Submitting form with ajax?

[Vernier]

Hey, I use this code:

<li><a href="#" id="jobs"></a>
                        <ul>
                        <li><a href="javascript:ajaxpage('/staffpanel/_frontend/jobApps.php', 'content_container');">Open Jobs</a></li> 
                        </ul> 
                        </li>

to display a job application form on my site. When they click "Open jobs" the job apps page loads into the content container. When I click submit on the form, it simply refreshes the page. My question is: How do I submit the form so that it doesn't just refresh the page?

Thanks

~ David

[abduraooft]

Where's the definition of function ajaxpage()?

[Vernier]

Originally Posted by abduraooft

Where's the definition of function ajaxpage()?

This is in my head tags if this is what you mean:

<script type="text/javascript">

/***********************************************
* Dynamic Ajax Content- © Dynamic Drive DHTML code library (www.dynamicdrive.com)
* This notice MUST stay intact for legal use
* Visit Dynamic Drive at http://www.dynamicdrive.com/ for full source code
***********************************************/

var bustcachevar=1 //bust potential caching of external pages after initial request? (1=yes, 0=no)
var loadedobjects=""
var rootdomain="http://"+window.location.hostname
var bustcacheparameter=""

function ajaxpage(url, containerid){
var page_request = false
if (window.XMLHttpRequest) // if Mozilla, Safari etc
page_request = new XMLHttpRequest()
else if (window.ActiveXObject){ // if IE
try {
page_request = new ActiveXObject("Msxml2.XMLHTTP")
}
catch (e){
try{
page_request = new ActiveXObject("Microsoft.XMLHTTP")
}
catch (e){}
}
}
else
return false
page_request.onreadystatechange=function(){
loadpage(page_request, containerid)
}
if (bustcachevar) //if bust caching of external page
bustcacheparameter=(url.indexOf("?")!=-1)? "&"+new Date().getTime() : "?"+new Date().getTime()
page_request.open('GET', url+bustcacheparameter, true)
page_request.send(null)
}

function loadpage(page_request, containerid){
if (page_request.readyState == 4 && (page_request.status==200 || window.location.href.indexOf("http")==-1))
document.getElementById(containerid).innerHTML=page_request.responseText
}

function loadobjs(){
if (!document.getElementById)
return
for (i=0; i<arguments.length; i++){
var file=arguments[i]
var fileref=""
if (loadedobjects.indexOf(file)==-1){ //Check to see if this object has not already been added to page before proceeding
if (file.indexOf(".js")!=-1){ //If object is a js file
fileref=document.createElement('script')
fileref.setAttribute("type","text/javascript");
fileref.setAttribute("src", file);
}
else if (file.indexOf(".css")!=-1){ //If object is a css file
fileref=document.createElement("link")
fileref.setAttribute("rel", "stylesheet");
fileref.setAttribute("type", "text/css");
fileref.setAttribute("href", file);
}
}
if (fileref!=""){
document.getElementsByTagName("head").item(0).appendChild(fileref)
loadedobjects+=file+" " //Remember this object as being already added to page
}
}
}

</script>

Thanks

~ David

[Vernier]

-bump- any1 know if this is possible? Thanks

~ David

[Vernier]

Sorry for bump again, anyone know how this can be done? I can provide the ajax page's code if that would help?

Cheers,

~ David

[Vernier]

Hey guys, been searching for a while now and maybe ajaxform() , not sure if this would work or even how to implement it. Any ideas?

Thanks guys

~ David

[defencedog]

hey I have a similar problem but quiet imple can anyone help:
http://www.codingforums.com/showthread.php?t=250621

[DaveyErwin]

not ajax but this is how to
post a form

document.forms[0].onsubmit = function () {
        var a = this.elements;
        var b = "";
        for (var i = a.length; i--; ) if (a[i].name) b += a[i].name + "=" + encodeURIComponent(a[i].value) + "&";
        var req = new XMLHttpRequest;
        req.open("POST", "whatever.php", false);
        req.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        req.setRequestHeader("Content-length", b.length);
        req.setRequestHeader("Connection", "close");
        req.send(b);
        elDiv.innerHTML =  req.responseText;
        return false;
    };

[Vernier]

Originally Posted by defencedog

hey I have a similar problem but quiet imple can anyone help:
http://www.codingforums.com/showthread.php?t=250621

Originally Posted by DaveyErwin

not ajax but this is how to
post a form

document.forms[0].onsubmit = function () {
        var a = this.elements;
        var b = "";
        for (var i = a.length; i--; ) if (a[i].name) b += a[i].name + "=" + encodeURIComponent(a[i].value) + "&";
        var req = new XMLHttpRequest;
        req.open("POST", "whatever.php", false);
        req.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        req.setRequestHeader("Content-length", b.length);
        req.setRequestHeader("Connection", "close");
        req.send(b);
        elDiv.innerHTML =  req.responseText;
        return false;
    };

The way I got round it was use ajax to load the contents of the page, but have an iframe in a seperate document.


~ David