c# - 使用 LINQ 解析 XML：获取附近节点值的最佳方法是什么

Question

我正在尝试解析以下 XML：

<bs:School>
    <bs:Class number="1">
        <bs:Student>
            <bs:Gender>Male</bs:Gender>
            <bs:Name>Jason</bs:Name>
        </bs:Student>
        <bs:Student>
            <bs:Gender>Male</bs:Gender>
            <bs:Name>Danny</bs:Name>
        </bs:Student>
        <bs:Student>
            <bs:Gender>Female</bs:Gender>
            <bs:Name>Jessica</bs:Name>
        </bs:Student>
        <bs:Student>
            <bs:Gender>Male</bs:Gender>
            <bs:Name>Will</bs:Name>
        </bs:Student>
        <bs:Student>
            <bs:Gender>Female</bs:Gender>
            <bs:Name>Betty</bs:Name>
        </bs:Student>
    </bs:Class>
    <bs:Class number="2">
        <bs:Student>
            <bs:Gender>Male</bs:Gender>
            <bs:Name>Sean</bs:Name>
        </bs:Student>
        <bs:Student>
            <bs:Gender>Female</bs:Gender>
            <bs:Name>Jessica</bs:Name>
        </bs:Student>
        <bs:Student>
            <bs:Gender>Male</bs:Gender>
            <bs:Name>Dexter</bs:Name>
        </bs:Student>
    </bs:Class>
</bs:School>

如何选择所有男学生的名字？

我尝试使用以下 LINQ 查询：

 var resources = from c in doc.Descendants("Gender")
                 where (c.Value.ToString().Contains("Male"))
                 select c.Parent.Descendants("Name");

它总是返回一个空对象。

Answer 1

注意：我在根节点中添加了命名空间声明xmlns:bs = ""MyNamespace""。<bs:School xmlns:bs = "MyNamespace">所以在我的示例中 ，根节点看起来像<bs:School>

var xml = //your xml

var doc = XDocument.Parse(xml);

XNamespace bs = "MyNamespace";

string[] malesNames = 
                doc.Descendants(bs + "Student")
                   .Where(stud => stud.Element(bs + "Gender").Value == "Male")
                   .Select(stud => stud.Element(bs + "Name").Value)
                   .ToArray();

Console.WriteLine(string.Join(Environment.NewLine, malesNames));

将打印：

Jason 
Danny 
Will 
Sean 
Dexter 

---

下面给出了搜索男学生姓名的与命名空间无关的解决方案：

doc.Descendants()
   .Where(node => node.Name.LocalName == "Student")
   .Where(node => node.Elements().First(child => child.Name.LocalName == "Gender").Value == "Male") 
   .Select(node => node.Elements().First(child => child.Name.LocalName == "Name").Value);

---

随意将常见的查询片段提取到单独的扩展方法中，例如：

public static class XElementExtensions
{
    public static XElement GetFirstChild(this XElement node, string nodeName)
    {
        return node.Elements().First(child => child.Name.LocalName == nodeName);
    }
}

现在你可以：

doc.Descendants()
   .Where(node => node.Name.LocalName == "Student")
   .Where(node => node.GetFirstChild("Gender").Value == "Male") 
   .Select(node => node.GetFirstChild("Name").Value);