VOTING/RATING SYSTEM: HOW TO ADD POINTS AND UPLOAD SQL VALUE?

Sep 2, 2006

I am trying to build a voting system. Website visitors will rate a pictures from 0-10.
This value should update the amount of points already stored is SQL.
I think the way to do it is passing the value of a field to a varible, performing the operation and updating the database, but I don't know how to capture a value in Sql and pass it to variable...
Does anybody there could post a piece of code or point to a link where I could find this information?

thanks

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Rating System

Jul 20, 2005

this is probably a simple sql solution, but i could use some help.i've got one table that has rows of documents, and another table withany number of ratings for each of the documents. The rating table islinked with an id column. the issue i am running up against isreturning a single average rating for a document.SELECT document.title, (SELECT AVG(scale)FROM Rating, documentWHEREdocument.pkDocumentId = rating.documentId)FROM Documentreturns the same avg for all the ratings.So i'd ideally like to return the values of the document columns(tableA.*), with an additional column containing the average rating ofthat document (avg(tableB.Scale) where pkDocumentId = documentId).thanks in advance.

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Jun 18, 2008

I've managed to get my query to this: Product | Color | Votes
======== ======= =======
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Thanks, Pete.

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May 5, 2007

i currently work with a microsoft crm which works over an sql database. however i need to pull out some reports but i am having a hard time.

Sales Person Leadname adress rating
------------ -------- ----- ------
john doe mary doe 12 1st street hot
jane doe example 21 mary st cool

how do i pull out the total of hot or cold ratings per sales person and present all the information on one table.
Please help



Melvin Felicien
IT Manager
DCG Properties Limited

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Jun 22, 2007

Greetings everyone, I'm posting this thread with the hope that someone will notice it and might offer me a helping hand regarding one of my problems.I have the database named "DBEXAMPLE" with the table MEMB_INFO that contains two important columns that are named cur_points and points where cur_points column contains the total available points that a member can use/spend on a game and the points column stores the total of the points that a member used so far.So as i plan to wipe all data of the database, i need to keep the member login,password and total points that each member purchased.So i will somehow need to update the cur_points column with the total of current cur_points+points columns and then wipe points column.I've personally asked a friend regarding this and he said that this should be something complicated and it might require php also.Really appreciate if someone could help me regarding this.small schema:Database: DBEXAMPLETable: MEMB_INFOColumn cur_points -> total available points that a member can spendColumn points -> total points that the member has already spentColumn memb___id -> member login aka account idTODO -> update cur_points column with the total of cur_points+points columns for each member(buyer)With best regards.

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Dim connectionString As String = "Data Source=localhostSQLEXPRESS;Initial Catalog=TimeSheet;Integrated Security=SSPI"



I am not running the webpage in a virtual directory but in

C:Inetpubwwwrootusercontrol

and I have a simple index.html that tries to read from an sql db but throws

the error

System.Security.SecurityException: Request for the permission of type 'System.Data.SqlClient.SqlClientPermission, System.Data, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089' failed.
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at System.Data.Common.DbConnectionOptions.DemandPermission()
at System.Data.SqlClient.SqlConnection.PermissionDemand()
at System.Data.SqlClient.SqlConnectionFactory.PermissionDemand(DbConnection outerConnection)
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etc etc

The action that failed was:
Demand
The type of the first permission that failed was:
System.Data.SqlClient.SqlClientPermission
The Zone of the assembly that failed was:
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any help is appreciate. I have tried books online etc.. cannot find a good
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Dear experts,

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DECLARE @EmployeeNumber nvarchar(50)
Select
@EmployeeNumber = EmployeeID
from
Employee
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+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++




Code Snippetcreate function dbo.Distance( @lat1 float , @long1 float , @lat2 float , @long2 float)
returns float

as

begin

declare @DegToRad as float
declare @Ans as float
declare @Miles as float

set @DegToRad = 57.29577951
set @Ans = 0
set @Miles = 0

if @lat1 is null or @lat1 = 0 or @long1 is null or @long1 = 0 or @lat2 is
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begin

return ( @Miles )

end

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set @Miles = CEILING(@Miles)

return ( @Miles )

end

DECLARE @RC float
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PRINT @RC /* in miles */

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Regards

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Here's an image, stolen from technet's articles.

http://www.microsoft.com/library/media/1033/technet/images/prodtechnol/sql/2000/maintain/ppc1106_big.gif

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Apr 18, 2008

DECLARE @EffLevels TABLE (ChangePoint int, Value Int)

INSERT@EffLevels
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SELECT'1000', '675' UNION ALL
SELECT'1001', '600' UNION ALL--Changed
SELECT'1001', '545' UNION ALL
SELECT'1001', '765' UNION ALL
SELECT'1000', '673' UNION ALL--Changed
SELECT'1002', '343' UNION ALL--Changed
SELECT'1002', '413' UNION ALL
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-- My Result should be
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-- 1000Null767
-- 1001 1000 675
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Any suggestion ?

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Nov 1, 2006

This sounds so simple but yet don't know why it doesn't work.

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closingUnits = 25093.53640

closingAmt = 59110.33

i use a derived column transformation to generate a string column which is the division of those two above.

mystrfield = closingAmt /closingUnits = 2.355599

what i want is cast it to just 4 decimal points when i run the expression below it cast it to 4 decimal points but it doesn't do any rounding.

the value i get is 2.3555 when i should get 2.3556

(ISNULL(closingAmt ) || closingUnits == 0) ? "0.0000" : ((closingAmt / closingUnits) < 1 && (closingAmt / closingUnits) > -1 ? "0" : "") + (DT_WSTR,15)(DT_DECIMAL,4)(closingAmt / closingUnits)

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Sep 4, 2007

Hi all,
I am seeking your expertise to create SQL codes (SQL server 2005) that can help me to answer the problem below.

I have two tables (points and station), presented in form of SQL codes below. I€™d like to find the 6 closest panels for each of the station. As can be seen in the result table below, the 6 closest panel names are arranged from the first closest (P1) to the sixth closest (P6). Similar procedure also applies for the distance column arrangement. This distance column (D1 €“ D6) is the distance of panels P1 €“ P6 to the station. The distance between two points (with x-y coordinates) can be calculated using a simple Cartesian formula:
Distance = ( (X1 €“ X2)2 + (Y1 - Y2)2 ) 0.5 . As the sample, distance between station €˜A€™ and panel €˜P19-04W€™ is = ((737606.383 - 737599.964)2 + (9548850.844 - 9548856.856)2) 0.5 = 8.79.
The expected result of the work is presented in the table below:



















Panel
distance

Station
P1
P2
P3
P4
P5
P6
D1
D2
D3
D4
D5
D6

A
P19-04W
P19-06E
P19-05E
P19-05W
P19-03W
P19-07E
8.79
12.00
13.54
19.02
21.00
27.43

B
P19-07W
P19-09E
P19-08E
P19-06W
P19-08W
P19-07E
9.50
11.58
12.92
21.15
24.85
29.11

C
P19-11E
P19-10W
P19-12E
P19-09W
P19-10E
P19-11W
8.45
11.42
16.06
16.38
23.15
25.30

Table 1:
create table 1 (
Panels varchar(20),
X_Coord float,
Y_Coord float
)
go
set nocount on

insert into 1 values('P19-03E','737640.722','9548882.875')
insert into 1 values('P19-04E','737630.166','9548868.3')
insert into 1 values('P19-05E','737619.611','9548853.726')
insert into 1 values('P19-06E','737609.054','9548839.15')
insert into 1 values('P19-07E','737598.495','9548824.571')
insert into 1 values('P19-08E','737587.941','9548809.998')
insert into 1 values('P19-09E','737577.386','9548795.425')
insert into 1 values('P19-10E','737563.359','9548776.163')
insert into 1 values('P19-11E','737552.795','9548761.578')
insert into 1 values('P19-12E','737542.256','9548746.919')
insert into 1 values('P19-13E','737531.701','9548732.345')
insert into 1 values('P19-14E','737521.146','9548717.772')
insert into 1 values('P19-03W','737610.519','9548871.43')
insert into 1 values('P19-04W','737599.964','9548856.856')
insert into 1 values('P19-05W','737589.404','9548842.275')
insert into 1 values('P19-06W','737578.849','9548827.702')
insert into 1 values('P19-07W','737568.294','9548813.128')
insert into 1 values('P19-08W','737554.274','9548793.77')
insert into 1 values('P19-09W','737543.718','9548779.195')
insert into 1 values('P19-10W','737533.157','9548764.614')
insert into 1 values('P19-11W','737522.603','9548750.041')

set nocount off
go


Table 2:
create table 2 (
Station varchar(20),
X_Coord float,
Y_Coord float
)
go
set nocount on

insert into 2 values('A','737606.383','9548850.844')
insert into 2 values('B','737575.41','9548806.838')
insert into 2 values('C','737544.437','9548762.832')

set nocount off
go

Thanks alot in advance!

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E:

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SQLSHARED local folder -sql shared tools for all db instances

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SQLSNAP1 Drive F -db snapshot files

SQLTEMPDB1 Drive H -tempdb main data file

SQLWORK Drive D - DBA work area

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SQLDATA2 Future Drive -if SQLDATA1 is too large for any direct attached drive, or to get more I/O throughput.

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KTMD

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