Question

我需要帮助，我有两个XML文件，它们具有相同的id（“ name”），我希望两者之间有所不同。如果我在Xml file_1中有多余的节点，那么它将转到diff。仅根据ID的xml文件。我已经编写了som代码，但我不知道如果怎么做；我知道我们可以在Microsoft Diff and Patch工具中执行操作，但是我想要一个不同的代码：这是我的代码：

    public static void Main()
    {
           //fields
         const string XML1 = @"File_1.xml";
         const string XML2 = @"File_2.xml";
         const string ResultFile = @"ResultFile.xml";
         XmlDocument doc1 = new XmlDocument();
         doc1.Load(XML1);
         XmlDocument doc2 = new XmlDocument();
         doc2.Load(XML2);

         for(var d = 0; d < doc1.SelectNodes("root/data").Count; d++)
         {
             var child = doc1.SelectNodes("root/data")[d];
            if (I don't know …..) {

           }
         }

XML fil_1

<root>
  <data name="senChangePassword" xml:space="preserve">
    <value>Byt lösenord</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
  <data name="senChangesWereSuccessfullySaved" xml:space="preserve">
    <value>Ändringarna är sparade</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
  <data name="senChangeUserSettings" xml:space="preserve">
    <value>Ändra uppgifter</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
  <data name="senCompareWith" xml:space="preserve">
    <value>Jämför</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
  <data name="senCreatedQuestions" xml:space="preserve">
    <value>Skapade frågor</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
</root>

** XML文件_2 **

<root>
  <data name="senChangePassword" xml:space="preserve">
    <value>Change Password</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
  <data name="senChangesWereSuccessfullySaved" xml:space="preserve">
    <value>Saved changes</value>
    <comment>Sprint 02  Jessica</comment>
  </data>
  <data name="senCompareWith" xml:space="preserve">
    <value>Compare</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
</root>

XML file_result

<root>
  <data name="senChangeUserSettings" xml:space="preserve">
    <value>Ändra uppgifter</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
  <data name="senCreatedQuestions" xml:space="preserve">
    <value>Skapade frågor</value>
    <comment>Sprint 02 - Jessica</comment>
  </data>
</root>

Answer 1

简单版本（如果您不关心性能或内存大小）：

using System.Linq;

// ....
var nodes1 = doc1.SelectNodes("root/data");
var children1 = new List<XmlNode>();
int i = 0;
foreach(XmlNode a in nodes1) { children1[i++] = a; }
var nodes2 = doc2.SelectNodes("root/data");
var children2 = new List<XmlNode>();
i = 0;
foreach(XmlNode b in nodes2) { children2[i++] = b; }
var result = new List<XmlNode>();

for(var d = 0; d < children1.Count; d++)
         {
             var child = children1[d];
             var match = children2.FirstOrDefault(x=>
                 XElement.Parse(x.OuterXml).ToString() == XElement.Parse(child.OuterXml).ToString());
            if (match == null) {
               result.Add(child);
           }
         }

将结果节点尽可能地转储到结果文件中

比较两个XML文件，并在新的XML文件中获得它们之间的区别

1 个答案: