Warning: Division by zero in ... on line 133

Question

Basically, i've made a script that login to a website, get elements from the website, and divide one element with another. The problem is, i have already checked plenty of times, and the divisor is NOT zero. here is the snippet:

if($num==0)
{
    echo "<td>".$estate_income."/".$num."</td>";
}
else
{
    echo "<td>".$estate_income/$num."</td>";
}

this would output something like 50/2000, which means the if statement is true, which means that $num is somehow equal to 0. if i try to divide the two variables with each other, it would output php warning diivision by zero:

echo "<td>".$estate_income/$num."</td>";

What i ask now is, for a solution, maybe some error detection methods that could tell me what i am doing wrong. it's probably something very obvious that i have overlooked.

Thanks in advance!

Ahmad Albayati

Edit:

            try
            {
                echo "<td>".$estate_income/$num."</td>";
            }
            catch(Exception $e)
            {
                echo 'Caught exception: ',  $e->getMessage(), "\n";
            }

Outputs Warning: Division by zero in . . . on line 133

The variables i am trying to divide is within a foreach loop.

The $num variable is defined in a function that gets included:

function num_format($n)
{
    $n=str_replace(",","",str_replace(" ","",$n));
    if(strpos($n,".")===FALSE)
    {
        if(strpos($n,"K")!==FALSE)
        {
            $n=str_replace("$","",str_replace("K","000",$n));
        }
        elseif(strpos($n,"mil")!==FALSE)
        {
            $n=str_replace("$","",str_replace("mil","000000",$n));
        }
        elseif(strpos($n,"bil")!==FALSE)
        {
            $n=str_replace("$","",str_replace("bil","000000000",$n));
        }
        elseif(strpos($n,"tril")!==FALSE)
        {
            $n=str_replace("$","",str_replace("tril","000000000000",$n));
        }
        else
        {
            $n=str_replace("$","",$n);
        }
    }
    else
    {
        $n=str_replace(".","",$n);
        if(strpos($n,"K")!==FALSE)
        {
            $n=str_replace("$","",str_replace("K","00",$n));
        }
        elseif(strpos($n,"mil")!==FALSE)
        {
            $n=str_replace("$","",str_replace("mil","00000",$n));
        }
        elseif(strpos($n,"bil")!==FALSE)
        {
            $n=str_replace("$","",str_replace("bil","00000000",$n));
        }
        elseif(strpos($n,"tril")!==FALSE)
        {
            $n=str_replace("$","",str_replace("tril","00000000000",$n));
        }
        else
        {
            $n=str_replace("$","",$n);
        }
    }
    global $num;
    $num=$n;
}

Answer 1

Found the problem. As I have already mentioned, I got the values from a website. When I use var_dumb on the string, it outputs string(149) "202000". This actually had confused me, because the string/number was only 6 characters. so I decided to look at the source code, and I found this:

<spanstyle="white-space:nowrap;"><imgsrc="http: staticstorm8com="" vl="" images="" bloodpng?v="330&quot;width=&quot;9&quot;height=&quot;12&quot;style=&quot;padding-right:2px&quot;">202000<br></imgsrc="http:></spanstyle="white-space:nowrap;">

I got some other HTML elements in my variable when I was taking the values from the website.