1

如何在 WCF 的帮助下通过 post 发送 xml 数据？

例如我有一些代码：

public interface IServiceForILobby
{
    [OperationContract]
    [WebInvoke(Method = "POST")] 
    string SendXml(string response);
}

//这是主机

static void Main(string[] args)
{
    Console.WriteLine("*Console Based Rest HOST*");

    using (WebServiceHost serviceHost = new WebServiceHost(typeof(ServiceForILobby)))
    {
        serviceHost.Open();
        Console.ReadLine();                
    }

/*这是客户*/

using (ChannelFactory<IServiceForILobby> cf = new   ChannelFactory<IServiceForILobby>(new WebHttpBinding(), "http://192.168.1.103:50000/RestService/SendXml?response={x}"))
{
    cf.Endpoint.EndpointBehaviors.Add(new WebHttpBehavior());
    IServiceForILobby channel = cf.CreateChannel();
    string s;
    // s = channel.SendXml("http://192.168.1.103:50000/RestService/SendXml?response={x}");
    string a;

    using (StreamReader sr = new StreamReader("simplexml.txt"))
    {
        string xmlCode = sr.ReadToEnd();
        s = channel.SendXml(xmlCode);
    }

我想将 XML 从客户端发送到主机，然后像这样解析它如何解析 XML 文件？

4

2 回答 2

10

为了通过 POST 发送 xml 数据，您需要根据 WCF 服务正确构造数据。这基本上是您需要的：

1）WCF服务接口

[OperationContract]
[WebInvoke(Method = "POST",
    UriTemplate = "GetData",
    RequestFormat = WebMessageFormat.Xml,
    BodyStyle = WebMessageBodyStyle.Bare)]
string GetData(DataRequest parameter);

2）WCF服务实现

public string GetData(DataRequest parameter)
{
    //Do stuff
    return "your data here";
}

3) WCF 服务中的数据协定（在本例中为 DataRequest）

[DataContract(Namespace = "YourNamespaceHere")]
public class DataRequest
{
    [DataMember]
    public string ID{ get; set; }
    [DataMember]
    public string Data{ get; set; }
}

4) 发送数据的客户端必须正确构造数据！（本例中为 C# 控制台应用程序）

static void Main(string[] args)
{
    ASCIIEncoding encoding = new ASCIIEncoding();
    string SampleXml = "<DataRequest xmlns=\"YourNamespaceHere\">" +
                                    "<ID>" +
                                    yourIDVariable +
                                    "</ID>" +
                                    "<Data>" +
                                    yourDataVariable +
                                    "</Data>" +
                                "</DataRequest>";

    string postData = SampleXml.ToString();
    byte[] data = encoding.GetBytes(postData);

    string url = "http://localhost:62810/MyService.svc/GetData";

    string strResult = string.Empty;

    // declare httpwebrequet wrt url defined above
    HttpWebRequest webrequest = (HttpWebRequest)WebRequest.Create(url);
    // set method as post
    webrequest.Method = "POST";
    // set content type
    webrequest.ContentType = "application/xml";
    // set content length
    webrequest.ContentLength = data.Length;
    // get stream data out of webrequest object
    Stream newStream = webrequest.GetRequestStream();
    newStream.Write(data, 0, data.Length);
    newStream.Close();

    //Gets the response
    WebResponse response = webrequest.GetResponse();
    //Writes the Response
    Stream responseStream = response.GetResponseStream();

    StreamReader sr = new StreamReader(responseStream);
    string s = sr.ReadToEnd();

    return s;
}
于 2013-11-11T20:19:02.763 回答
2

以下来自c#-corner的代码片段提供了一个很好的示例：

    string postData = SampleXml.ToString();
    // convert xmlstring to byte using ascii encoding
    byte[] data = encoding.GetBytes(postData);
    // declare httpwebrequet wrt url defined above
    HttpWebRequest webrequest = (HttpWebRequest)WebRequest.Create(url);
    // set method as post
    webrequest.Method = "POST";
    // set content type
    webrequest.ContentType = "application/x-www-form-urlencoded";
    // set content length
    webrequest.ContentLength = data.Length;
    // get stream data out of webrequest object
    Stream newStream = webrequest.GetRequestStream();
    newStream.Write(data, 0, data.Length);
    newStream.Close();
    // declare & read response from service
    HttpWebResponse webresponse = (HttpWebResponse)webrequest.GetResponse();
于 2013-11-11T14:01:29.790 回答