MySQL into PHP Variables

willgooders

Hi,

I am trying to use information from a database to do some simple calculations. I am struggling with turning the information into variables and then using these variables in some calculations.

Here is the code i have so far. If i delete calc 1, 2 and 3 it will display all the data.

<?PHP

$host="localhost";
$username="root"; 
$password="root"; 
$db_name="research"; 
$tbl_name="users"; 

mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$query  = "SELECT windows, heating FROM users WHERE username ='test'";
$result = mysql_query($query);

while($row = mysql_fetch_row($result))
{
    $windows = $row[0];
    $heating = $row[1];

$calc1 = ($heating*0.1);
$calc2 = ($windows*100);
$calc3 = (($windows*100)/$heating*0.1));

echo 	        "<h2>Double Glazing Savings Calculator</h2>" .
		"<p>Number of Windows: $windows</p>" .
		"<p>Yearly Heating Cost: £$heating</p>" .
		"<p>Cost of Replacement: £$calc2</p>" .
		"<p>Annual Saving: £$calc1</p>" .
		"<p>Payback Period: $calc3 years</p>"
}

?>

Can anyone please help me?

Will

laserlight

What does it display now?

willgooders

It currently displays a blank screen. I am assuming my code is not working somewhere.

bradgrafelman

willgooders wrote:
I am assuming my code is not working somewhere.

Although your assumption is correct, you shouldn't have to assume in the first place. Turn on display_errors (or log_errors) and set error_reporting to E_ALL. You'll then see that PHP is trying to tell you that you have a parse error (missing semicolon).

willgooders

Thanks for that, it is still saying there is an error in line 85 of the code which does not exist.

I probs need help from someone who can in simple terms explain how to tax the two results from the MySQL table, turn them into variables of some sort and then use these variables in calculations, with the results of these being displayed.

bradgrafelman

Did you fix the parse error (missing semicolon after the echo statement)?

willgooders

Yes i did that, that was a silly error. Thanks for pointing it out. I am now unsure why it is not working. I no it will display the information so its not about getting the info from the database.

bradgrafelman

Does it show the echo() statement without the calculated data, or does it not show the echo statement at all (possibly indicating that no row(s) were matched in the D😎?

willgooders

It does not show the echo statment.

Rows are selected from the database as when i remove $calc1, $calc2 and $calc3 and their associated bits and bobs it works. What i am wanting is to take the values from the database (numbers) and put them through some simple calculations. Its the one small thing holding my dissertation up and im sure it will be some simple fix and hopefully a ureka moment.

Thanks for the help!!!

Weedpacket

You're a bit overboard on your parentheses:

    $calc1 = ($heating*0.1);
    $calc2 = ($windows*100);
    $calc3 = (($windows*100)/$heating*0.1));

None of the parentheses there are necessary (especially since $calc3 can be calculated given $calc1 and $calc2); and in fact you're missing one, which would lead to a parse error.

But those could be calculated in the database.