Average

nathan2oo3

I was just wondering, I have a website for reviewing DVD's and use radio buttons to post the reviews by each user into a table. Is there any way I could use this information to show the average reviews of each movie...?

Thanks

dagon

nathan2oo3;10947366 wrote:
I was just wondering, I have a website for reviewing DVD's and use radio buttons to post the reviews by each user into a table. Is there any way I could use this information to show the average reviews of each movie...?

Thanks

assuming mysql, it has an average function

nathan2oo3

apologies, I posted that earlier with a different problem but couldnt work out how to take the post down..

nathan2oo3

actually was the right post, ignore that last comment :S been one of those days, will look into mysql averages, thanks 🙂

nathan2oo3

Okay, so I came up with the following code however when I echo it ( I have put in some test values from user accounts) it doesnt show the value.

$dbQuery2="SELECT dvdref, AVG(rating) FROM ratings GROUP BY dvdref";					
$result2 = mysql_query($dbQuery2,$db);

<?php echo"$result2";?>

Any ideas guys what im doing wrong?

bradgrafelman

[man]mysql_query/man returns a MySQL resource (if the query was successfully executed, of course - otherwise it returns boolean FALSE).

You need to use one of the mysql_fetch_*() functions to actually retrieve the data from the result set.

nathan2oo3

Ok.. I have the information in two different tables though using a while loop to return the values of the dvd's the ratings is in a different table to the dvds themselves. Would I be best using a join between the two tables with the above avg statement or using seperate queries?

Thanks

nathan2oo3

THe code I have got so far is as below

$dbQuery2="SELECT dvdref, AVG(rating) FROM ratings GROUP BY dvdref";					
					$result2 = mysql_query($dbQuery2,$db);	
					$rating = mysql_result($result2); 

				$dbQuery1="SELECT * FROM dvd order by title";
				$result1 = mysql_query($dbQuery1,$db);	
				while($dbRow=mysql_fetch_array($result1))
					{	
					$dvdref=$dbRow["dvdref"];  
					$dvdLink=$dbRow["dvdLink"]; 
					$title=$dbRow["title"];

However this gives me the error message:

Warning: mysql_result() expects at least 2 parameters, 1 given in N:\home\scie\d07cn\COM570\loginCentre.php on line 119 which is assigning the value to the rating variable...

dagon

mysql_result()