C＃：如何使用LINQ从XML读取类似的节点 [英] C#: how to read similar nodes from XML with LINQ

问题描述

我有一个更大的XML文件，我想读到我的应用程序并将一些值存储在自定义类中。在该文件中，有几个< tu>节点。 （新的TUs类/ tu部分）。有问题的部分是< seg>节点。 （每个< tu>具有两个具有相同父节点但具有不同属性（语言）的< seg>节点）。

如您所见，我的基本脚本将捕获相同的（第一个）< ; SEG>节点两次。

I have a bigger XML file what I wish to read to my app and store some of the values in a custom class. In the file, there are several <tu> nodes. (new TUs class / tu section). The Problematic part is with <seg> node. (each <tu> has two <seg> nodes with the same parent node but with different attribute (language)).
As you can see my basic script will catch the same (first) <seg> node twice.

<tu creationdate="20130619T135814Z" creationid="ANA" changedate="20130619T135814Z" changeid="ANA" lastusagedate="20130619T135814Z">
      <prop type="x-LastUsedBy">ANA</prop>
      <prop type="x-Origin">TM</prop>
      <prop type="x-OriginalFormat">TradosTranslatorsWorkbench</prop>
      <tuv xml:lang="de-DE">
        <seg>Die Teile werden im Querdurchlauf transportiert.<ph x="1" type="1" /></seg>
      </tuv>
      <tuv xml:lang="en-GB">
        <seg>The parts are transported in a cross conveyance.<ph x="1" type="1" /></seg>
      </tuv>
    </tu>

我尝试过：

What I have tried:

XDocument xDoc;
           xDoc = XDocument.Load(path);

           var result = from q in xDoc.Descendants("tu")
                        select new TUs
                        {
                            SOURCE = q.Element("tuv").Element("seg").Value,
                            TARGET = q.Element("tuv").Element("seg").Value,
                        };

           foreach (TUs TU in result)
           {
               Console.WriteLine(TU.SOURCE);
               Console.WriteLine(TU.TARGET);
           }

推荐答案

这只有在您的XML文件格式正确时才有效：

This can only work if your XML file is correctly formatted:

XDocument xDoc;
xDoc = XDocument.Load(path);

foreach (var node in xDoc.Root.Descendants().Where(n => n.Name == "seg"))
{
    Console.WriteLine(node.Name + " = " + node.Value);
}

XML数据：

XML data:

<tu creationdate="20130619T135814Z" creationid="John" changedate="20130619T135814Z" changeid="John" lastusagedate="20130619T135814Z">
  <prop type="x-LastUsedBy">
    John
    <prop type="x-Origin">
      TM
      <prop type="x-OriginalFormat">
        TestTool
        <tuv xml:lang="de-DE">
          <seg>This is the German text</seg>
        </tuv>
        <tuv xml:lang="en-GB">
          <seg>This is the English text</seg>
        </tuv>
      </prop>
    </prop>
  </prop>
</tu>

这是一个适用于多个tu节点的解决方案：

Here is a solution that works better with multiple tu nodes:

XDocument xDoc = XDocument.Load(path);

foreach (XElement tu in xDoc.Root.Descendants("tu"))
{
    Console.WriteLine("tu = " + tu.Attribute("creationid").Value);

    foreach (XElement tuv in tu.Descendants("tuv"))
    {
        if (tuv.ToString().Contains("\"en-GB\""))
        {
            Console.WriteLine("seg = " + tuv.Descendants("seg").First().Value);
        }
    }

    Console.WriteLine();
}

XML数据必须具有Root元素：

XML data must have a Root element:

<Root>
<tu creationdate="20130619T135814Z" creationid="John" changedate="20130619T135814Z" changeid="John" lastusagedate="20130619T135814Z">
  <prop type="x-LastUsedBy">John</prop>
  <prop type="x-Origin">TM</prop>
  <prop type="x-OriginalFormat">TestTool</prop>
  <tuv xml:lang="de-DE">
    <seg>This is the German text</seg>
  </tuv>
  <tuv xml:lang="en-GB">
    <seg>This is the English text</seg>
  </tuv>
</tu>
<tu creationdate="20130619T135814Z" creationid="Rick" changedate="20130619T135814Z" changeid="Rick" lastusagedate="20130619T135814Z">
  <prop type="x-LastUsedBy">Rick</prop>
  <prop type="x-Origin">TM</prop>
  <prop type="x-OriginalFormat">TestTool</prop>
  <tuv xml:lang="de-DE">
    <seg>This is Ricks German text</seg>
  </tuv>
  <tuv xml:lang="en-GB">
    <seg>This is Ricks English text</seg>
  </tuv>
</tu>
</Root>

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