Viewing 15 posts - 46 through 60 (of 95 total)
Never used this..but check BOL on this..
DBCC CHECKIDENT ('table_name', RESEED, new_reseed_value)
HTH
February 6, 2006 at 12:02 pm
"I also checked the records in Tbl_Accounts - there are a few records with a null value in Basecurrency but all other records have a matching record in Tbl_Currencies."
Why do...
February 6, 2006 at 11:59 am
Something like this...
create table #Temp
(
Field1 some_date_type,
Field2 some_date_type,
Field3 some_date_type,
)
If [Condition1]
insert #temp( Field1,Field2, Field3)
Select Field1,Field2, Field3
from Table
order by Field3,Field2, Field1
Else
If [Condition2]
insert #temp( Field1,Field2, Field3)
Select Field1,Field2, Field3
from Table
order by Field3,Field2, Field1
Else
insert #temp( Field1,Field2,...
February 6, 2006 at 10:04 am
Pivoting the table works as Matthew mentions...but I'd probably do it something as follows..
declare @possible_answers table
(
possible_answer char(1)
)
declare @questions table
(
question char(1)
)
declare @results table
(
question char(1),
answer char(1),
tally int default 0
)
insert @questions
select question
from...
February 6, 2006 at 9:59 am
Create a DTS package..and create a batch file with a dtsrun acommand. Have the user run the batch file as needed.
HTH
February 6, 2006 at 7:55 am
If I'm understanding you...
select Distinct [ID],CustomerName,CustomerNumber,SalesmanName,Min(ItemDescription) AS ItemDescription,OrigionalDateofService,InsuranceCarrier, Diag1,Diag2,Diag3,Diag4 from O2
Group By
[ID],
CustomerName,
CustomerNumber,
SalesmanName,
OrigionalDateofService,
InsuranceCarrier,
Diag1,
Diag2,
Diag3,
Diag4
By the way if {id] is trully an identity column ..you shouldn't need distinct at all.
HTH
February 6, 2006 at 7:53 am
Well one thing is you'd have to determine what "id" you want to return. Do you just return one of them for each distinct set you have or ..do...
February 6, 2006 at 7:47 am
Still not sure exactly what you need. What is your DDL, and what is an example result you are expecting ?
February 4, 2006 at 2:17 pm
I'd do it like this...
declare @table table
(
[id] int primary key identity(1,1),
value float
)
declare @results table
(
[id] varchar(50) primary key,
value float
)
declare @low int,
@high int
set @low = 0
set @high = 1000
while @low <= @high
begin
insert...
February 4, 2006 at 2:14 pm
Your DDL would be helpful.
February 4, 2006 at 12:47 pm
something like this...
insert results_table(User_id,LessonId,Num_Abset)
select user_id,lesson_id,count(*) Num_absent
from SomeTable T
where t.status = 'Absent'
group by t.user_id,t.lesson_id
having count(*) = 5
February 4, 2006 at 5:51 am
If I'm understanding you correctly..a simple join..
Select M.*,D.*
from MasterTable M
join DetailTable D
on M.ID = D.ID
HTH
February 4, 2006 at 5:41 am
try ..
SELECT TOP 10
S.id_solicitud,
CONVERT(NCHAR(10), S.fecha_creacion, 120) AS fecha_creacion,
S.nombre,
S.clave_material,
IsNull(CONVERT(NCHAR(10), S.fecha_ini_pruebas, 120), 'N/A') AS fecha_ini_pruebas,
IsNull(CONVERT(NCHAR(10), S.fecha_fin_pruebas, 120), 'N/A') AS fecha_fin_pruebas,
IsNull(CONVERT(NCHAR(10), S.fecha_emision_reporte, 120), 'N/A') AS fecha_emision_reporte
FROM
dbo.tblSolsInstrum_Solicitudes S
WHERE
id_usuario = 'car6527'
ORDER BY
2...
February 3, 2006 at 11:29 am
Have your script file update a log file. Before executing the next step in your program, check the log file to make sure log says the script ended sucessfully....
February 3, 2006 at 10:53 am
Check out the books that helped me pass on my blog.
Good luck.
HTH
February 3, 2006 at 9:28 am
Viewing 15 posts - 46 through 60 (of 95 total)