Question

我有一个字符串属性，其中包含带换行符的文本。此文本具有HTML文本的一些属性，因为空格被忽略。

如果我使用XML序列化序列化此类型，则新行被正确序列化，但缩进是“错误的”。我希望序列化过程缩进行以保持XML的格式，因为无论如何这些空格字符都会被忽略。

这是一个示例LINQPad程序：

void Main()
{
    var d = new Dummy();
    d.Text = @"Line 1
Line 2
Line 3";

    var serializer = new XmlSerializer(typeof(Dummy));
    var ns = new XmlSerializerNamespaces();
    ns.Add("", "");

    using (var writer = new StringWriter())
    {
        serializer.Serialize(writer, d, ns);
        writer.ToString().Dump();
    }
}

[XmlType("dummy")]
public class Dummy
{
    [XmlElement("text")]
    public string Text
    {
        get;
        set;
    }
}

实际输出：

<?xml version="1.0" encoding="utf-16"?>
<dummy>
  <text>Line 1
Line 2
Line 3</text>
</dummy>

期望的输出：

<?xml version="1.0" encoding="utf-16"?>
<dummy>
  <text>
    Line 1
    Line 2
    Line 3
  </text>
</dummy>

这可能吗？如果是这样，怎么样？我宁愿不做只是在我自己添加空白的hackish方式。

原因是这个XML将由人们查看和编辑，所以我希望初始输出能够更好地为他们开箱即用。

Answer 1

我碰到了同样的问题。最后，我找到了一位自定义作家：

public class IndentTextXmlWriter : XmlTextWriter
{
    private int indentLevel;
    private bool isInsideAttribute;

    public IndentTextXmlWriter(TextWriter textWriter): base(textWriter)
    {
    }

    public bool IndentText { get; set; }

    public override void WriteStartAttribute(string prefix, string localName, string ns)
    {
        isInsideAttribute = true;
        base.WriteStartAttribute(prefix, localName, ns);
    }

    public override void WriteEndAttribute()
    {
        isInsideAttribute = false;
        base.WriteEndAttribute();
    }

    public override void WriteStartElement(string prefix, string localName, string ns)
    {
        indentLevel++;
        base.WriteStartElement(prefix, localName, ns);
    }

    public override void WriteEndElement()
    {
        indentLevel--;
        base.WriteEndElement();
    }

    public override void WriteString(string text)
    {
        if (String.IsNullOrEmpty(text) || isInsideAttribute || Formatting != Formatting.Indented || !IndentText || XmlSpace == XmlSpace.Preserve)
        {
            base.WriteString(text);
            return;
        }

        string[] lines = text.Split(new[] { Environment.NewLine }, StringSplitOptions.RemoveEmptyEntries);
        string indent = new string(IndentChar, indentLevel * Indentation);
        foreach (string line in lines)
        {
            WriteRaw(Environment.NewLine);
            WriteRaw(indent);
            WriteRaw(line.Trim());
        }

        WriteRaw(Environment.NewLine);
        WriteRaw(new string(IndentChar, (indentLevel - 1) * Indentation));
    }
}

你可以像这样使用它：

[TestMethod]
public void WriteIndentedText()
{
    var result = new StringBuilder();
    using (var writer = new IndentTextXmlWriter(new StringWriter(result)){Formatting = Formatting.Indented, IndentText = true})
    {
        string text = @" Line 1
Line 2
    Line 3  ";
        // some root
        writer.WriteStartDocument();
        writer.WriteStartElement("root");
        writer.WriteStartElement("child");

        // test auto-indenting
        writer.WriteStartElement("elementIndented");
        writer.WriteString(text);
        writer.WriteEndElement();

        // test space preserving
        writer.WriteStartElement("elementPreserved");
        writer.WriteAttributeString("xml", "space", null, "preserve");
        writer.WriteString(text);

        writer.WriteEndDocument();
    }

    Debug.WriteLine(result.ToString());
}

输出：

<?xml version="1.0" encoding="utf-16"?>
<root>
  <child>
    <elementIndented>
      Line 1
      Line 2
      Line 3
    </elementIndented>
    <elementPreserved xml:space="preserve"> Line 1
Line 2
    Line 3  </elementPreserved>
  </child>
</root>

在字符串属性的xml序列化中缩进文本？

1 个答案: