Forwarding HTTP POST: how to return correct HTTP status?

[google]

I'm building a small ASPX page in C# to forward HTTP POST requests to
another site. When the other site returns an HTTP error such as 404 or
405, I want to return that same HTTP status value to my caller. But when
I do so, my caller always reports HTTP error 500. How can I return the
actual HTTP status that I received?

Here is my code:

private void Page_Load(object sender, System.EventArgs e)
{
bool failed = false; // No exceptions caught yet.
Exception copyOfException = null;
try {
string UrlForThisQuery = Global.NextServerUrl() + "BAD";

System.Net.WebClient client = new System.Net.WebClient();
byte [] byXmlOut = null;

if ( 0 == Request.Form.Count ) {

// First case: no parameters; this must be an HTTP GET.
// Perform an HTTP GET.

byXmlOut = client.DownloadData(UrlForThisQuery);
} else {

// Second (main) case: one or more parameters;
// this is a normal XML query in the form of HTTP POST.
// Just forward the POST.

byXmlOut = client.UploadValues(UrlForThisQuery,
Request.Form);
}

// Return the content of the page to the caller.
Response.BinaryWrite(byXmlOut);
}
catch (WebException ex) {
HttpWebResponse r = (HttpWebResponse)ex.Response;
copyOfException = new WebException(
ex.Message,
ex,
(WebExceptionStatus) r.StatusCode, // 404
ex.Response);
failed = true;
}

// Relay any exception to the caller.
if (failed) {
throw copyOfException; // But the caller reports 500.
}

// Normal exit. We already performed Response.BinaryWrite() above.
}

-- Phil Davidson, Oakland, CA

[Phil Davidson]

Here's the answer: When I throw an exception from my ASPX page, it's
always reported as HTTP error 500. Instead I need to put the HTTP status
information into the Response object and return normally from the
Page_Load function:

catch (WebException ex) {
HttpWebResponse r = (HttpWebResponse)ex.Response;
copyOfException = new WebException(
ex.Message,
ex,
(WebExceptionStatus) r.StatusCode,
ex.Response);
Response.StatusCode = (int) r.StatusCode;
Response.StatusDescription =
"HTTP status returned from " + UrlForThisQuery + "\r\n"
+ ex.Message;

// CopyStream is a simple function that copies from the first
// stream to the second stream in blocks.
CopyStream(ex.Response.GetResponseStream(),
Response.OutputStream);
failed = true;
re_throw = false;
}
...
// Relay non-HTTP exceptions to the caller.
if (re_throw) {
throw copyOfException;
}

-- Phil Davidson

In microsoft.public.dotnet.framework.webservices,
go****@phildavidson.com wrote:

I'm building a small ASPX page in C# to forward HTTP POST requests to
another site. When the other site returns an HTTP error such as 404 or
405, I want to return that same HTTP status value to my caller. But when
I do so, my caller always reports HTTP error 500. How can I return the
actual HTTP status that I received?

Here is my code:

private void Page_Load(object sender, System.EventArgs e)
{
bool failed = false; // No exceptions caught yet.
Exception copyOfException = null;
try {
string UrlForThisQuery = Global.NextServerUrl() + "BAD";

System.Net.WebClient client = new System.Net.WebClient();
byte [] byXmlOut = null;

if ( 0 == Request.Form.Count ) {

// First case: no parameters; this must be an HTTP GET.
// Perform an HTTP GET.

byXmlOut = client.DownloadData(UrlForThisQuery);
} else {

// Second (main) case: one or more parameters;
// this is a normal XML query in the form of HTTP POST.
// Just forward the POST.

byXmlOut = client.UploadValues(UrlForThisQuery,
Request.Form);
}

// Return the content of the page to the caller.
Response.BinaryWrite(byXmlOut);
}
catch (WebException ex) {
HttpWebResponse r = (HttpWebResponse)ex.Response;
copyOfException = new WebException(
ex.Message,
ex,
(WebExceptionStatus) r.StatusCode, // 404
ex.Response);
failed = true;
}

// Relay any exception to the caller.
if (failed) {
throw copyOfException; // But the caller reports 500.
}

// Normal exit. We already performed Response.BinaryWrite() above.
}

-- Phil Davidson, Oakland, CA