Rest API c# -如何传递流参数

Question

我需要调用以下REST API

[OperationContract]
[WebInvoke(Method = "POST",
UriTemplate = "/{subscriptionId}/servers?op=ChangeSubscription")]
Stream ChangeAllServersSubscription(string subscriptionId, Stream requestStream);

如何创建并传递Stream requestStream ?流将包含JSON格式的字符串

Answer 1

在使用任何类型的Stream之前，您应该询问API供应商他希望接收哪个流。

下面是一个使用MemoryStream的例子，在您创建JSON之后:

// Random objects, just to make the code clear
var myObject = new {Name = "Yakov"};
var myId = "1";
var myJson = JsonConvert.SerializeObject(myObject);
// Create a MemoryStream (Encoding should be as needed)
var memoryStream = new MemoryStream(Encoding.UTF8.GetBytes(myJson));
// Call your code
var responseStream = ChangeAllServersSubscription(myId, memoryStream);
// And dispose the MemoryStream after you're done.
memoryStream.Dispose();

Answer 2

您可以在客户端创建一个HttpWebRequest，设置url指向您的服务，设置适当的方法-在您的情况下POST。例如:

HttpWebRequest request = (HttpWebRequest)HttpWebRequest.Create("http://server/id/servers?op=ChangeSubscription");
request.Method = "POST";

然后使用GetRequestStream()在客户端获取流，将json写入该流，关闭流，然后通过调用GetResponse()执行请求。

    Stream stream= request.GetRequestStream();
    //stream.Write write contents of your file here!
    stream.Close();
    HttpWebResponse response= (HttpWebResponse)request.GetResponse();
    //process the response from server

如果一切正常，你的web服务将会收到你的文件。