我正在尝试为ServiceStack服务创建RSS源。我尽可能地遵循各种例子。我的问题是我没有输出,我不知道如何解决问题。我怀疑我在序列化上做错了什么。这是(我的简化版)
我的DTO是
using System.Collections.Generic;
using ServiceStack;
using Library;
[Route("/MyCollection/Tomorrow/{ID}", "GET, POST")]
[Api("MyCollections Delivery")]
public class MyCollectionTomorrow
: IReturn<MyCollectionTomorrowResponse>
{
public long ID { get; set; }
}
public class MyCollectionTomorrowResponse : IHasResponseStatus
{
public long ID { get; set; }
public List<MyCollection> Result { get; set; }
public ResponseStatus ResponseStatus { get; set; }
}
public class MyCollection
{
public string Description { get; set; }
public string MyCollectionDayOfWeek { get; set; }
public DateTime MyCollectionDate { get; set; }
public bool Assisted { get; set; }
public string RoundType { get; set; }
public string Description { get; set; }
}
我的服务是
using System;
using Library;
using ServiceStack;
using ServiceStack.Configuration;
using System;
using Library;
using ServiceStack;
using ServiceStack.Configuration;
using MyCollection.Tomorrow;
using MyCollections.Tomorrow;
public class MyCollectionTomorrowService : Service
{
public object Any(WasteCollectionTomorrow request)
{
int id;
var param = new CollectionTomorrow();
param.ID = ID;
var response = client.Get<CollectionTomorrowResponse>(param);
return response;
}
catch (Exception ex)
{
var response = new CollectionTomorrowResponse();
response.Result = null
var status = new ResponseStatus { Message = ex.Message, StackTrace = ex.StackTrace };
response.ResponseStatus = status;
return response;
}
}
}
我的媒体类型是
namespace DataFeedServices
{
using System;
using System.IO;
using System.ServiceModel.Syndication;
using System.Text;
using System.Xml;
using ServiceStack;
using ServiceStack.Data;
using ServiceStack.Web;
using MyCollections.Tomorrow;
public class RssFormat
{
private const string RssContentType = "application/rss+xml";
public static void Register(IAppHost appHost)
{
appHost.ContentTypes.Register(RssContentType, SerializeToStream, DeserializeFromStream);
}
public static void SerializeToStream(IRequest req, object response, Stream stream)
{
StreamWriter sw = null;
try
{
var syndicationFeedResponse = response as MyCollectionResponse;
sw = new StreamWriter(stream);
if (response != null)
{
WriteRssCollectionFeed(sw, syndicationFeedResponse);
}
}
finally
{
if (sw != null)
{
sw.Dispose();
}
}
}
public static void WriteRssCollectionFeed(StreamWriter sw, MyCollectionResponse Mycollections)
{
const string Baseuri = "example.com";
try
{
var uri = new Uri(Baseuri);
var syndicationFeed = new SyndicationFeed(
"MyCollection Service",
"Mycollections " ,
uri);
syndicationFeed.Authors.Add(new SyndicationPerson("email@mysite.com"));
if (Mycollections.Result != null)
{
foreach (var cats in Mycollections.Result)
{
syndicationFeed.Categories.Add(new SyndicationCategory(cats.RoundID));
}
}
syndicationFeed.Generator = "MyApp";
syndicationFeed.Copyright = new TextSyndicationContent("Copyright 2015");
syndicationFeed.LastUpdatedTime = DateTime.Now;
if (Mycollections.Result != null)
{
// set items
foreach (var coll in Mycollections.Result)
{
var item = new SyndicationItem { Title = new TextSyndicationContent(coll.CollectionDate) };
item.Links.Add(new SyndicationLink(uri));
item.Authors.Add(new SyndicationPerson("email@mysite.com"));
var itemContent = new StringBuilder();
itemContent.Append("My Item content");
item.Content = new TextSyndicationContent(
itemContent.ToString(),
TextSyndicationContentKind.Plaintext);
}
}
Rss20FeedFormatter rssFeed = syndicationFeed.GetRss20Formatter();
var xwriter = XmlWriter.Create(sw);
rssFeed.WriteTo(xwriter);
}
catch (Exception)
{
throw new Exception("Something bad happened");
}
}
public static object DeserializeFromStream(Type type, Stream stream)
{
throw new NotImplementedException();
}
}
}
答案 0 :(得分:1)
由于您的ContentType不可重用并且与特定MyCollectionResponse
耦合,因此使用RSS XML返回原始字符串更容易:
[AddHeader(ContentType = "application/rss+xml")]
public object Any(WasteCollectionTomorrow request)
{
//..
return rssXml;
}
您也可以使用以下内容将其直接写入响应输出流:
public object Any(WasteCollectionTomorrow request)
{
//..
base.Response.ContentType = "application/rss+xml";
RssFormat.SerializeToStream(response, Response.OutputStream);
base.Response.EndRequest();
return null;
}