Displays commas in type FLOAT

I'm writing a SQL script. I'd like to display the value I have in a
column of type FLOAT so that it appears with the commas in the correct
place. It displays now as 10000000.0; I'd like it to display as
10,000,000. Thanks for any help."Rick Charnes" <rickxyz--nospam.zyxcharnes@.thehartford.com> wrote in message
news:MPG.1cffedcae86a60059898e0@.msnews.microsoft.com...
> I'm writing a SQL script. I'd like to display the value I have in a
> column of type FLOAT so that it appears with the commas in the correct
> place. It displays now as 10000000.0; I'd like it to display as
> 10,000,000. Thanks for any help.
Your front-end application should really be doing this type of work, not the
database itself.
Rick Sawtell
MCT, MCSD, MCDBA|||Formatting should be done client side, not in SQL.
"Rick Charnes" <rickxyz--nospam.zyxcharnes@.thehartford.com> wrote in message
news:MPG.1cffedcae86a60059898e0@.msnews.microsoft.com...
> I'm writing a SQL script. I'd like to display the value I have in a
> column of type FLOAT so that it appears with the commas in the correct
> place. It displays now as 10000000.0; I'd like it to display as
> 10,000,000. Thanks for any help.|||Do this in the reporting tool / client application, or cast the value to
money (not good) and then use function "convert" to cast it to varchar with
style 1.
Example:
use northwind
go
select
convert(varchar(25), cast(cast(orderid as float) as money), 1)
from
dbo.orders
AMB
"Rick Charnes" wrote:

> I'm writing a SQL script. I'd like to display the value I have in a
> column of type FLOAT so that it appears with the commas in the correct
> place. It displays now as 10000000.0; I'd like it to display as
> 10,000,000. Thanks for any help.
>

displaying found duplicate - both records

I would like to look at both records returned as duplicates. I am using the
following script to find the duplicates but can it only returns one record
not both. I would like to see the entire record so that I can make a
determination on more than just the criteria that I selected as the basis of
the record being a duplicate.
SELECT Patient.Last_Name,
Patient.First_Name,
Patient.Birthdate
COUNT(*) AS Dupes
FROM Patient
WHERE Patient_Key IN (SELECT DISTINCT Patient_Key FROM Patient_Elg WHERE
Payor_Key = 36)
GROUP BY Last_Name, First_Name, Birthdate
HAVING (COUNT(*) > 1)
Message posted via webservertalk.com
http://www.webservertalk.com/Uwe/Forum...amming/200510/1In order to do that, you may need to put your current query as "Result
Table", something like:
SELECT P1.Last_Name,
P1.First_Name,
P1.Birthdate,
Dupes
FROM Patient P1,
(SELECT P2.Last_Name P2LastName,
P2.First_Name P2FirstName,
P2.Birthdate P2Birthday
COUNT(*) AS Dupes
FROM Patient P2
WHERE P2.Patient_Key IN (SELECT DISTINCT Patient_Key FROM
Patient_Elg WHERE
P2.Payor_Key = 36)
GROUP BY P2.Last_Name, P2.First_Name, P2.Birthdate
HAVING (COUNT(*) > 1)) DupCount
where P1.Last_Name = P2LastName
and P1.First_Name = P2FirstName
and P1.Birthdate = P2Birthday
This code has not been tested yet but just an idear.
Perayu
"Jay via webservertalk.com" <u7124@.uwe> wrote in message
news:5685c45d8a0c6@.uwe...
>I would like to look at both records returned as duplicates. I am using the
> following script to find the duplicates but can it only returns one record
> not both. I would like to see the entire record so that I can make a
> determination on more than just the criteria that I selected as the basis
> of
> the record being a duplicate.
> SELECT Patient.Last_Name,
> Patient.First_Name,
> Patient.Birthdate
> COUNT(*) AS Dupes
> FROM Patient
> WHERE Patient_Key IN (SELECT DISTINCT Patient_Key FROM Patient_Elg WHERE
> Payor_Key = 36)
> GROUP BY Last_Name, First_Name, Birthdate
> HAVING (COUNT(*) > 1)
>
> --
> Message posted via webservertalk.com
> http://www.webservertalk.com/Uwe/Forum...amming/200510/1|||This did work for me after I made some tweaks. Thanks allot.
Perayu wrote:
>In order to do that, you may need to put your current query as "Result
>Table", something like:
>SELECT P1.Last_Name,
> P1.First_Name,
> P1.Birthdate,
> Dupes
>FROM Patient P1,
> (SELECT P2.Last_Name P2LastName,
> P2.First_Name P2FirstName,
> P2.Birthdate P2Birthday
> COUNT(*) AS Dupes
> FROM Patient P2
> WHERE P2.Patient_Key IN (SELECT DISTINCT Patient_Key FRO
M
>Patient_Elg WHERE
> P2.Payor_Key = 36)
> GROUP BY P2.Last_Name, P2.First_Name, P2.Birthdate
> HAVING (COUNT(*) > 1)) DupCount
>where P1.Last_Name = P2LastName
> and P1.First_Name = P2FirstName
> and P1.Birthdate = P2Birthday
>This code has not been tested yet but just an idear.
>Perayu
>
>[quoted text clipped - 13 lines]
Message posted via http://www.webservertalk.com