c# - 如何使用没有声明元素的 xml 编写器创建 xml

Question

我正在使用XmlWriter.Create()获取编写器实例然后编写 XML，但结果有<?xml version="1.0" encoding="utf-16" ?>，我如何告诉我的 xml 编写器不生成它？

Answer 1

使用XmlWriterSettings.OmitXmlDeclaration.

不要忘记设置XmlWriterSettings.ConformanceLevel为ConformanceLevel.Fragment.

Answer 2

您可以子类XmlTextWriter化并覆盖该WriteStartDocument()方法以不执行任何操作：

public class XmlFragmentWriter : XmlTextWriter
{
    // Add whichever constructor(s) you need, e.g.:
    public XmlFragmentWriter(Stream stream, Encoding encoding) : base(stream, encoding)
    {
    }

    public override void WriteStartDocument()
    {
       // Do nothing (omit the declaration)
    }
}

用法：

var stream = new MemoryStream();
var writer = new XmlFragmentWriter(stream, Encoding.UTF8);
// Use the writer ...

参考：Scott Hanselman的这篇博文。

Answer 3

你可以使用XmlWriter.Create()：

new XmlWriterSettings { 
    OmitXmlDeclaration = true, 
    ConformanceLevel = ConformanceLevel.Fragment 
}

    public static string FormatXml(string xml)
    {
        if (string.IsNullOrEmpty(xml))
            return string.Empty;

        try
        {
            XmlDocument document = new XmlDocument();
            document.LoadXml(xml);
            using (MemoryStream memoryStream = new MemoryStream())
            using (XmlWriter writer = XmlWriter.Create(
                memoryStream, 
                new XmlWriterSettings { 
                    Encoding = Encoding.Unicode, 
                    OmitXmlDeclaration = true, 
                    ConformanceLevel = ConformanceLevel.Fragment, 
                    Indent = true, 
                    NewLineOnAttributes = false }))
            {
                document.WriteContentTo(writer);
                writer.Flush();
                memoryStream.Flush();
                memoryStream.Position = 0;
                using (StreamReader streamReader = new StreamReader(memoryStream))
                {
                    return streamReader.ReadToEnd();
                }
            }
        }
        catch (XmlException ex)
        {
            return "Unformatted Xml version." + Environment.NewLine + ex.Message;
        }
        catch (Exception ex)
        {
            return "Unformatted Xml version." + Environment.NewLine + ex.Message;
        }
    }