PHP - Php Linking Question

I found this code which makes a BMI Calculator (Form) for me however when I click on submit it takes the user back to the index page ie. domain.com/index.php. How do I change it to go to, say, domain.com/calculator.php ?    The code is below:

<?

/**
 * @package Module Body Mass Index Calculator for Joomla! 1.5
 * @version $Id: mod_bodymassindexcalculator.php 599 2010-03-20 23:26:33Z you $
**/


defined( '_JEXEC' ) or die( 'Restricted access' );

$heightcm=$_POST["heightcm"];
$weightkg=$_POST["weightkg"];

if ($heightcm!="" && $weightkg!="") {

$heightm = $heightcm / 100;
$bmi=round($weightkg / ($heightm*$heightm),1);

echo "Heigth, m: ".$heightm."<br />";
echo "Weigth, kg: ".$weightkg."<br />";
echo "Body Mass Index (BMI): ".$bmi."<br />";
echo "<strong>";

if ($bmi<16.5) {echo "Severely Underweight</strong><br />";}
if ($bmi>=16.5 && $bmi<=18.4) {echo "Underweight</strong><br />";}
if ($bmi>=18.5 && $bmi<=24.9) {echo "Normal</strong><br />";}
if ($bmi>=25 && $bmi<=29.9) {echo "Overweight</strong><br />";}
if ($bmi>=30 && $bmi<=34.9) {echo "Obese Class I</strong><br />";}
if ($bmi>=35 && $bmi<=39.9) {echo "Obese Class II</strong><br />";}
if ($bmi>=40) {echo "Obese Class III</strong><br />";}
echo "<br />";

}


$domain = $_SERVER['HTTP_HOST'];
$path = $_SERVER['SCRIPT_NAME'];
$queryString = $_SERVER['QUERY_STRING'];
$url = "http://" . $domain . $path;
$url3 = "http://" . $domain . $_SERVER['REQUEST_URI'];
$mystring1="?";
$s1=strpos($url3,$mystring1);
if($s1==0) {$url2=$url3;}
if($s1!=0) {$url2=substr($url3,0,$s1);}
$path = $url2;

//1 foot = 0.3048 meters
//1 inch = 2.54 centimeters
//1 pound = 0.45359237 kilograms

$n1=230;


echo "<table style=\"width: 100%\" cellspacing=\"0\" cellpadding=\"0\" align=\"center\"><tr><td valign=\"top\">";
//echo "<h3>BMI Calculator</h3>";
echo "<form action=\"".$path."\" method=\"post\" >";
echo "<strong>Height</strong><br />";
echo "<select name=\"heightcm\" >";
for ($i=30; $i<=$n1; $i++){ echo "<option value=\"$i\">".$i." cm / ".floor($i / 30.48)." ft ".round(($i-(floor($i / 30.48)*30.48)) / 2.54, 1)." in </option>";}
echo "</select>";
echo "<br />";
echo "<strong>Weight</strong><br />";
echo "<select name=\"weightkg\" >";
for ($i=30; $i<=$n1; $i++){ echo "<option value=\"$i\">".$i." kg / ".round($i / 0.45359237,2)." pounds </option>";}
echo "</select>";
echo "<br />";
//echo "<input name=\"searchterm\" type=text size=\"27\" class=\"ns1\">";
echo "<br />";
echo "<input type=\"submit\" value=\"Calculate\" name=\"B1\">";
echo "</form><br />";

//DON'T REMOVE THIS LINK - DO NOT VIOLATE GNU/GPL LICENSE!!!
echo "<a href=\"http://nutritioncaloriecounter.com\">Nutrition Calorie Counter</a>";
//DON'T REMOVE THIS LINK - DO NOT VIOLATE GNU/GPL LICENSE!!!
echo "</td></tr></table>";




?>

I have been trying to figure out how this is done?  One php file and all that changes is the name of the brand and the logo.  This has to be clickable from a menu and also if a user changes the brand name it changes to the appropriate name and logo.  Any ideas as to how this is done?  If you can guide me in the right direction or give me an example of how this is done would be greatly appreciated.  Thanks.


http://www.drivermanager.com/en/download-confirmation.php?brand=compaq&logo=

Hi,

I have an issue with a banner I integrated in a page (http://hostelsuites.com) if you go to the page you'll see a banner on the left side , called "Combo Andes". The problem I am encountering is that this banner is linking to inside pages, and it works all fine for me but does not on some other computers, although they are using the same explorer or firefox versions....

I have no idea as to what could trigger this type of error, any help would be muche welcome.

Some details that might help :
The link is winthin the flash movie
The code that brings the banner up depending on the chosen language is

<div class="subtitulo"><?= ucfirst($lang["combomendoza"]) ?></div>

 

Thank you very much in advance

is there a way in php to link from the root dir ?

like in html you just use the '/' at the start for the link " <a herf="/link.php" ></a> "

but i noticed this does not work when using php like include or require.

so is there anywey to tell a link to start from root dir? without using the ../../link.php

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=326928.0