f1r3fl3x Posted November 4, 2007 Share Posted November 4, 2007 Can someone help me. I started learning ajax a week ago and i don't know much. I tryed to create a submit form with POST but it didn't work. Please someone help me. Here's the code. form.html <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <link rel="stylesheet" type="text/css" href="style.css"> <script type="text/javascript" language="javascript"> var http_request = false; function show_hint ( p_hint_text, p_span ) { document.getElementById(p_span).innerHTML = p_hint_text ; } function makePOSTRequest(url, parameters, SpanName) { http_request = false; if (window.XMLHttpRequest) { // Mozilla, Safari,... http_request = new XMLHttpRequest(); if (http_request.overrideMimeType) { http_request.overrideMimeType('text/xml'); } } else if (window.ActiveXObject) { // IE try { http_request = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try { http_request = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e) {} } } if (!http_request) { alert('Cannot create XMLHTTP instance'); return false; } http_request.onreadystatechange = function() { if (http_request.readyState == 4) { if (http_request.status == 200) { //alert(http_request.responseText); result = http_request.responseText; document.getElementById(SpanName).innerHTML = result; document.getElementById('status').innerHTML = 'Ready'; } else { alert('There was a problem with the request.'); } } }; http_request.open('POST', url, true); http_request.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); http_request.setRequestHeader("Content-length", parameters.length); http_request.setRequestHeader("Connection", "close"); http_request.send(parameters); } function Contact(obj,SpanName) { var curDateTime = new Date(); //For IE var poststr = "name=" + encodeURI( document.getElementById("name").value ) + "&email=" + encodeURI( document.getElementById("email").value ) + "&uniqueID=" + curDateTime + "&msg=" + encodeURI( document.getElementById("msg").value ) ; var SpanName = SpanName; //alert (SpanName); makePOSTRequest('Process.php', poststr, SpanName); } </script> </head> <body> <span id="myspan"> <table width="316" border="0" cellpadding="4" cellspacing="0"> <form name="ContactForm" id="ContactForm" action="javascript:Contact(document.getElementById('ContactForm'),'myspan');show_hint('Sending Data... Please wait!', 'status');" method="post"> <tr> <td width="141" align="right" >Name:</td> <td width="157"><input name="name2" type="text" id="name" size="20" maxlength="50" /></td> </tr> <tr> <td align="right" >Email:</td> <td><input name="email2" type="text" id="email" size="20" maxlength="50" /></td> </tr> <tr> <td align="right">Message:</td> <td><input name="msg2" type="text" id="msg" size="20" maxlength="50" /></td> </tr> <tr> <td align="right"> </td> <td><input name="Submit2" type="submit" value="Submit" /></td> </tr> </form> </table> </span> </body> </html> Process.php <?php $name = stripslashes($_POST['name']); $email = stripslashes($_POST['email']); $msg = stripslashes($_POST['msg']); echo " <table width=\"316\" border=\"0\" cellpadding=\"4\" cellspacing=\"0\"> <tr> <td align=\"left\">Your Information has been processed</td> </tr> <tr> <td align=\"left\">Name: $name <br> Email: $email <br> Message: $msg</td> </tr> </table>"; ?> :'( Link to comment https://forums.phpfreaks.com/topic/76001-ajax-post-request-help/ Share on other sites More sharing options...
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