49

我想要以下格式的 XML：

<configuration><!-- Only one configuration node -->
  <logging>...</logging><!-- Only one logging node -->
  <credentials>...</credentials><!-- One or more credentials nodes -->
  <credentials>...</credentials>
</configuration>

我正在尝试创建一个Configuration具有该[Serializable]属性的类。要序列化凭据节点，我有以下内容：

[XmlArray("configuration")]
[XmlArrayItem("credentials", typeof(CredentialsSection))]
public List<CredentialsSection> Credentials { get; set; }

但是，当我将其序列化为 XML 时，XML 采用以下格式：

<configuration>
  <logging>...</logging>
  <configuration><!-- Don't want credentials nodes nested in a second
                      configuration node -->
    <credentials>...</credentials>
    <credentials>...</credentials>
  </configuration>
</configuration>

如果我删除该[XmlArray("configuration")]行，我会得到以下信息：

<configuration>
  <logging>...</logging>
  <Credentials><!-- Don't want credentials nodes nested in Credentials node -->
    <credentials>...</credentials>
    <credentials>...</credentials>
  </Credentials>
</configuration>

如何以我想要的方式序列化这个，<credentials>在单个根节点中有多个节点<configuration>？我想这样做而不必实现IXmlSerializable和进行自定义序列化。这就是我的班级的描述方式：

[Serializable]
[XmlRoot("configuration")]
public class Configuration : IEquatable<Configuration>
4

1 回答 1

80

以下应该按照您想要的方式正确序列化。线索[XmlElement("credentials")]在名单上。我通过获取您的 xml，在 Visual Studio 中从中生成架构 (xsd) 来做到这一点。然后在架构上运行 xsd.exe 以生成一个类。（还有一些小的修改）

public class CredentialsSection
{
    public string Username { get; set; }
    public string Password { get; set; }
}

[XmlRoot(Namespace = "", IsNullable = false)]
public class configuration
{
    /// <remarks/>
    public string logging { get; set; }

    /// <remarks/>
    [XmlElement("credentials")]
    public List<CredentialsSection> credentials { get; set; }

    public string Serialize()
    {
        var credentialsSection = new CredentialsSection {Username = "a", Password = "b"};
        this.credentials = new List<CredentialsSection> {credentialsSection, credentialsSection};
        this.logging = "log this";
        XmlSerializer s = new XmlSerializer(this.GetType());
        StringBuilder sb = new StringBuilder();
        TextWriter w = new StringWriter(sb);
        s.Serialize(w, this);
        w.Flush();
        return sb.ToString();
    }
}

给出以下输出

<?xml version="1.0" encoding="utf-16"?>
<configuration xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <logging>log this</logging>
  <credentials>
    <Username>a</Username>
    <Password>b</Password>
  </credentials>
  <credentials>
    <Username>a</Username>
    <Password>b</Password>
  </credentials>
</configuration>
于 2010-07-21T20:15:37.973 回答