PHP MySQL read from a table to use in a graph

Question

I have a graph that auto fills based on a few values. The values are the goal (total) the current (where it is at now) and in the total height of the graph. I am trying to pull the goal and current out of the database format it into money and it should fill the graph to the point it needs to be.

I had it working with a URL get which was ?current=584559096&goal=1000000000 and I just used $_GET. For easier maintenance for others around I want to pull it from a database which can be updated a post.

The code I have so far is:

    <? php function formatMoney($number, $fractional=false) {
        if ($fractional) {
            $number = sprintf('%.0f', $number);
        }
        while (true) {
            $replaced = preg_replace('/(-?\d+)(\d\d\d)/', '$1,$2', $number);
            if ($replaced != $number) {
                $number = $replaced;
            } else {
                break;
            }
        }
        return $number;
    }

        $goal = $row['goal'];
        $current = $row['current'];
        $percent = round($current / $goal * 100);
        $totalheight = 216;
        $ourheight = $totalheight * $percent / 100;

        $halfofarrowheight = 12;
            $arrowheight = $totalheight-$ourheight-$halfofarrowheight; 
            if($arrowheight < 0)
                $arrowheight = 0;
    mysql_close($con);
?>

Here is my query

mysql_select_db("databasename", $con);

$result = mysql_query("select * from `dbname`.`tablename");

This is my error Warning: Division by zero in E:\inetpub\wwwroot\test.website.net\web\includes\globalfoot.php on line 36

Line 36 is the $percent = round($current / $goal * 100);

So, I been fooling with this thing for 4 days now trying to just get the numbers out of the goal column and current column format them to money and have the chart fill. For test sake lets say the goal is 1000000000000 and the current is 5000000000.

Where is $row getting set? Can you show that code?

Brendan Bullen
– Brendan Bullen

2011-07-29 16:23:25 +00:00
Commented Jul 29, 2011 at 16:23 — Brendan Bullen
– Brendan Bullen, Commented Jul 29, 2011 at 16:23

totallyNotLizards · Accepted Answer · 2011-07-29 16:00:33Z

2

$row['goal'] = $goal;
$row['current'] = $current;

Please forgive me if I haven't understood your code, but shouldn't the above be:

$goal=$row['goal'];
$current=$row['current'];

answered Jul 29, 2011 at 16:00

totallyNotLizards

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8 Comments

EMAW and Cougs Over a year ago

Yeah, I switched them around to see if it would do anything and forgot to switch it back. No change on the error.

totallyNotLizards Over a year ago

again, not sure if i'm teaching my granny to suck eggs but have you tried just echoing the values of those two vars prior to the calculation? a division by zero warning suggests that $goal is zero or empty, you should verify that it gets set

EMAW and Cougs Over a year ago

Yes I did, I did they values come up on with an while ($row = mysql_fetch_array($result)) echo $row['goal'] and $row['current '].

totallyNotLizards Over a year ago

but what about just doing echo $row['goal']?

EMAW and Cougs Over a year ago

@ Dmitriy I don't understand, through out the rest it is just used as an <?php echo ($arrowheight)); ?>, <?php echo($totalheight-$ourheight);?>, <?php echo($ourheight);?>, <?php echo(formatMoney($current)); ?>, in the css and on top of the arrow. But the error is saying that it is not reading it from the table.... even though it will echo it on test.

|

lazyhammer · Accepted Answer · 2011-07-29 16:43:04Z

0

Your code should look like this:

mysql_select_db("databasename", $con);
$result = mysql_query("select * from `dbname`.`tablename");
if (!$result || !($row = mysql_fetch_assoc($result))) {
    echo "DB error: ".mysql_error();
    exit;
}

$goal = $row['goal'];
$current = $row['current'];
// ...

edited Jul 29, 2011 at 16:43

answered Jul 29, 2011 at 16:36

lazyhammer

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4 Comments

lazyhammer Over a year ago

So, maybe there is something with your query? See updated code

EMAW and Cougs Over a year ago

@ Dmitriy yeah it worked, I had missed a ; which is always a pain with PHP miss one little thing and it ruins your day.

EMAW and Cougs Over a year ago

Wish I could check both of the answers because you both helped.

EMAW and Cougs Over a year ago

Thank you very much for your help! I like the feeling when stuff actually works.

Mr Griever · Accepted Answer · 2011-07-29 16:12:42Z

0

Always, always double-check your variables when doing division.

$percent = ($goal > 0 || $goal < 0) ? round($current / $goal * 100) : 0;

answered Jul 29, 2011 at 16:12

Mr Griever

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2 Comments

EMAW and Cougs Over a year ago

Thank you Matt that got rid of the error. For some reason though my graph is not working still. Does not even echo the row column on the arrow. It is like it does not even read the db, even though the test echo shows it will pull the numbers out. Do I need to close the connection at the bottom of the page?

EMAW and Cougs Over a year ago

Wish I could check both of the answers because you both helped.

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