http文件上载到wcf rest服务时出现问题

我想使用如下表单上传文件：

<form action="http://127.0.0.1:8090/MyService/" 
      method="post" enctype="multipart/form-data">
  <p>Select File<br>
     <input name="Datei" type="file" size="50">
     <input type="submit" value="Submit">
  </p>
</form>

[ServiceContract]
public interface IDocumentConverter
{
    [OperationContract, WebInvoke(UriTemplate = "/", Method = "POST")]
    void UploadFile(Stream fileContents);         
}

ÿØÿà...

[ServiceContract]
public interface IDocumentConverter
{
    [OperationContract, WebInvoke(UriTemplate = "/{filename}", Method = "POST")]
    void UploadFile(string filename, Stream fileContents);         
}

我有两个问题。当我将流写回文件时，它不仅包含文件数据，还包含一些http内容。这似乎很奇怪。是否有很好的方法获取原始文件数据

-----------------------------67832811011758
Content-Disposition: form-data; name="Datei"; filename="haha.jpg"
Content-Type: image/jpeg

ÿØÿà...

原始文件如下所示：

<form action="http://127.0.0.1:8090/MyService/" 
      method="post" enctype="multipart/form-data">
  <p>Select File<br>
     <input name="Datei" type="file" size="50">
     <input type="submit" value="Submit">
  </p>
</form>

[ServiceContract]
public interface IDocumentConverter
{
    [OperationContract, WebInvoke(UriTemplate = "/", Method = "POST")]
    void UploadFile(Stream fileContents);         
}

ÿØÿà...

[ServiceContract]
public interface IDocumentConverter
{
    [OperationContract, WebInvoke(UriTemplate = "/{filename}", Method = "POST")]
    void UploadFile(string filename, Stream fileContents);         
}

其次，我想在wcf的文件名。理想情况如下：

<form action="http://127.0.0.1:8090/MyService/" 
      method="post" enctype="multipart/form-data">
  <p>Select File<br>
     <input name="Datei" type="file" size="50">
     <input type="submit" value="Submit">
  </p>
</form>

[ServiceContract]
public interface IDocumentConverter
{
    [OperationContract, WebInvoke(UriTemplate = "/", Method = "POST")]
    void UploadFile(Stream fileContents);         
}

ÿØÿà...

[ServiceContract]
public interface IDocumentConverter
{
    [OperationContract, WebInvoke(UriTemplate = "/{filename}", Method = "POST")]
    void UploadFile(string filename, Stream fileContents);         
}

---

但这不起作用，因为我不知道如何使用表单/输入内容将文件名集成到url中。

我已经编辑了您的标题。请参阅“”，其中的共识是“不，他们不应该”。