com - How can I retrieve 3rd & 4th Parameter of MSXML2.ServerXMLHTTP' Open method? -


i using following code:

 set objxmlhttp = createobject("msxml2.serverxmlhttp")  objxmlhttp.open "post", url + "/" + method, false,uid,password  objxmlhttp.send ' not sending custom parameters  wscript.sleep 500  loop  msgbox objxmlhttp.responsetext

is there way retrieve uid , password sent inside url (i have got code web service)?

ps: information seek not in httprequest.current.request.headers.

afaik need set header content-type: application/x-www-form-urlencoded post request, try changing code this:

set objxmlhttp = createobject("msxml2.serverxmlhttp") objxmlhttp.open "post", url + "/" + method, false, uid, password objxmlhttp.setrequestheader "content-type", "application/x-www-form-urlencoded" objxmlhttp.send

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