如何在C#中从XmlNode获取参数值

Question

如何获取XmlNode标签中参数的值。例如：

<weather time-layout="k-p24h-n7-1">
    <name>Weather Type, Coverage, and Intensity</name>
    <weather-conditions weather-summary="Mostly Sunny"/>
</weather>

我想要获取节点'weather-conditions'中参数'weather-summary'的值。

- Brian

Answer 1

var node = xmldoc.SelectSingleNode("weather/weather-conditions");
var attr = node.Attributes["weather-summary"];

- Dean Harding

Answer 2

为了完整起见，应该同时提供 .Net 3.5 的方式:

假设

XDocument doc = XDocument.Parse(@"<weather time-layout='k-p24h-n7-1'>
    <name>Weather Type, Coverage, and Intensity</name>
    <weather-conditions weather-summary='Mostly Sunny'/></weather>");

然后选择其中一种方式

return doc.Element("weather").Element("weather-conditions").Attribute("weather-summary").Value;

或者

return doc.Descendants("weather-conditions").First().Attribute("weather-summary").Value;

会给你相同的答案。

- Dathan