dawg1 Posted April 21, 2011 Share Posted April 21, 2011 <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! Link to comment https://forums.phpfreaks.com/topic/234378-how-to-submit-triple-dropdown-result-on-same-page/ Share on other sites More sharing options...
dawg1 Posted April 21, 2011 Author Share Posted April 21, 2011 --------EDIT FOR INITIAL POST-------------(not sure why I can't modify the original post) echo "<form method=post name=f1 action='dd3.php'>"; is actually echo "<form method=post name=f1 action='dd3ck.php'>"; in my current code <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='dd3ck.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! Link to comment https://forums.phpfreaks.com/topic/234378-how-to-submit-triple-dropdown-result-on-same-page/#findComment-1204594 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.