dawg1

<SCRIPT language=JavaScript> function reload(form) { var val=form.price.options[form.price.options.selectedIndex].value; self.location='dd3.php?price=' + val ; } function reload3(form) { var val=form.price.options[form.price.options.selectedIndex].value; var val2=form.weight.options[form.weight.options.selectedIndex].value; self.location='dd3.php?price=' + val + '&weight=' + val2 ; } </script> <? --GET DATA HERE-- echo "<form method=post name=f1 action='dd3.php'>"; --DROP DOWN LISTS HERE-- echo "<input type=submit value='View'></form>"; ?> How can I change the script above so that the results will be returned on the same page? Below is an AJAX function that can return the results to a div on the same page AJAX Script: <script language="javascript" type="text/javascript"> <!-- //Browser Support Code function ajaxFunction(){ var ajaxRequest; // The variable that makes Ajax possible! try{ // Opera 8.0+, Firefox, Safari ajaxRequest = new XMLHttpRequest(); } catch (e){ // Internet Explorer Browsers try{ ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try{ ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e){ // Something went wrong alert("Your browser broke!"); return false; } } } // Create a function that will receive data sent from the server ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxDiv'); ajaxDisplay.innerHTML = ajaxRequest.responseText; } } var price = document.getElementById('price').value; var weight = document.getElementById('weight').value; var taxable = document.getElementById('taxable').value; var queryString = "?price=" + price + "&weight=" + weight + "&taxable=" + taxable; ajaxRequest.open("GET", "ajax_example.php" + queryString, true); ajaxRequest.send(null); } //--> </script> <div id='ajaxDiv'>Your result will display here</div> I know there has to be a way to get what I need from these two functions, but I can't put it together! Any help would be much appreciated!!! Thanks!

GOT IT <a href=\"page/D/PROD/$row[code]\"> Thanks SO MUCH!

K, I did that, and made sure the case was correct, but still no luck. The data displays just fine when I don't try to make it a link. <?php $dbhost = ""; $dbuser = ""; $dbpass = ""; $dbname = ""; //Connect to MySQL Server mysql_connect($dbhost, $dbuser, $dbpass); //Select Database mysql_select_db($dbname) or die(mysql_error()); // Retrieve data from Query String $price = $_GET['price']; $taxable = $_GET['taxable']; $weight = $_GET['weight']; $code = $_GET['code']; // Escape User Input to help prevent SQL Injection $price = mysql_real_escape_string($price); $taxable = mysql_real_escape_string($taxable); $weight = mysql_real_escape_string($weight); $code = mysql_real_escape_string($code); //build query $query = "SELECT * FROM Products WHERE taxable = '$taxable'"; if(is_numeric($price)) $query .= " AND price <= $price"; if(is_numeric($weight)) $query .= " AND weight <= $weight"; //Execute query $qry_result = mysql_query($query) or die(mysql_error()); //Build Result String $display_string = "<table>"; $display_string .= "<tr>"; $display_string .= "<th>Code</th>"; $display_string .= "<th>Price</th>"; $display_string .= "<th>Taxable</th>"; $display_string .= "<th>Weight</th>"; $display_string .= "</tr>"; // Insert a new row in the table for each product returned while($row = mysql_fetch_array($qry_result)){ $display_string .= "<tr>"; $display_string .= "<td><a href="www.dawgwear.net/page/D/PROD/$code">$row[code]</a></td>"; $display_string .= "<td>$row[price]</td>"; $display_string .= "<td>$row[taxable]</td>"; $display_string .= "<td>$row[weight]</td>"; $display_string .= "</tr>"; } echo "Query: " . $query . "<br />"; $display_string .= "</table>"; echo $display_string; ?> I am using javascript on my webpage that displays the results...might there be something there I need to adjust as well? <script language="javascript" type="text/javascript"> <!-- //Browser Support Code function ajaxFunction(){ var ajaxRequest; // The variable that makes Ajax possible! try{ // Opera 8.0+, Firefox, Safari ajaxRequest = new XMLHttpRequest(); } catch (e){ // Internet Explorer Browsers try{ ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try{ ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e){ // Something went wrong alert("Your browser broke!"); return false; } } } // Create a function that will receive data sent from the server ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxDiv'); ajaxDisplay.innerHTML = ajaxRequest.responseText; } } var price = document.getElementById('price').value; var weight = document.getElementById('weight').value; var taxable = document.getElementById('taxable').value; var queryString = "?price=" + price + "&weight=" + weight + "&taxable=" + taxable; ajaxRequest.open("GET", "ajax_example.php" + queryString, true); ajaxRequest.send(null); } //--> </script>

Thanks for responding so fast! I tried your suggestion but when I add the html code for the link no results display. Did I set it up correctly? <?php $dbhost = ""; $dbuser = ""; $dbpass = ""; $dbname = ""; //Connect to MySQL Server mysql_connect($dbhost, $dbuser, $dbpass); //Select Database mysql_select_db($dbname) or die(mysql_error()); // Retrieve data from Query String $price = $_GET['price']; $taxable = $_GET['taxable']; $weight = $_GET['weight']; $code = $_GET['code']; // Escape User Input to help prevent SQL Injection $price = mysql_real_escape_string($price); $taxable = mysql_real_escape_string($taxable); $weight = mysql_real_escape_string($weight); $code = mysql_real_escape_string($code); //build query $query = "SELECT * FROM Products WHERE taxable = '$taxable'"; if(is_numeric($price)) $query .= " AND price <= $price"; if(is_numeric($weight)) $query .= " AND weight <= $weight"; //Execute query $qry_result = mysql_query($query) or die(mysql_error()); //Build Result String $display_string = "<table>"; $display_string .= "<tr>"; $display_string .= "<th>Code</th>"; $display_string .= "<th>Price</th>"; $display_string .= "<th>Taxable</th>"; $display_string .= "<th>Weight</th>"; $display_string .= "</tr>"; // Insert a new row in the table for each person returned while($row = mysql_fetch_array($qry_result)){ $display_string .= "<tr>"; $display_string .= "<td><a href="www.storefront.net?$Code">$row[code]</a></td>"; $display_string .= "<td>$row[price]</td>"; $display_string .= "<td>$row[taxable]</td>"; $display_string .= "<td>$row[weight]</td>"; $display_string .= "</tr>"; } echo "Query: " . $query . "<br />"; $display_string .= "</table>"; echo $display_string; ?>

I've tried to find an answer in the forum and although there have been posts similar I am unable to covert the answer to fit what I've got (mostly because I know very very little about php - sorry). What I need is one of the results to be added to the end of a url in order to create a link. Basically I have a storefront (www.storefront.com/) and the result NAME can be added onto the end of our url to create a link to the product page. (www.storefront.com/NAME) <?php $dbhost = ""; $dbuser = ""; $dbpass = ""; $dbname = ""; //Connect to MySQL Server mysql_connect($dbhost, $dbuser, $dbpass); //Select Database mysql_select_db($dbname) or die(mysql_error()); // Retrieve data from Query String $price = $_GET['price']; $taxable = $_GET['taxable']; $weight = $_GET['weight']; // Escape User Input to help prevent SQL Injection $age = mysql_real_escape_string($price); $sex = mysql_real_escape_string($taxable); $wpm = mysql_real_escape_string($weight); //build query $query = "SELECT * FROM Products WHERE taxable = '$taxable'"; if(is_numeric($price)) $query .= " AND price <= $price"; if(is_numeric($weight)) $query .= " AND weight <= $weight"; //Execute query $qry_result = mysql_query($query) or die(mysql_error()); //Build Result String $display_string = "<table>"; $display_string .= "<tr>"; $display_string .= "<th>Name</th>"; $display_string .= "<th>Price</th>"; $display_string .= "<th>Taxable</th>"; $display_string .= "<th>Weight</th>"; $display_string .= "</tr>"; // Insert a new row in the table for each person returned while($row = mysql_fetch_array($qry_result)){ $display_string .= "<tr>"; $display_string .= "<td>'<a href="www.storefront.com/">' . $row[name] . ' </a>'/td>"; $display_string .= "<td>$row[price]</td>"; $display_string .= "<td>$row[taxable]</td>"; $display_string .= "<td>$row[weight]</td>"; $display_string .= "</tr>"; } echo "Query: " . $query . "<br />"; $display_string .= "</table>"; echo $display_string; ?> I know the way it is setup in the code is wrong. Can anyone help me out? Would really appreciate it. Thanks - and again, sorry for my lack of php knowledge.