Outputting a percentage

hunty

I'm trying to output a percentage, but when I run this code all I get is a blank screen. I know there's errors somewhere so if somebody could point them out to me it would be great thanks!

<html>

<body>

<?php
$con = mysql_connect("mysql360int.cp.blacknight.com","u1041802_ahunty","andrew23");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("db1041802_buyer1", $con);

$query1="Select COUNT(rate) AS PositiveRating FROM feedback WHERE rate
= 'positive'";
$result1 =mysql_query($query1);

if(mysql_num_rows($result1))
{
while($row = mysql_fetch_assoc($result1))
{
$PositiveRating = $row['PositiveRating'] ;
/echo ("$PositiveRating");/
}
}
else
{
echo "No matches were found in the database for your query.\n";
}

$query="Select COUNT(rate) AS NegativeRating FROM feedback WHERE rate
= 'negative'";
$result =mysql_query($query);

echo " ";

if(mysql_num_rows($result))
{
while($row = mysql_fetch_assoc($result))
{
$NegativeRating = $row['NegativeRating'] ;
/echo ("$NegativeRating");/
}
}
else
{
echo "No matches were found in the database for your query.\n";
}

function percent($result1, $result) {

$num_total = $result1 + $result;

$count1 = $result1 / $num_total;

$count2 = $count1 * 100;

return $count2;
}
echo $count2;

if (!mysql_query($query,$con))
{
die('Error: ' . mysql_error());
}

mysql_close($con)
?>
</body>

</html>

hunty

The output for $per is always 40, even though $result1 and $result are 5 and 4 respectively. I have no idea why this is, please help!

<html>

<body>

<?php
$con = mysql_connect("mysql360int.cp.blacknight.com","u1041802_ahunty","andrew23");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("db1041802_buyer1", $con);

$query1="Select COUNT(rate) AS PositiveRating FROM feedback WHERE rate
= 'positive'";
$result1 =mysql_query($query1);

if(mysql_num_rows($result1))
{
while($row = mysql_fetch_assoc($result1))
{
$PositiveRating = $row['PositiveRating'] ;
echo ("$PositiveRating");
}
}
else
{
echo "No matches were found in the database for your query.\n";
}

$query="Select COUNT(rate) AS NegativeRating FROM feedback WHERE rate
= 'negative'";
$result =mysql_query($query);

echo " ";

if(mysql_num_rows($result))
{
while($row = mysql_fetch_assoc($result))
{
$NegativeRating = $row['NegativeRating'] ;
echo ("$NegativeRating");
}
}
else
{
echo "No matches were found in the database for your query.\n";
}

echo " ";

/4 + 4/
$num_total = $result1 + $result;

$per = number_format(($result1 * 100) / $num_total);
echo $per.'%';

if (!mysql_query($query,$con))
{
die('Error: ' . mysql_error());
}

mysql_close($con)
?>

</body>

</html>

NogDog

In the section where you calculate and display the percentage, you are using the variables $result and $result1. Those variables are not database values, but rather are PHP resource IDs (pointing to mysql_query() result sets, in this case). You want to use values from the arrays returned by the mysql_fetch operations.

PS: Please use [noparse]

...

[/noparse] tags around your PHP code sections.

ixalmida

You're missing a semi-colon in your last statement.

Also, there is no $count2 variable outside the scope of your function (where you echo it). But that should give you an actual error, not a blank page.

NogDog

[Threads have been merged.]

Please post your code within [noparse]

...

[/noparse] tags, preferably with some consistent indenting, so that we can read your code more easily. Also, there's no reason to post in multiple sub-forums or start new threads for the same problem.