/03) 500
> 2 6 (on 6/15/03) 900
> 3 8 (on 9/1/03) 600
> 4 13 (on 9/10/03) 100
> ========
> Result => 2100
>|||Thanks! this works. However its too slow to run on the actual table with
thousands of rows but i've managed to eliminate most of the rows by creating
a temporary table and then I run this query on the temporary table.

"oj" <nospam_ojngo@.home.com> wrote in message
news:uqjta41dDHA.2320@.TK2MSFTNGP12.phx.gbl...
> select customerid,paymentid,[date],amount
> from payments p1
> where paymentid=(select top 1 paymentid from payments p2 where
> p2.customerid=p1.customerid order by [date] desc, paymentid desc)
>
> --
> -oj
> RAC v2.2 & QALite!
> http://www.rac4sql.net
>
> "MAB" <dsfoalsdfsdfadouisdf@.yahoo.com> wrote in message
> news:bjmcjb$klvrf$1@.ID-31123.news.uni-berlin.de...
> > I want the sum of the last payments (amount) for all customers. The last
> > payment is with one with most recent date. And if there are more than
one
> > payment on the most recent date then the one with the higher paymentid
is
> > the last payment. for example in the given data the insert statement
that
> > starts with capital I is the last payment of that customer. The correct
> > answer should be 2100 as given below. both queries by Erland and Anith
> give
> > the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause
from
> > both queries since right now I want current sum (not till some date). So
> > what should be the right query.
> > Thanks again for the help.
> > create table payments (
> > paymentid int,
> > customerid int,
> > amount int,
> > date datetime
> > )
> > insert payments values (1, 1, 100, '1/1/03')
> > insert payments values (2, 1, 200, '2/28/03')
> > Insert payments values (3, 1, 500, '5/15/03')
> > insert payments values (4, 2, 400, '1/16/03')
> > insert payments values (9, 2, 800, '4/30/03')
> > insert payments values (5, 2, 200, '6/15/03')
> > Insert payments values (6, 2, 900, '6/15/03')
> > insert payments values (7, 3, 700, '3/1/03')
> > insert payments values (10,3, 300, '7/10/03')
> > Insert payments values (8, 3, 600, '9/1/03')
> > insert payments values (11,4, 300, '8/1/03')
> > insert payments values (12,4, 400, '9/10/03')
> > Insert payments values (13,4, 100, '9/10/03')
> > customerid lastpayment amount
> > 1 3 (on 5/15/03) 500
> > 2 6 (on 6/15/03) 900
> > 3 8 (on 9/1/03) 600
> > 4 13 (on 9/10/03) 100
> > ========
> > Result => 2100|||MAB (fkdfjdierkjflafdafa@.yahoo.com) writes:
> Thanks! this works. However its too slow to run on the actual table with
> thousands of rows but i've managed to eliminate most of the rows by
> creating a temporary table and then I run this query on the temporary
> table.

You might need to review your indexes. It is difficult to test
performance on the small sample, but I would try somthing like:

CREATE CLUSTERED INDEX payments_index ON
payments(customerid, date DESC, paymentid DESC)

--
Erland Sommarskog, SQL Server MVP, sommar@.algonet.se

Books Online for SQL Server SP3 at
http://www.microsoft.com/sql/techin.../2000/books.asp|||As Erland has suggested you should visit your indexing strategy. There is a
cost for creating the temp table and inserting data into it. With proper
index, this should be a breeze.

--
-oj
RAC v2.2 & QALite!
http://www.rac4sql.net

"MAB" <fkdfjdierkjflafdafa@.yahoo.com> wrote in message
news:bjoghs$l7pi0$1@.ID-31123.news.uni-berlin.de...
> Thanks! this works. However its too slow to run on the actual table with
> thousands of rows but i've managed to eliminate most of the rows by
creating
> a temporary table and then I run this query on the temporary table.
> "oj" <nospam_ojngo@.home.com> wrote in message
> news:uqjta41dDHA.2320@.TK2MSFTNGP12.phx.gbl...
> > select customerid,paymentid,[date],amount
> > from payments p1
> > where paymentid=(select top 1 paymentid from payments p2 where
> > p2.customerid=p1.customerid order by [date] desc, paymentid desc)
> > --
> > -oj
> > RAC v2.2 & QALite!
> > http://www.rac4sql.net
> > "MAB" <dsfoalsdfsdfadouisdf@.yahoo.com> wrote in message
> > news:bjmcjb$klvrf$1@.ID-31123.news.uni-berlin.de...
> > > I want the sum of the last payments (amount) for all customers. The
last
> > > payment is with one with most recent date. And if there are more than
> one
> > > payment on the most recent date then the one with the higher paymentid
> is
> > > the last payment. for example in the given data the insert statement
> that
> > > starts with capital I is the last payment of that customer. The
correct
> > > answer should be 2100 as given below. both queries by Erland and Anith
> > give
> > > the result 100 ( I removed the "WHERE p1.date <= '20030301' " Clause
> from
> > > both queries since right now I want current sum (not till some date).
So
> > > what should be the right query.
> > > > Thanks again for the help.
> > > > create table payments (
> > > paymentid int,
> > > customerid int,
> > > amount int,
> > > date datetime
> > > )
> > > > insert payments values (1, 1, 100, '1/1/03')
> > > insert payments values (2, 1, 200, '2/28/03')
> > > Insert payments values (3, 1, 500, '5/15/03')
> > > > insert payments values (4, 2, 400, '1/16/03')
> > > insert payments values (9, 2, 800, '4/30/03')
> > > insert payments values (5, 2, 200, '6/15/03')
> > > Insert payments values (6, 2, 900, '6/15/03')
> > > > insert payments values (7, 3, 700, '3/1/03')
> > > insert payments values (10,3, 300, '7/10/03')
> > > Insert payments values (8, 3, 600, '9/1/03')
> > > > insert payments values (11,4, 300, '8/1/03')
> > > insert payments values (12,4, 400, '9/10/03')
> > > Insert payments values (13,4, 100, '9/10/03')
> > > > > customerid lastpayment amount
> > > > 1 3 (on 5/15/03) 500
> > > 2 6 (on 6/15/03) 900
> > > 3 8 (on 9/1/03) 600
> > > 4 13 (on 9/10/03) 100
> > > > ========
> > > Result => 2100
> > > >

HelpTotals within Totals

I have a report with say two fields, market, amount. I need a report to present like this, but to do totals after line item presentation...example report looks like this. Question is, how do I do the totals part seperate from the line items??

Market amount
-- --
Texas 10.00
Texas 20.00
Texas 30.00
Boston 10.00
Boston 20.00
Boston 90.00

Totals Texas 60.00
Totals Boston 120.00You can set it up like the following by adding a Group on Market and adding a Running Total Field which Sums Amount to the Group Footer.

Market amount
-- --
Texas 10.00
Texas 20.00
Texas 30.00

Totals Texas 60.00

Boston 10.00
Boston 20.00
Boston 90.00

Totals Boston 120.00

- or -

If you know what Market is going to be, you can set up Formula Fields with the following Formula:

NumberVar intTotal
IF {Market} = 'Texas' THEN
NumberVar := NumberVar + {Amount}

- or -

There's a Formula that looks like: Sum(fld, condFld, cond). I have never used it, but I assume you can set it up simialr to:

Sum ( {Amount}, {Market}, 'Texas' )

(I didn't test this code, so you may have to tweak it a bit to work out any bugs.)|||Thanks malleyo but there's a subtle difference in the report...the footer has and entries for each market. The list of markets is dynamic (a group by field)...thus, option 1 and 2-3 aren't quite it...I have a feeling I have to use subreports.

Let me know if you have any other thoughts and thanks for your help.|||Hi Mallego
How to sum amount large than 10 ?|||ReportDude,
I guess you are right there is no option otherthan subreport, I had the same issue before and I used sub report.
But if you find better solution other than SubReport please post the solution.

Thanks,
Gragi