字符串转义为XML-Attribute [英] string escape into XML-Attribute

问题描述

我看了一下字符串转义为XML ，发现它非常有用。 p>

我想做一个类似的事情：逃避一个要在XML属性中使用的字符串。

字符串可能包含\r\\\
。
XmlWriter类产生类似\r\\\
- >& #xD;& #xA;

我目前正在使用的解决方案包括XmlWriter和一个StringBuilder，而且很丑陋。

任何提示？

Edit1： / strong>

对不起，LarsH，购买我的第一种方法是

  public static string XmlEscapeAttribute（string unescaped）
 {
 XmlDocument doc = new XmlDocument（）; 
 XmlAttribute attr = doc.CreateAttribute（attr）; 
 attr.InnerText = unescaped; 
 return attr.InnerXml; 
} 
  

它不工作。 XmlEscapeAttribute（Foo\r\\\
Bar）将导致Foo\r\\\
Bar

我使用.NET Reflector来了解XmlTextWriter如何转义属性。它使用内部的XmlTextEncoder类...

我的方法我目前是usig这样的信息：

  public static string XmlEscapeAttribute（string unescaped）
 {
 if（String.IsNullOrEmpty（unescaped））return unescaped; 
 
 XmlWriterSettings settings = new XmlWriterSettings（）; 
 settings.OmitXmlDeclaration = true; 
 StringBuilder sb = new StringBuilder（）; 
 XmlWriter writer = XmlWriter.Create（sb，settings）; 
 
 writer.WriteStartElement（a）; 
 writer.WriteAttributeString（a，unescaped）; 
 writer.WriteEndElement（）; 
 writer.Flush（）; 
 sb.Length  -  =\/>\".Length; 
 sb.Remove（0，< a a = \。 
 
 return sb.ToString（）; 
} 
  

它的丑陋可能很慢，但它的工作： XmlEscapeAttribute（Foo\r\\\
Bar）将导致Foo& #xD;& #xA; Bar / p>

Edit2：

  SecurityElement.Escape （未逸出）; 
  

不起作用。

Edit3（final）：

使用Lars的所有非常有用的注释，我的最终实现如下所示：

注意： .Replace（\r，& #xD;）替换（\\\
& #xA;）; 不需要有效的XMl。这只是一个美容方法！

  public static string XmlEscapeAttribute（string unescaped）
 {
 
 XmlDocument doc = new XmlDocument（）; 
 XmlAttribute attr = doc.CreateAttribute（attr）; 
 attr.InnerText = unescaped; 
 //替换是*不*必填！ 
 return attr.InnerXml.Replace（\r，& #xD;）。Replace（\\\
，& #xA;）; 
} 
  

事实证明，这是有效的XML，将被任何标准兼容XMl-parser：

 < response message =谢谢，
 LarsH！ /> 
  

解决方案

修改您引用的解决方案， >

  public static string XmlEscape（string unescaped）
 {
 XmlDocument doc = new XmlDocument（）; 
 var node = doc.CreateAttribute（foo）; 
 node.InnerText = unescaped; 
 return node.InnerXml; 
} 
  

我所做的只是将CreateElement（）更改为CreateAttribute（）。
属性节点类型具有InnerText和InnerXml属性。

我没有环境来测试这个，但是我很好奇地知道如果它有效。

更新：或者更简单的说，使用 SecurityElement.Escape（） ，如您所链接的问题的另一个答案所建议的。这将排除引号，因此适用于属性文本。

更新2：请注意回车和换行不需要在属性值中转义，以使XML格式正确。如果您希望他们因为其他原因被转义，可以使用String.replace（）来执行，例如

  SecurityElement.Escape （未转义）.Replace（\r，& #xD;）。Replace（\\\
，& #xA;）; 
  

或

  return node.InnerXml.Replace（\r，& #xD;）。Replace（\\\
，& #xA;）; 
  

I had a look at string escape into XML and found it very useful.

I would like to do a similar thing: Escape a string to be used in an XML-Attribute.

The string may contain \r\n. The XmlWriter class produces something like \r\n -> 


The solution I'm currently using includes the XmlWriter and a StringBuilder and is rather ugly.

Any hints?

Edit1:
Sorry to disappoint LarsH, buy my first approach was

public static string XmlEscapeAttribute(string unescaped)
{
    XmlDocument doc = new XmlDocument();
    XmlAttribute attr= doc.CreateAttribute("attr");
    attr.InnerText = unescaped;
    return attr.InnerXml;
}

It does not work. XmlEscapeAttribute("Foo\r\nBar") will result in "Foo\r\nBar"

I used the .NET Reflector, to find out how the XmlTextWriter escapes Attributes. It uses the XmlTextEncoder class which is internal...

My method I'm currently usig lokks like this:

public static string XmlEscapeAttribute(string unescaped)
{
    if (String.IsNullOrEmpty(unescaped)) return unescaped;

    XmlWriterSettings settings = new XmlWriterSettings();
    settings.OmitXmlDeclaration = true;
    StringBuilder sb = new StringBuilder();
    XmlWriter writer = XmlWriter.Create(sb, settings);

    writer.WriteStartElement("a");
    writer.WriteAttributeString("a", unescaped);
    writer.WriteEndElement();
    writer.Flush();
    sb.Length -= "\" />".Length;
    sb.Remove(0, "<a a=\"".Length);

    return sb.ToString();
}

It's ugly and probably slow, but it does work: XmlEscapeAttribute("Foo\r\nBar") will result in "Foo
Bar"

Edit2:

SecurityElement.Escape(unescaped);

does not work either.

Edit3 (final):

Using all the very useful comments from Lars, my final implementation looks like this:

Note: the .Replace("\r", "
").Replace("\n", "
"); is not required for valid XMl. It is a cosmetic measure only!

    public static string XmlEscapeAttribute(string unescaped)
    {

        XmlDocument doc = new XmlDocument();
        XmlAttribute attr= doc.CreateAttribute("attr");
        attr.InnerText = unescaped;
        // The Replace is *not* required!
        return attr.InnerXml.Replace("\r", "
").Replace("\n", "
");
    }

As it turns out this is valid XML and will be parsed by any standard compliant XMl-parser:

<response message="Thank you,
LarsH!" />

解决方案

Modifying the solution you referenced, how about

public static string XmlEscape(string unescaped)
{
    XmlDocument doc = new XmlDocument();
    var node = doc.CreateAttribute("foo");
    node.InnerText = unescaped;
    return node.InnerXml;
}

All I did was change CreateElement() to CreateAttribute(). The attribute node type does have InnerText and InnerXml properties.

I don't have the environment to test this in, but I'd be curious to know if it works.

Update: Or more simply, use SecurityElement.Escape() as suggested in another answer to the question you linked to. This will escape quotation marks, so it's suitable for using for attribute text.

Update 2: Please note that carriage returns and line feeds do not need to be escaped in an attribute value, in order for the XML to be well-formed. If you want them to be escaped for other reasons, you can do it using String.replace(), e.g.

SecurityElement.Escape(unescaped).Replace("\r", "
").Replace("\n", "
");

or

return node.InnerXml.Replace("\r", "
").Replace("\n", "
");

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