将对象序列化为XML，包括“xsi：schemaLocation”和属性顺序

Question

我有这门课

public class EnvioDTE
{
    [XmlAttribute]
    public string version { get; set; }

    public EnvioDTE()
    {
        this.version = "1.0";
    }
}

我使用以下代码将其序列化为XML：

EnvioDTE envioDTE = new EnvioDTE();

string xml = "";

var serializer = new XmlSerializer(typeof(EnvioDTE));
var settings = new XmlWriterSettings
{
    Indent = true,
    NewLineChars = "\n",
    OmitXmlDeclaration = false,
    Encoding = Encoding.GetEncoding("ISO-8859-1")
};

using (var stream = new MemoryStream())
{
    using (var xmlWriter = XmlWriter.Create(stream, settings))
    {
        serializer.Serialize(xmlWriter, this);
        xml = Encoding.GetEncoding("ISO-8859-1").GetString(stream.ToArray());
    }
}

Console.WriteLine(xml);

这给了我以下XML：

<?xml version="1.0" encoding="iso-8859-1"?>
<EnvioDTE version="1.0">
</EnvioDTE>

我需要向代码添加什么才能获得以下属性？

<?xml version="1.0" encoding="iso-8859-1"?>
<EnvioDTE xmlns="http://www.sii.cl/SiiDte" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.sii.cl/SiiDte EnvioDTE_v10.xsd" version="1.0">
</EnvioDTE>

属性的顺序对我很重要。

我尝试了两个更改：

改变＃1

[XmlRoot(Namespace = "http://www.sii.cl/SiiDte")] // *** NEW
public class EnvioDTE
{
    [XmlAttribute("schemaLocation", Namespace = "http://www.w3.org/2001/XMLSchema-instance")] // *** NEW
    public string schemaLocation { get; set; } // *** NEW
    [XmlAttribute]
    public string version { get; set; }

    public EnvioDTE()
    {
        this.version = "1.0";
    }
}

改变＃2

EnvioDTE envioDTE = new EnvioDTE();

string xml = "";

var namespaces = new XmlSerializerNamespaces(); // *** NEW
namespaces.Add("xsi", "http://www.w3.org/2001/XMLSchema-instance"); // *** NEW

var serializer = new XmlSerializer(typeof(EnvioDTE));
var settings = new XmlWriterSettings
{
    Indent = true,
    NewLineChars = "\n",
    OmitXmlDeclaration = false,
    Encoding = Encoding.GetEncoding("ISO-8859-1")
};

using (var stream = new MemoryStream())
{
    using (var xmlWriter = XmlWriter.Create(stream, settings))
    {
        serializer.Serialize(xmlWriter, this, namespaces); // *** NEW
        xml = Encoding.GetEncoding("ISO-8859-1").GetString(stream.ToArray());
    }
}

Console.WriteLine(xml);

随着这些变化，我得到了这个：

<?xml version="1.0" encoding="iso-8859-1"?>
<EnvioDTE xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.sii.cl/SiiDte EnvioDTE_v10.xsd" version="1.0" xmlns="http://www.sii.cl/SiiDte">
</EnvioDTE>

这不是我想要的属性的顺序......

Answer 1

所以你只是想读出XML？我会这样做：

XmlReader reader = XmlReader.Create(path, settings);
StringBuilder sb = new StringBuilder();
if (reader != null)
{                           
    while (reader.Read())
    {
        sb.AppendLine(reader.ReadOuterXml());
    }
    doc = new XmlDocument();
    doc.LoadXml(Convert.ToString(sb));

    XmlNodeList xmlNodeList;

    xmlNodeList = doc.GetElementsByTagName("Whatever your tag name");

    for (i = 0; i < xmlNodeList.Count; i++)
    {
        if (xmlNodeList[i].InnerXml.Length > 0)
        {
            foo = xmlNodeList[i].InnerXml;
        }
    }
}

好的，你只是想输出？然后这样做。我知道这有点像黑客，可能有一百万种方法可以做到：

string myXMLstring = "<?xml version=\"1.0\" encoding="iso-8859-1\"?>" +
<EnvioDTE xmlns=\"http://www.sii.cl/SiiDte\"" + 
xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\"" +
"xsi:schemaLocation=\"http://www.sii.cl/SiiDte EnvioDTE_v10.xsd\"" + 
"version=\"1.0\">" +
"</EnvioDTE>";

我已经将引号转义并将其构建为字符串。您甚至可以在知道需要换行时随时添加\ r \ n。我希望有所帮助。快乐的编码！