Question

我有以下xml架构，我想使用linq查询2个属性。我一直在寻找，但没有找到正确的解决方案。

<Object class="MA" Name="Sample">
  <bist name="act">false</bist>
  <bist name="Dynamic">1234</bist>
  <bist name="Fast">false</bist>
  <bist name="plane">false</bist>
  <bist name="Tnl">2232</bist>
</Object>

对于上面的xml，获取＆＃34; Dynamic＆＃34;的值后并将它与oldTnl变量进行比较，如果相等，我想选择或得到＆＃34; Tnl＆＃34;价值（2232）。

目前，我正在使用此代码进行测试，并成功获得＆＃34; Dynamic＆＃34;但我真的想要＆＃34; Tnl＆＃34;。

的价值

 private void mGetTnlFromXML(string oldTnl)
    {
        XDocument doc = XDocument.Load("sample.xml");

        var bt = from p in doc.Descendants()
                  where (string)p.Attribute("name") == "Dynamic"
                  select p;

        foreach (string b in bt)
        {
            if (b == oldTnl)
            {
                MessageBox.Show(b.ToString());
            }
        }
    }

类似的东西：

private void mGetTnlFromXML(string oldTnl)
    {
        XDocument doc = XDocument.Load("sample.xml");

        var bt = from p in doc.Descendants()
                  where (string)p.Attribute("name") == "Dynamic"
                  //is there are way i can also find "Tnl" here and use 
                  //later?
                  select p; //or select "Tnl" here.

        foreach (string b in bt)
        {
            if (b == oldTnl)
            {
                //select "Tnl" value (2232)
                //use "Tnl" value (2232)
                //do something....
            }
        }
    }

提前谢谢你......我还在学习LinQ：）。

更新了XML：

<Root>
  <Data>
    <Object class="MA" Name="Sample">
      <bist name="act">false</bist>
      <bist name="Dynamic">1234</bist>
      <bist name="Fast">false</bist>
      <bist name="plane">false</bist>
      <bist name="Tnl">2232</bist>
    </Object>
  </Data>
</Root>

Answer 1

在我看来，你想要这个：

private void mGetTnlFromXML(string oldTnl)
{
    XDocument doc = XDocument.Load("sample.xml");
    var dynamic = doc.Root.Elements("bist").Where(x => x.Attribute("name").Value == "dynamic").First().Value;
    if (dynamic == oldTnl)
    {
        var tnl = doc.Root.Elements("bist").Where(x => x.Attribute("name").Value == "Tnl").First().Value;
        MessageBox.Show(tnl);
    };
}

假设您在问题中显示了整个XML，这是假设的。如果它只是一个更大的XML文件的一部分，那么你需要展示整个文件。

鉴于您的XML更像是这样：

<Root>
  <Data>
    <Object class="MA" Name="Sample">
      <bist name="act">false</bist>
      <bist name="Dynamic">1234</bist>
      <bist name="Fast">false</bist>
      <bist name="plane">false</bist>
      <bist name="Tnl">2232</bist>
    </Object>
  </Data>
</Root>

然后代码可能会更像这样：

private void mGetTnlFromXML(string oldTnl)
{
    XDocument doc = XDocument.Load("sample.xml");
    foreach (var element in doc.Root.Element("Data").Elements("Object"))
    {
        var dynamic = element.Elements("bist").Where(x => x.Attribute("name").Value == "dynamic").First().Value;
        if (dynamic == oldTnl)
        {
            var tnl = element.Elements("bist").Where(x => x.Attribute("name").Value == "Tnl").First().Value;
            MessageBox.Show(tnl);
        };
    }
}

Answer 2

使用Xml提供的Enigmativity，我们可以将其减少到一行XPath：

XDocument xml = XDocument.Parse(@"
<Root>
  <Object class=""MA"" Name=""Sample"">
    <bist name = ""act"" > false </bist >
    <bist name = ""Dynamic"" > 1234 </bist >
    <bist name = ""Fast"" > false </bist >
    <bist name = ""plane"" > false </bist >
    <bist name = ""Tnl"" > 2232 </bist >
  </Object >
  <Object class= ""MA"" Name = ""Sample"" >
      <bist name = ""act"" > false </bist >
      <bist name = ""Dynamic"" > 1234 </bist >
      <bist name = ""Fast"" > false </bist >
      <bist name = ""plane"" > false </bist >
      <bist name = ""Tnl"" > 2232 </bist >
  </Object >
</Root >");

xml.XPathSelectElements("//Object[bist[@name='Dynamic']]/bist[@name='Tnl']").Dump();

转换为：

//Object[...] - 找到有......的“对象”....
bist[@name='Dynamic'] - 一个“bist”元素，其中“name”等于“Dynamic”
/bist[...] - 然后在“对象”下面加上“bist”元素，其中包含......
@name='Tnl' - 属性“name”等于“tnl”

如何选择多个属性

2 个答案: