HttpWebRequest in VB6 2

[kennedymr2]

I have some sample code as below, which i need to emulate in vb6
-----------------------
HttpWebRequest httpRequest = (HttpWebRequest)WebRequest.Create("

https://xxxxxxx.xxx.com/serv/wserv.ashx");

httpRequest.Method = "POST";
httpRequest.ContentType = "application/x-

http://www-form-urlencoded";

byte[] bytePostData = Encoding.UTF8.GetBytes(GetPostData());
httpRequest.ContentLength = bytePostData.Length;

using (Stream datastream = httpRequest.GetRequestStream())
{
datastream.Write(bytePostData, 0, bytePostData.Length);
datastream.Close();
}

HttpWebResponse httpResponse = (HttpWebResponse)httpRequest.GetResponse();

string responseString = "";
using (Stream dataStream = httpResponse.GetResponseStream())
{
StreamReader reader = new StreamReader(dataStream);
responseString = reader.ReadToEnd();
reader.Close();
dataStream.Close();
}

Console.WriteLine(responseString);
}

private static string GetPostData()
{
StringBuilder sUrl = new StringBuilder();

sUrl.AppendFormat("u={0}&p={1}&method={2}", 5XXXXXX, XXXX, "Method1");
sUrl.Append("&data=xxxxxxxxxx" }]}");


return sUrl.ToString();
}
}
}



??????????? How do i perform this function in vb6


Really appreciate some advice

Regards Kennedymr2

 

[strongm]

Untested, given I don't have access to a relevant web service and also the apparant redaction in your example:

[blue]' Requires a reference to Microsoft XML library
Dim sURL As String
Dim strXML As String
Dim xmlhttp As MSXML2.xmlhttp

Set xmlhttp = New xmlhttp

sURL = "[URL unfurl="true"]https://xxxxxxx.xxx.com/serv/wserv.ashx"[/URL]
strXML = "u=5XXXXXX&p=XXXX&method=Method1&data=xxxxxxxxxx" 'probably - but suspect all the Xs are the result of redaction or placeholding

xmlhttp.Open "POST", sURL, False
xmlhttp.setRequestHeader "Content-Type", "application/x-[URL unfurl="true"]www-form-urlencoded"[/URL]
xmlhttp.send strXML
MsgBox xmlhttp.responseXML.XML 'displays any response string[/blue]

 

[kennedymr2]

Thanks strongm for coming to the rescue again..


The server is off line for 12 hours from now... i will give it a test in the morning.

I used the xxx's to blank out the server address , as i felt i should keep it private...

I really need to get this going.. and will appreciate the help once again


Regards Kennedymr2

 

[kennedymr2]

strongm..

I have done some testing ... the send does not come up with any error.. but i seem to get a blank response.

I have tried a few things..

Dim ll
For ll = 1 To 888888
DoEvents
Next ll

MsgBox ("a")

MsgBox (xmlhttp.responseXML.xml) 'displays any response string
MsgBox (xmlhttp.ResponseText)


Later on today i can actually check if the send is getting to the server properly.

I notice on searching the web a few other people have had this problem. but none with a "solved"

Appreciate any ideas

Regards Kennedymr2

 

[kennedymr2]

Have done a little more

MsgBox (xmlhttp.ReadyState) 'result = 4 looks ok ???


MsgBox (xmlhttp.responseXML.xml) ' get nothing
MsgBox (xmlhttp.ResponseText)'get nothing
MsgBox (xmlhttp.GetAllResponseHeaders) 'get server info etc
MsgBox (xmlhttp.ResponseBody) 'nothing
MsgBox (xmlhttp.StatusText) ' get ok

Have also converted code to vb2008 and get the same as above

Maybe its a server thing...


Regards Kennedymr2

 

[kennedymr2]

Have found another thing...


getallheaders... transfer-encoding chunked

is chunked the problem ????????????


regards kennedymr2

 

[kennedymr2]

Server: Microsoft-IIs/6.0
X-Powerwed-By : ASP.net
X-ASpNet-Version :2.0.50727
Transfer-Encoding: chunked
Content-Type :text/html; charset = utf-8

 

[strongm]

>Maybe its a server thing...


I'm guessing so. If I replace sURL and strXML with known good values the code works (although on reflection I'd be using xmlhttp.ResponseText rather than the XML versiokn I originally suggested). Or perhaps the strXML is malformed - although even so I'd expect a response from the server

 

[kennedymr2]

Yes i agree... i think strxml may be 'deformed'
The instruction manual is rather weak.!!!!!

I will have to speak to the server programmers..

Will let you know what the outcome is.


Thanks for your advice


Regards Kennedymr2

 

[kennedymr2]

StrongM

Yes... it was one small error in the strXml.......!!!!!!!@!@

Sorry to have caused the fuss.

The string was very long and it was hard to see the potential problem..


Really appreciate the help offered.

Regards Kennedymr2

 

[strongm]

Glad you've got it working