Saving XML HELP!!!

busher · November 5th, 2004, 04:00 PM

I am posting form information to a site that returns an xml page with data that our company needs. I am very new to the XML world, and I don't understand how to automatically save the returned xml page locally.

Since the returned page requires form data, I cannot do the following:

xml.Open "GET", "http://www.somesite.com", False

which is common on all of the examples I have seen on the net.
I would like our intranet users to be able to click on a page; it sends the form data, saves the returned page, and then runs a parser to consume the data.

Can someone please help me?

Ajay Swamy · November 23rd, 2004, 12:27 PM

This is what you do:

Use your appropriate MSXML Object, then use store the returned XML (from get or post) into a variable. Use the FileSystem Object to write the XML locally (hence cached XML). You can they reference the file into your transform code for XSL., You can also directly parse the stored XML in the variable. Warning: its a little slow!!

Sample code:

        Dim file_out1, objXMLHttp1, response_xml1
        Set objXmlHttp1 = Server.CreateObject("Msxml2.ServerXMLHTTP")
        objXmlHttp1.open "GET", YOUR_url & "YOUR PARAMETERS", False
        objXmlHttp1.send
        response_xml1 = objXmlHttp1.responseXML

NOW transform with XSL:

    Set objXML = Server.CreateObject("Microsoft.XMLDOM")
    objXML.async = False
    objXML.Load(response_xml1)

    Set objXSL = Server.CreateObject("Microsoft.XMLDOM")
    objXSL.async = false
    objXSL.load("YOUR location of XSL file - absolute URL")
    xmlResponseDoc = objXML.transformNode(objXSL)

and thats it.

joefawcett · November 23rd, 2004, 12:34 PM

This function retrieves any sort of file from the web:

Code:

function saveRemoteFile(From, To)
{
  var oXmlHttp = new ActiveXObject("Msxml2.XmlHttp.4.0");  //Change to version 3 if necessary
  oXmlHttp.open("GET", From, false);
  oXmlHttp.send();
  if (oXmlHttp.status == 200)
  {
    var oStream = new ActiveXObject("Adodb.Stream");
    oStream.type = 1; //Binary
    oStream.open();
    oStream.write(oXmlHttp.responseBody);
    oStream.saveToFile(To, 1 && 2); 
    oStream.close();
  }
  else
  {
    throw new Error(oXmlHttp.statusText);  //Needs improving
  }
}

saveRemoteFile("http://www.google.com", "C:\\myFolder\\myGooglePage.htm");

If you know the data is going to be XML then once you have opened and sent you can just do:

Code:

//create msxml2.xmlhttp.4.0, open and send
oXmlHttp.responseXML.save(<path to save to>);

--

Joe (Microsoft MVP - XML)