2

我有以下 RSS 提要。我想阅读特定描述标签中的信息。例如，当标题标签包含当前日期时，我想获取描述标签中的信息。我不知道该怎么做。请帮忙

<item>
<title>Forecast for Saturday as of Jul. 14 5:30 AM IST</title> //If today is Saturday get information in description tag
<link>http://www.wunderground.com/global/stations/43466.html</link>    
  <description>
Thunderstorm. Low:26 &amp;deg; C.
  </description>
  <pubDate>Sat, 14 Jul 2012 00:00:00 GMT</pubDate>
  <guid isPermaLink="false">1342267200-1-night</guid>
</item>
<item>
<title>Forecast for Sunday as of Jul. 14 5:30 AM IST</title>
<link>http://www.wunderground.com/global/stations/43466.html</link>
  <description>
Chance of a Thunderstorm. High:30 &amp;deg; C.
  </description>
  <pubDate>Sat, 14 Jul 2012 00:00:00 GMT</pubDate>
  <guid isPermaLink="false">1342353600-2-day</guid>
 </item>

我能够使用以下方法获得当前日期：

string datenow = DateTime.Today.ToString("dddd / M / yyyy");
string[] words= datenow.Split(' ');
string day = words[0];

这就是我阅读 RSS 提要的方式：

 public class RssReader
    {
        public static List<RssNews> Read(string url)
        {
            var webClient = new WebClient();

            string result = webClient.DownloadString(url);

            XDocument document = XDocument.Parse(result);

            return (from descendant in document.Descendants("item")
                    select new RssNews()
                    {
                        Description = descendant.Element("description").Value,
                        Title = descendant.Element("title").Value,
                        PublicationDate = descendant.Element("pubDate").Value
                    }).ToList();
        }
    }
4

2 回答 2

1

根据您的要求，您只想获取当天的描述。您有带有标题、描述等的新闻提要列表现在您可以更改您的课程（在您的课程中添加一个方法）如果当天是当天

公共静态字符串 GetCurrentDayDescription(){

   var lst = Read("url");
   var resDescription = (from x in lst where x.Title.Contains(day) 
                          select x.Description).ToArray() ;
    return resDescription[0] ;
}
于 2012-07-14T22:15:09.717 回答
1

您应该能够使用 XmlSerializer 直接反序列化 rss 提要，无需手动映射。您需要修改RssNews对象以正确映射，例如：

[XmlRoot(ElementName="item")]
public class RssNews
{
    [XmlElement(ElementName = "title")]
    public string Title { get; set; }

    [XmlElement(ElementName = "pubDate")]
    public string PublicationDate  { get; set; }

    [XmlElement(ElementName = "description")]
    public string Description { get; set; }

    [XmlElement(ElementName = "link")]
    public string Link { get; set; }

    [XmlElement(ElementName = "guid")]
    public string Description { get; set; }
}

现在您应该可以使用反序列化器了：

    var feed = new List<RssNews>();
    using (var webClient = new WebClient())
    {

        string result = webClient.DownloadString(url);
        using (var stringReader = new StringReader(result))
        {
            var serializer = new XmlSerializer(feed.GetType());
            feed = (List<RssNews>)serializer.Deserialize(stringReader);
        }
    }
    return feed;
于 2012-07-14T21:44:18.623 回答