simple rating system

[ashley526]

I am in need of a very simple rating system that works off of numbers 1.00 to 10.00. I have a database already set up that has the items I wanted rated.

My question is can I use that same database or do I have to create a seperate one for rating/voting?

I'd ideally like to have average rate and # of votes. I saw the 5 star rating system thing, but I think that is a little more then I am wanting.

Thanks for all the help in advance!

Ash

[tekky]

ratings table
user_id
rated_id (foreign key to table containing the ID of what you are rating...)
rating

SELECT SUM(rating) AS sum, COUNT(rating) AS total WHERE rated_id = $id GROUP BY rated_id

should give you sum and total and then you can divide sum/total to get the avg... Shrug thats what comes to my mind...

[ashley526]

So is this a new table created? within the database?

[TheDefender]

PHP Freaks has a decent tutorial on their site on how to create a rating system you might find handy. I know you said the 5 star rating system may be more than you want, but you may be able to use it to figure out how to do what you want. Their tutorials are good.

[tekky]

Originally posted by ashley526
So is this a new table created? within the database?

yes it would be (it has to be from all my rationalization... -- cant think of a way it could be done otherwise...)

[ashley526]

I am going to read through that star rating system at php freaks, and go over what tekky said. I am very new to php and such so the star rating system really confused me. lol

I will be back with questions I am sure.

[ashley526]

Okay this is what I have....

<?php

if (!isset($mode))
{
$mode = 'index';
}

switch($mode) {
case 'rate':

?><p align="center"><b>Rate Item</b></p><?php

// Database Connection Variables, adjust it for your server
// MySQL Connection Variables 
$user="user";
$pass="pw";
$hostname="localhost";
$dbase='db';


$connection = mysql_connect("$hostname" , "$user" , "$pass");
$db = mysql_select_db($dbase , $connection);

$query = "SELECT * from names where nameid='$nameid' ";
$result = mysql_query($query);

while ($row=mysql_fetch_array($result))
{
$rateid = $row["rateid"];
$nameid = $row["nameid"];
$totalcount = $row["totalcount"];
$totalsum = $row["totalsum"];

// compute new total ratings

$totalsum = $totalsum + $rateitem;

// Increment total count by 1 since 1 more user rate item
$totalcount = ++$totalcount;

// Compute new average, take note of division by zero error

$newavg = $totalsum / $totalcount;

// round new average to 2 dec place.
$newavg = round($newavg,2);

// Now update rating table with new values.

$newquery = "UPDATE rate set avgrate = '$newavg', totalsum='$totalsum', totalcount='$totalcount' 
where nameid='$nameid' ";

$newresult = mysql_query($newquery);

if ($newresult)
{
echo "Thank you, Item has been rated.";
}

} 
mysql_free_result($result); 
break;

case 'index':

?><p align="center"><b>Rate Item</b></p>
<form action="<?php echo "rate.php?mode=rate&itemid=$itemid"; ?>" method="post">
<SELECT name="rateitem" >
<option value="10">10</option>
<option value="9">9</option>
<option value="8">8</option>
<option value="7">7</option>
<option value="6">6</option>
<option value="5">5</option>
<option value="4">4</option>
<option value="3">3</option>
<option value="2">2</option>
<option value="1">1</option>
</select><input type="submit" value="Rate"></form>

<?php


break;
}
?>

However I REALLY don't think that it's doing it right... I have two tables names and rate. The nameid is on the names table and the rest is on the rate table.

I am really confusing myself...

Okay so this is what I need to know. First, I already have an established database with some names in it. I want to be able to rate those.

Here is my database info if you wanna see what I am talking about:

http://www.simply-baby.net/forum/vbaindex.php?page=babynames

And here is where they enter the names. Ignore the rating thing that is there because obviously it doesn't work! lol

http://www.simply-baby.net/forum/vbaindex.php?page=entername