是什么让缩进的XML与XmlDocument的换行符最简单的方法是什么？ [英] What is the simplest way to get indented XML with line breaks from XmlDocument?

问题描述

当我与的XmlDocument 构建XML从头开始，在 OuterXml 属性已有一切都很好地缩进与换行。但是，如果我叫 loadXML的一些非常COM pressedXML（没有换行符或缩进），则输出 OuterXml 保持这种方式。所以......

When I build XML up from scratch with XmlDocument, the OuterXml property already has everything nicely indented with line breaks. However, if I call LoadXml on some very "compressed" XML (no line breaks or indention) then the output of OuterXml stays that way. So ...

什么是得到美化XML输出​​从的XmlDocument ？

What is the simplest way to get beautified XML output from an instance of XmlDocument?

推荐答案

根据其他答案，我看着的XmlTextWriter ，并提出了以下辅助方法：

Based on the other answers, I looked into XmlTextWriter and came up with the following helper method:

static public string Beautify(this XmlDocument doc)
{
    StringBuilder sb = new StringBuilder();
    XmlWriterSettings settings = new XmlWriterSettings
    {
        Indent = true,
        IndentChars = "  ",
        NewLineChars = "\r\n",
        NewLineHandling = NewLineHandling.Replace
    };
    using (XmlWriter writer = XmlWriter.Create(sb, settings)) {
        doc.Save(writer);
    }
    return sb.ToString();
}

这是一个有点更code不是我希望的，但它只是桃色。

It's a bit more code than I hoped for, but it works just peachy.

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