WCF Restful post Services返回BAD REQUEST(400)

时间:2013-04-30 13:28:12

标签: c# xml wcf rest

我正在尝试从几天开始消费wcf宁静的服务,但它给了我糟糕的要求。请帮帮我。

这是我的配置文件

<system.serviceModel>

<bindings>
  <webHttpBinding>
    <binding name="state" allowCookies="true">
      <security mode="None"></security>

    </binding>
  </webHttpBinding>
</bindings>
<serviceHostingEnvironment aspNetCompatibilityEnabled="true"/>
<behaviors>
  <serviceBehaviors>
    <behavior name="ServiceBehaviour">
      <serviceMetadata httpGetEnabled="True"/>
      <serviceDebug includeExceptionDetailInFaults="True"/>
    </behavior>
  </serviceBehaviors>
  <endpointBehaviors>
    <behavior name="web">
      <webHttp/>
    </behavior>
  </endpointBehaviors>
</behaviors>
<services>
  <service behaviorConfiguration="ServiceBehaviour" name="RESTFUL_DEMO.Web.services.Calc">
    <endpoint address="" bindingConfiguration="state" binding="webHttpBinding" name="Http" contract="RESTFUL_DEMO.Web.services.ICalc"/>
    <endpoint address="abcd" binding="wsHttpBinding" name="wsHttp" contract="RESTFUL_DEMO.Web.services.ICalc"/>

    <endpoint address="mex" binding="mexHttpBinding" name="MEX" contract="IMetadataExchange"/>

  </service>
</services>

我的服务合同和数据合同界面如下。

[ServiceContract(SessionMode = SessionMode.Allowed)]
[XmlSerializerFormat]
public interface ICalc
{
    [OperationContract]
    [WebInvoke(UriTemplate = "dowork", Method = "POST", RequestFormat = WebMessageFormat.Xml, ResponseFormat = WebMessageFormat.Xml, BodyStyle = WebMessageBodyStyle.Bare)]
    int DoWork(Enroll a);
}


[DataContract]
public class Enroll
{
    public Enroll()
    {

    }
    public Enroll(string Avalue)
    {
        this.Avalue = Avalue;
    }
    [DataMember(IsRequired = true)]
    public string Avalue
    {
        get;
        set;
    }


}

我使用该服务的代码如下

HttpWebRequest request = WebRequest.Create("http://localhost/RESTFUL_DEMO.Web/services/Calc.svc/dowork") as HttpWebRequest;
XmlDocument doc = new XmlDocument();
        doc.Load(@"d:\test.xml");
        string sXML = doc.InnerXml;
        request.ContentLength = sXML.Length;
        request.ContentType = "test/xml; charset=utf-8";
        var sw = new StreamWriter(request.GetRequestStream());
        sw.Write(sXML);
        sw.Close();
        WebResponse response = request.GetResponse();
        StreamReader stream = new StreamReader(response.GetResponseStream());
        String result = stream.ReadToEnd();

3 个答案:

答案 0 :(得分:2)

您在使用Rest服务时犯了一个小错误。您已将请求的ContentType指定为 test / xml; charset = utf-8但它应该是text / xml或application / xml。

request.ContentType = "text/xml; charset=utf-8";

或者它应该是

request.ContentType = "application/xml";

答案 1 :(得分:0)

在我的例子中,我在服务接口IBookService.cs中的方法如下所示

[OperationContract]
[WebInvoke(Method = "PUT", UriTemplate = "/Book", RequestFormat = WebMessageFormat.Json, ResponseFormat = WebMessageFormat.Json)]
IList<Book> UpdateBook(Book book);

在我的客户端我正在给予

client.Headers[HttpRequestHeader.ContentType] = "text/xml";  

而不是

client.Headers[HttpRequestHeader.ContentType] = "text/json";

这解决了我的问题。请参阅下面的完整解决方案

 private void btnUpdateBook_Click(object sender, EventArgs e)
    {
        try
        {
            using(WebClient client = new WebClient())
            {
                client.Headers[HttpRequestHeader.ContentType] = "text/json";                   
                Uri uri = new Uri(@"http://localhost:8085/BookService/Book");

                Book updateBook = new Book() { Id = 3, Name = "UpdateBook Name 3", Price = 77.77f };

                MemoryStream requestStream = new MemoryStream();
                DataContractJsonSerializer requestSerializer = new DataContractJsonSerializer(typeof(Book));
                requestSerializer.WriteObject(requestStream, updateBook);

                client.UploadDataCompleted += OnUpdateBookCompleted;
                client.UploadDataAsync(uri, "PUT",requestStream.ToArray());
            }
        }
        catch (Exception ex)
        {

        }
    }

    void OnUpdateBookCompleted(object sender, UploadDataCompletedEventArgs e)
    {
        byte[] result = e.Result as byte[];
        MemoryStream responseStream = new MemoryStream(result);
        DataContractJsonSerializer serializer = new DataContractJsonSerializer(typeof(IList<Book>));
        IList<Book> books = (IList<Book>)serializer.ReadObject(responseStream);

        bindingSource1.DataSource = books;
        dvBooks.DataSource = bindingSource1;
    }

答案 2 :(得分:-2)

在Visual Studio实例中启动并运行该服务,然后使用Test Client确保服务正常运行。

打开VS的新实例并添加一个服务引用,它将为您构建客户端代码,然后使用此客户端来调用该服务。