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我编写了一个控制台应用程序,它调用一个 RESTful Web 服务,通过 HTTP 发送 XML 请求,返回 XML 响应。这是我目前正在使用的代码:

WebRequest wrq = WebRequest.Create("<?xml version=1.0 encoding=ISO-8859-1?>
<Request xmlns=https://sapiqa.overstock.com/api><MerchantKey>"+MerchantKey+"
</MerchantKey><AthenticationKey>"+AuthenticationKey+"</AuthenticationKey><"+APIMethod+"
/></Request>");
wrq.Method = "POST";
// Create POST data and convert it to a byte array. Strip out unnecessary text. 
byte[] byteArray = Encoding.UTF8.GetBytes(URL.Replace(APIMethod, ""));
// Set the ContentType property of the WebRequest. 
wrq.ContentType = "text/xml";
// Set the ContentLength property of the WebRequest. 
wrq.ContentLength = byteArray.Length;

Stream dataStream = wrq.GetRequestStream();
// Write the data to the request stream. 
dataStream.Write(byteArray, 0, byteArray.Length);
// Close the Stream object. 
dataStream.Close();
// Get the response. 
WebResponse response = wrq.GetResponse();
// Display the status. 
Console.WriteLine(((HttpWebResponse)response).StatusDescription);
// Get the stream containing content returned by the server. 
dataStream = response.GetResponseStream();
// Open the stream using a StreamReader for easy access. 
StreamReader reader = new StreamReader(dataStream);
// Read the content. 
string responseFromServer = reader.ReadToEnd();
// Display the content. 
GetOrders2Response = responseFromServer;
Console.WriteLine(responseFromServer);

// Clean up the streams. 
reader.Close();
dataStream.Close();
response.Close();
Console.ReadKey();

当我运行此代码时,我收到错误说明:

Invalid URI: The URI scheme is not valid.

如何通过 WebRequest 进行补救以发送格式正确的 XML 并接收 XML 响应?

4

1 回答 1

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正如错误和文档明确指出的那样,WebRequest.Create采用 URL,而不是 POST 有效负载。

您需要将 URL 传递给Create(),并将 XML 有效负载写入请求流。

于 2013-06-14T14:20:43.130 回答