Question

Before this question gets stamped as a duplicate, I am sorry! I’ve read ALL the duplicate questions and if anything, it has confused me even more. So maybe this question is slightly different.

I’ve written a little Javascript library that makes ajax calls and fetches and parses information from the graph facebook API.

This enables me to pretty much show all my page status’ on my web page. However I’m just about to launch, and I have done as much testing as I can.

However. I’m sure errors will occur, and I’ve written many error catches blah blah blah.

What I want to do, is save all my errors in a xml file.

So when an error occurs, I want the javascript to load the xml file from the server, add the errors, then save the changes.

I know how to load the xml doc using XmlHttpRequests, And I’m sure I can figure out how to modify the xml just by using dom manipulation.

All i really want to know is. How do i save these changes? does it save automatically?

Or do i have to “somehow” pass the updated xml version to php and get that to save it?

Im not quite sure how to go about it.

I would use mySQL and php but that means “somehow” passing the error information to php, then saving it.

However id much prefer XML seeing as I’m the only person that will be reading the xml file.

Thanks very much.

Alex

Answer 1

Or do i have to “somehow” pass the updated xml version to php and get that to save it?

Yes, you’ll want to use an XML HTTP request to send the XML DOM to the server where PHP can save it:

function postXML(xmlDOM, postURL, fileName, folderPath){
try{     
  // Create XML HTTP Request Object
  oXMLReq = new ActiveXObject("MSXML2.XMLHTTP.3.0");
  // Issue Synchronous HTTP POST 
  oXMLReq.open("POST",postURL,false);
  // Set HTTP Request Headers
  if(fileName != null){ oXMLReq.setRequestHeader("uploadFileName", fileName); }        // What should file be named when saved on server?
  if(folderPath != null){ oXMLReq.setRequestHeader("uploadDir", folderPath); }   // What folder should file be saved in on server?
  // SEND XML
  ///WScript.Echo(xmlDOM.xml);
  oXMLReq.send(xmlDOM.xml);
  return oXMLReq.responseText;
 }catch(e){
  return "postXML failed - check network connection to server";
 }
}