nezbo Posted March 19, 2008 Share Posted March 19, 2008 Hi all please can some one help me... I dont know javascript to well and i know aJax even less, i am looking to call a function that gets info from a database and then puts it in to a text box. i can do the textbox thing once have the info in Javascript but i dont know how to get it from PHP/MySql to javascript. i have looked all over the internet for a turotial and it has goven me a spliting headake... :'( here is my code so far : the javascript/ajax bit <script type="text/javascript"> function getStaffHours() { var theTotalHours1 = ajax.getElementById('thetotalhours', 'aJaxUserQ.php'); document.getElementById("staffHours1").value = theTotalHours1; } </script> the form bit echo "<select name='theUser' id='nameSelected' onchange='getStaffHours(this.value)'><option value=\"\"></option>"; the PHP bit (ajaxUserQ.php) <?php $getTheUser = mysql_query("SELECT * FROM person WHERE CallID = '" . $_COOKIE['user'] . "'"); while ($getTheUser2 = mysql_fetch_array($getTheUser)) { $userHours = $getTheUser2['contHours']; } ?> <div id='thetotalhours'><?php echo $userHours; ?></div> Quote Link to comment Share on other sites More sharing options...
nezbo Posted March 19, 2008 Author Share Posted March 19, 2008 my function now looks like this : <script type="text/javascript"> function ajaxFunction(AJAXpage) { var xmlHttp; try { // Firefox, Opera 8.0+, Safari xmlHttp=new XMLHttpRequest(); } catch (e) { // Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } catch (e) { alert("Your browser does not support AJAX!"); return false; } } } xmlHttp.onreadystatechange=function() { if(xmlHttp.readyState==4) { document.getElementById("nameSelected").value=xmlHttp.responseText; } } xmlHttp.open("GET","AJAXpage",true); xmlHttp.send(null); } function getStaffHours(AJAXpage) { ajaxFunction("aJaxUserQ.php"); } </script> and it still dose nothing :'( I thin it might be some thing to do with the php page not giving anything out. Quote Link to comment Share on other sites More sharing options...
nezbo Posted March 19, 2008 Author Share Posted March 19, 2008 i think i may be nearly there.... I am not geting this comming back in to the text box... <?xml version="1.0" encoding="ISO-8859-1"?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"><html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"><head><title>Object not found!</title><link rev="made" href="mailto:admin@localhost" /><style type="text/css"><!--/*--><![CDATA[/*><!--*/ body { color: #000000; background-color: #FFFFFF; } a:link { color: #0000CC; } p, address {margin-left: 3em;} span {font-size: smaller;}/*]]>*/--></style></head><body><h1>Object not found!</h1><p> The requested URL was not found on this server. The link on the <a href="http://localhost/call_Log/staffHours.php">referring page</a> seems to be wrong or outdated. Please inform the author of <a href="http://localhost/call_Log/staffHours.php">that page</a> about the error. </p><p>If you think this is a server error, please contactthe <a href="mailto:admin@localhost">webmaster</a>.</p><h2>Error 404</h2><address> <a href="/">localhost</a><br /> <span>03/19/08 17:03:17<br /> Apache/2.2.4 (Win32) DAV/2 mod_ssl/2.2.4 OpenSSL/0.9.8d mod_autoindex_color PHP/5.2.1</span></address></body></html> and i am expacting just the number 37... here is my code again... function ajaxFunction(AJAXpage) { var xmlHttp; try { // Firefox, Opera 8.0+, Safari xmlHttp=new XMLHttpRequest(); } catch (e) { // Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } catch (e) { alert("Your browser does not support AJAX!"); return false; } } } xmlHttp.onreadystatechange=function() { if(xmlHttp.readyState==4) { document.getElementById("staffHours1").value=xmlHttp.responseText; } } xmlHttp.open("GET","AJAXpage",true); xmlHttp.send(null); } function getStaffHours() { ajaxFunction("/aJaxUserQ.php"); } </script> and my php (if i run this on it's own it brings back the no 37) <?php include ("include/dbcon.php"); $getTheUser = mysql_query("SELECT * FROM person WHERE CallID = '" . $_COOKIE['user'] . "'"); while ($getTheUser2 = mysql_fetch_array($getTheUser)) { echo $userHours = $getTheUser2['contHours']; } ?> Quote Link to comment Share on other sites More sharing options...
nezbo Posted March 19, 2008 Author Share Posted March 19, 2008 i think i have narrowed it down to not finding the file, but the file is in the same folder and is name correctly. i am going bolder by the second :'( here is my letest code : <script type="text/javascript"> function ajaxFunction(AJAXpage) { var xmlHttp; try { // Firefox, Opera 8.0+, Safari xmlHttp=new XMLHttpRequest(); } catch (e) { // Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } catch (e) { alert("Your browser does not support AJAX!"); return false; } } } xmlHttp.onreadystatechange=function() { if(xmlHttp.readyState==4) { document.getElementById("staffHours1").value=xmlHttp.responseText; } } xmlHttp.open("GET","AJAXpage",true); xmlHttp.send(null); } function getStaffHours() { var theSelectedUser = document.getElementById("nameSelected").value; ajaxFunction("aJaxUserQ.php&nameSelected =" . theSelectedUser); } </script>code] Quote Link to comment Share on other sites More sharing options...
korbinus Posted March 23, 2008 Share Posted March 23, 2008 Well, it's difficult to understand without trying it a little bit, but I just noticed a mistake in your last function. You wrote function getStaffHours() { var theSelectedUser = document.getElementById("nameSelected").value; ajaxFunction("aJaxUserQ.php&nameSelected =" . theSelectedUser); } but you should put a question mark (?) after your script extension, definitely not a paramter separator (&): function getStaffHours() { var theSelectedUser = document.getElementById("nameSelected").value; ajaxFunction("aJaxUserQ.php?nameSelected =" . theSelectedUser); } Quote Link to comment Share on other sites More sharing options...
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