Re: Passing Variables and Internal Variables

Dash McElroy wrote:

You are referencing the variables properly on the first if line, but then you're calling $var1 and $var2 w/o using $_GET. Add that or do this: $var1 = $_GET['var1']; $var2 = $_GET['var2']; You may also want to use !isset($varname) instead of checking to see if the variables are equal to a space char (" "). isset() should not be preferred for variables coming from $_POST, $_GET or $_REQUEST

cause empty form-fields ARE set, so isset() will return TRUE use empty() instead!

-Dash "I have made mistakes but I have never made the mistake of claiming that I have never made one." -- James Gordon Bennett On Wed, 29 Jan 2003, Wade wrote:

01292003 1540 CST When I run this script I get the html at the end but no $result. If the result is run on the same page as the output, there shouldnt be anything variable wise stopping this from running, right? Wade <?php

    if (($_POST["val1"] == " ") || ($_POST["val2"] == " ") ||

($_POST["calc"] == " "))

    {
        header("Location: http://localhost/Learning PHP/PHP Fast &

Easy/Ch_6/calculate_form.html");

        exit;
    }

    if ($_POST["calc"] == "add")
    {
        $result = $val1 + $val2;
    }

        else if ($_POST["calc"] == "subtract")
        {
             $result = $val1 - $val2;
        }

        else if ($_POST["calc"] == "multiply")
        {
             $result = $val1 *  $val2;
        }

        else if ($_POST["calc"] == "divide")
        {
            $result = $val1 / $val2;
        }

you should use switch()

switch ($_POST["calc"]) { case "add": $result = $_POST["val1"] + $_POST["val2"]; break; .... default: $result = 'SOME ERROR'; }

?> <html> <head> <title>Calculation Result</title> <head> <body>

    <p>The result of the calculation is: <?php "$result"; ?></p>

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-- Sebastian Mendel info@sebastianmendel.de www.sebastianmendel.de www.tekkno4u.de www.nofetish.com