从WCT Restful服务中获取FileStream

时间:2015-08-31 07:10:47

标签: c# wcf rest filestream wcf-rest

我创建了一个WCF Restful服务,其代码如下:

接口

[ServiceContract]
public interface ISIGService
{
    [OperationContract]
    [WebInvoke(Method = "GET", ResponseFormat = WebMessageFormat.Xml,
        BodyStyle = WebMessageBodyStyle.Bare,
        UriTemplate = "GetMultimedia/{id}")]
    Stream GetMultimedia(string id);
}

实现此服务的代码:

public Stream GetMultimedia(string id)
{
    string filePath = MultimediaBLL.GetFilePath(int.Parse(id));
    if (string.IsNullOrEmpty(filePath))
        return null;

    try
    {
        FileStream multimediaFileStream = File.OpenRead(filePath);
        return multimediaFileStream;
    }
    catch (Exception ex)
    {
        Console.WriteLine("Not able to get multimedia stream:{0}", ex.Message);
        throw ex;
    }
}

到目前为止,我需要使用此服务来获取多媒体文件(图像,音频或视频)的文件流:

RestClient client = new RestClient(urlService);
RestRequest request = new RestRequest
{
    Method = Method.GET,
    RequestFormat = DataFormat.Xml,
    Resource = "GetMultimedia/{id}"
};
request.AddParameter("id", idMultimedia, ParameterType.UrlSegment);

var response = client.Execute(request);
if (!response.StatusCode != HttpStatusCode.OK)
   return;

???

我不知道如何获取流并处理它以在处理程序中显示它,如下所示:

context.Response.ClearContent();
context.Response.ClearHeaders();
context.Response.AddHeader("Content-Length", bytes.Length.ToString());
context.Response.BinaryWrite(bytes);
context.Response.End();

我想我必须得到文件字节才能将它发送给处理程序。

1 个答案:

答案 0 :(得分:2)

我不知道你正在尝试做什么。

<强>服务

public Stream GetMultimedia(string id)
{ 
    string filePath = MultimediaBLL.GetFilePath(int.Parse(id));
    if (string.IsNullOrEmpty(filePath))
        return null;

    try
    {
        FileStream multimediaFileStream = File.OpenRead(filePath);

        WebOperationContext.Current.OutgoingResponse.ContentType = "Your/contentType";
        WebOperationContext.Current.OutgoingResponse.Headers["Content-Disposition"] = "attachment; filename=Your_file_name.png";

        return multimediaFileStream;
    }
    catch (Exception ex)
    {
        Console.WriteLine("Not able to get multimedia stream:{0}", ex.Message);
        throw ex;
    }
}

<强>客户端

RestClient client = new RestClient(urlService);
RestRequest request = new RestRequest
{
    Method = Method.GET,
    RequestFormat = DataFormat.Xml,
    Resource = "GetMultimedia/{id}"
};

request.AddParameter("id", idMultimedia, ParameterType.UrlSegment);

context.Response.Clear();
content.Response.ClearHeaders();

# write original responseStream to the context.Response.OutputStream
request.ResponseWriter = (responseStream) => responseStream.WriteTo(context.Response.OutputStream);

var response = client.DownloadData(request);

# maybe you need to add content type and file name as well
context.Response.AddHeader("Content-Type", "fetch from response");
context.Response.AddHeader("Content-Disposition", "fetch from response");

context.Response.Flush();
context.Response.Close();
context.Response.End();

if (!response.StatusCode != HttpStatusCode.OK)
    # do something