LINQ - XML 选择节点泛型

本文关键字：节点 泛型 选择 XML LINQ | 更新日期: 2023-09-27 18:37:02

我正在尝试剥离我的XML并仅保留我需要的节点

输入 XML 是 

 <Employees>
  <Employee>
    <EmpId>1</EmpId>
    <Name>Sam</Name>
    <Sex>Male</Sex>
    <Address>
      <Country>USA</Country>
      <Zip>95220</Zip>
    </Address>
    <Address2>
      <Country>UK</Country>
      <Zip>E157JQ</Zip>
    </Address2>
  </Employee>
  <Employee>
    <EmpId>2</EmpId>
    <Name>Lucy</Name>
    <Sex>Female</Sex>
    <Address>
      <Country>USA</Country>
      <Zip>95220</Zip>
    </Address>
    <Address2>
      <Country>UK</Country>
      <Zip>E184JQ</Zip>
    </Address2>
  </Employee>
</Employees>

我的代码如下。

private void button1_Click(object sender, EventArgs e)
    {
        Stream s = openFileDialog1.OpenFile();
        var xDoc = XDocument.Load(s);
        string keep = "EmpId,Sex,Address,Zip,Address2,Country"; '' I can change this format
        string desStr = "Employee";
            string[] strArr = keep.Split(',');
            var nodesToDelete = xDoc.Root.Descendants(desStr)
                .SelectMany(el => el.Descendants()
                                  .Where(a => !strArr.Contains(a.Name.ToString())));
            foreach (var node in nodesToDelete.ToList())
                node.Remove();
            richTextBox1.Text = xDoc.ToString();
    }

我从上面得到的输出是 

<Employees>
      <Employee>
        <EmpId>1</EmpId>
        <Sex>Male</Sex>
        <Address>
          <Country>USA</Country>
          <Zip>95220</Zip>
        </Address>
        <Address2>
          <Country>UK</Country>
          <Zip>E157JQ</Zip>
        </Address2>
      </Employee>
      <Employee>
        <EmpId>2</EmpId>
        <Sex>Female</Sex>
        <Address>
          <Country>USA</Country>
          <Zip>95220</Zip>
        </Address>
        <Address2>
          <Country>UK</Country>
          <Zip>E184JQ</Zip>
        </Address2>
      </Employee>
    </Employees>

我需要的输出是 

<Employees>
  <Employee>
    <EmpId>1</EmpId>
    <Sex>Male</Sex>
    <Address>
     <Zip>95220</Zip>
    </Address>
    <Address2>
      <Country>UK</Country>
    </Address2>
  </Employee>
  <Employee>
    <EmpId>2</EmpId>
    <Sex>Female</Sex>
    <Address>
      <Zip>95220</Zip>
    </Address>
    <Address2>
      <Country>UK</Country>
    </Address2>
  </Employee>
</Employees>

如何查询地址''邮政编码和地址 2''国家/地区 我需要它是通用的（因此可以更改字符串保留），所以我无法对节点名称进行硬编码。

谢谢

LINQ - XML 选择节点泛型

好吧，

这取决于您希望这是通用的。这是保持当前代码结构的一种有点黑客的方式。

string keep = @"EmpId,Sex,Address,Address'Zip,Address2,Address2'Country";
string desStr = "Employee";
string[] strArr = keep.Split(',');
var nodesToDelete = xDoc.Root.Descendants(desStr)
                .SelectMany(el => el.Descendants()
                                  .Where(a => 
                                    {
                                        if (a.Parent.Name == desStr)
                                        {
                                            return !strArr.Contains(a.Name.ToString());
                                        }
                                        else
                                        {
                                            return !strArr.Contains(a.Parent.Name + @"'" + a.Name);
                                        }
                                    }));
foreach (var node in nodesToDelete.ToList())
      node.Remove();

正确的方法是保留要保留的所有节点的完整路径。

我的方式：

string keep = "Employees,Employee,EmpId,Sex,Address2,Address,Address.Zip,Address2.Country"; // I can change this format
string[] strArr = keep.Split(',');
    foreach (var node in xDoc.Descendants().ToArray())
    {
        var path = Path(node);
        if (!strArr.Any(path.EndsWith))
        {
            node.Remove();
        }
    }
    var results = xDoc.ToString();
}
private static string Path(XElement x)
{
    if (x.Parent != null)
    {
        return Path(x.Parent) + "." + x.Name;
    }
    return x.Name.ToString();
}
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