[SOLVED] A 5 star rating thingy.

[Twister1004]

Hey everyone! I'm the loser who is trying to make a 5 star rating thingy. Yeah I know I know. Well, either I thought of this as being too simple and wrote it like that, or I just totally didn't think it though. Which I think it is both cases!

 

So, If I may ask for some help, It would be greatly appreciated!

 

Objective: Figure out a proper way to write this script to where it won't fail. As in it will never decrease to 0 all the time.

 

Problem: When I add a NEW rating to the script, or to the variable, IT will allways decrease to 0. Even if it's all 5's.

 

My Code will be in 2 pieces. I will provide any other information that is needed. Thank you to everyone in advance!

 

FYI: I don't know how to use OOP.

 

My functions page:

function getblogRating($id){
$sql = mysql_query("SELECT `rating` FROM `blog` WHERE `id` = '$id' LIMIT 1");
$getBlog = mysql_fetch_array($sql);
return $getBlog['rating'];
}
function postblogRating($id, $rating){
if($rating == 0 || $rating == NULL){
	return "No value entered";
}
$sql = mysql_query("SELECT * FROM `blog` WHERE `id` = '$id' LIMIT 1");
$rate = mysql_fetch_array($sql);

$voters = $rate['voted'] + 1;
$combine = ($rate['totalrate'] + $rating) / $voters;
echo $combine;
$totalRate = "$rate[totalrate] + $rating";

$update1 = mysql_query("UPDATE `blog` SET
`rating` = '".$combine."',
`voted` = '".$voters."',
`totalrate` = '".$totalRate."'
WHERE `id` = '".$id."'");
if($update1){
	$update2 = mysql_query("UPDATE `peoplerating` SET
	`blogid` = '$id',
	`user` = '".$_SESSION['name']."'");
	if($update2){
		return "Your rating has been successfully posted!";
	}
	else{
		echo "error";
	}
}
else{
	echo "error";
}
}
function peopleRating($blogid, $name){
if($name == "" || $name == NULL){
	return 2;
}
$Rated = mysql_query("SELECT * FROM `peoplerating` WHERE `blogid` = '$blogid' AND `user` = '$name' LIMIT 1");
if(mysql_num_rows($Rated) > 0){
	return 1;
}
else{
	return 0;
}
}

 

Page with the displaying information. PART 1

 

@$checkUser = peopleRating(mysql_real_escape_string($_GET['blog']), $_SESSION['name']);
			if($checkUser == 0){
				echo "<div style='position:absolute; top:40px; text-align:right; margin-left:137px;'>
					<form method='post' action=''
						<input type=\"radio\" name=\"thisRating\" value=\"1\" /> 1
						<input type=\"radio\" name=\"thisRating\" value=\"2\" /> 2
						<input type=\"radio\" name=\"thisRating\" value=\"3\" /> 3
						<input type=\"radio\" name=\"thisRating\" value=\"4\" /> 4
						<input type=\"radio\" name=\"thisRating\" value=\"5\" /> 5
						<input type=\"submit\" name=\"vote\" value=\"Vote!\" />
					</form>
				</div>";
			}
			elseif($checkUser == 2){
				echo "<div style='position:absolute; top:40px; text-align:right; margin-left:300px;'>Please sign in</div>";
			}
			elseif($checkUser == 1){
				echo "<div style='position:absolute; top:40px; text-align:right; margin-left:215px;'>You have voted here before</div>";
			}

 

Part 2

 

if(isset($_POST['vote'])){
			postblogRating(mysql_real_escape_string($_GET['blog']), $_POST['thisRating']);
			//echo "<meta http-equiv='refresh' content='0; url='''/>";
		}

[Daniel0]

Have a table like this:

 

votes

vote_id (primary key)

user_id (foreign key referencing the voter to prevent double votes)

blog_id (foreign key referencing the thing that's voted on)

rating (an integer in the range of 1 to 5)

 

 

So when the person with user id 10 votes on the blog post with id 31 and wants to give it 4 stars you do this (after checking they didn't already vote):

INSERT INTO votes (user_id, blog_id, rating) VALUES (10, 31, 4);

 

When you want to get the score and the number of voters for blog post id 31 you do like this:

SELECT SUM(v.rating)/COUNT(*) AS score, COUNT(*) AS num_votes
  FROM votes AS v
  INNER JOIN blog AS b
    ON b.id = v.blog_id
  WHERE b.id = 31
  GROUP BY b.id;

[Twister1004]

Thank you very much, Daniel!! It worked, but I was struggling about how to call it out. xD. but I ended up just doing this.

function getblogRating($id){
$sql = mysql_query("SELECT SUM(rating)/COUNT(*) AS score, 
COUNT(*) AS num_votes FROM blograting 
AS v INNER JOIN blog 
AS b ON b.id = v.blogid WHERE b.id = '$id' GROUP BY b.id");
$getRating = mysql_fetch_array($sql);
return $getRating[0];
}

 

I really appreciate your work. -hugs-