Creating a rating script

Sharif

Okay, I need to make a script where users can rate a picture from 1 - 10. Now this is what I was thinking, have a table that stores the rating number (1-10) and the picture id(pid). When I need to display the average rating, I can do simple mathematics right? Add up all the ratings and find the average. Is this a good way of doing this or are there other approaches?

barand

Assuming the table is something like

Userid |  pictureid | rating
   1               1             3
   2               1             6 
   3               1             8
   1               2             5
   4               2             7

Then

$sql = "SELECT pictureid,  AVE(rating) as picaverage FROM
            mytable GROUP BY pictureid ";

will calculate the averages for you.

Sharif

HEEEY, the sql query does the calculating for me...

I was gonna do this:

$sql = "SELECT rating FROM table WHERE userid = '$userid'";
$query = mysql_query($sql);
$row = mysql_fetch_array($query);
$total = implode("+", $row['rating']);
$ave = $total / mysql_num_rows($query);

echo 'The average rating is ' . $ave . '.';

That's just off the top of my head... wonder if that would even work....

barand

Originally posted by Sharif
wonder if that would even work....

Doubt it 🙂

Sharif

Ok so what makes you say that? the implode() ?

sid

how the heck do you store your rating data?

Sharif

That I took care of already 🙂

Sharif

how the heck do you do that thing in your signature?

Sharif

barand,

Would this work:

$sql = "SELECT AVE(rating) AS average FROM rating WHERE userid = '$id'";

barand

Originally posted by Sharif
Ok so what makes you say that? the implode() ?

There's the implode, plus you only fetch the first row returned by the query.

As for this query

$sql = "SELECT AVE(rating) AS average FROM rating WHERE userid = '$id'";

Should be OK but I thought you wanted the average rating for a picture, not for a user. So shouldn't it be

... where pid = '$pid'

seby

Originally posted by Sharif
barand,

Would this work:
$sql = "SELECT AVE(rating) AS average FROM rating WHERE userid = '$id'";
[/B]

yes that will work,... except it should be AVG(rating)
(typo probably) ..

you can do something like this:

$query = mysql_fetch_array(mysql_query("SELECT AVG(rating) AS average FROM rating WHERE userid = '".$id."'"));
$rating = number_format($query['average'], 1, '.', '');

echo $rating . ' / 10 average ';