1

我有以下 XML:

    <MovieRunTimes>
      <ShowDate>6/9/2012</ShowDate>
      <ShowTimesByDate xmlns:a="http://schemas.microsoft.com/2003/10/Serialization/Arrays">
        <a:string>12:25</a:string>
        <a:string>17:30</a:string>
        <a:string>22:35</a:string>
      </ShowTimesByDate>
      <TicketURI>http://www.fandango.com/tms.asp?t=AANCC&amp;m=112244&amp;d=2012-06-09</TicketURI>
    </MovieRunTimes>

以及以下 C# 类:

public class MovieRunTimes
{
    [XmlElement("ShowDate")]
    public string ShowDate { get; set; }

    [XmlElement("TicketURI")]
    public string TicketUri { get; set; }

    [XmlArray("ShowTimesByDate", Namespace = "http://schemas.microsoft.com/2003/10/Serialization/Arrays")]
    public List<string> ShowTimesByDate { get; set; }

}

不幸的是,我反序列化后 ShowTimesByDate 是空的。如果我从 ShowTimesByDate 元素中删除名称空间并从字符串元素中删除前缀,那么它可以很好地反序列化。如何正确使用命名空间来反序列化 XML?

4

2 回答 2

4

我发现了如何做到这一点。如果我将课程修改为:

public class MovieRunTimes
{
    [XmlElement("ShowDate")]
    public string ShowDate { get; set; }

    [XmlElement("TicketURI")]
    public string TicketUri { get; set; }

    [XmlArray("ShowTimesByDate")]
    [XmlArrayItem(Namespace = "http://schemas.microsoft.com/2003/10/Serialization/Arrays")]
    public List<string> ShowTimesByDate { get; set; }

}

它正确反序列化。

于 2012-06-26T16:09:42.557 回答
1

诀窍是在您的 Collection 包装器元素中添加一个命名空间前缀(在您的情况下为“a”):

<MovieRunTimes >
  <ShowDate>6/9/2012</ShowDate>
  <a:ShowTimesByDate xmlns:a="http://schemas.microsoft.com/2003/10/Serialization/Arrays">
    <a:string>12:25</a:string>
    <a:string>17:30</a:string>
    <a:string>22:35</a:string>
  </a:ShowTimesByDate>
  <TicketURI>http://www.fandango.com/tms.asp?t=AANCC&amp;m=112244&amp;d=2012-06-09</TicketURI>
</MovieRunTimes>

这就是使用此代码序列化后的结果:

        XmlSerializer xs = new XmlSerializer(typeof(MovieRunTimes));
        XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
        ns.Add("a", "http://schemas.microsoft.com/2003/10/Serialization/Arrays");
        string result = null;
        using(StringWriter writer = new StringWriter())
        {
            xs.Serialize(writer,mrt,ns);
            result = writer.ToString();
        }
于 2012-06-26T15:45:33.467 回答