JavaKinetic

I am wondering what the best way is to have my Apache/PHP server accept an XML file that is sent at random using an ASP Send function. My client is using ASP (POST and SEND) to send a 0.5 to 2MB XML string, which I must immediately accept and save to disk. I have built the web pages to do this with normal user (html) interaction (ie: click to add a file, and upload by clicking submit), but I am stumped on how to do this through an automated means. I must be missing something really simple. Any thoughts? Cheers! '----- MY CLIENT'S ASP CODE THAT WILL SEND XML TO OUR PHP SERVER... ----- ' *** Set up a command to send an XML file to a remove server *** ' *** - Includes MP3 file Base64 encoded RawDir=Server.MapPath("/UserContent/" & NMD("CampaignID")) & "\Voice\" RS.Open "SELECT * FROM UserContent WHERE ContentID=" & ContentID , objApp.DB,adOpenStatic,adLockOptimistic,adCmdText ' *** Create XML object *** Set objXMLDoc = Server.CreateObject("Msxml2.DOMDocument.3.0") objXMLDoc.loadXML "<InstallVoice />" Set oRoot = objXMLDoc.documentElement ' *** Set up a bunch of elements here .. / 8< snip 8< ' *** Read the file into the stream *** Set oRoot = objXMLDoc.documentElement Set objStream = Server.CreateObject("ADODB.Stream") objStream.Type = 1 objStream.Open objStream.LoadFromFile RawDir & RS("UploadFile") ' *** Do base64 encoding for an MP3 file *** Set oElement = objXMLDoc.createElement("FileContents") oElement.dataType = "bin.base64" oElement.nodeTypedValue = objStream.Read oRoot.appendChild oElement Set xmlhttp = Server.CreateObject("MSXML2.ServerXMLHTTP") xmlhttp.open "POST", "http//myhostsomewhere.com/uploader.php", false ' *** Send the file up *** xmlhttp.send objXMLDoc.XML ' *** Close down the effort....