以特定格式獲取XML c＃

Question

我正在使用MVC Web API，其中一個函數需要將SQL DB結果轉換爲XML文件。

private string DbToXml(int fileID) 
     { 
      DataSet ds = new DataSet("Requirements"); 
      string connetionString = ConfigurationManager.ConnectionStrings["DemoConnectionString"].ConnectionString; 
      string XML; 
      using (SqlConnection connection = new SqlConnection(connetionString)) 
      { 
       string sql = "SELECT RequirementLabel as ID, VagueWord, Suggestion, Explanation, VaguePhrase, ContentText, VagueTypeText FROM [Test].[dbo].[Vague_Terms_View] where FileID=" + fileID; 
       string XML_Output_Path = System.Configuration.ConfigurationManager.AppSettings.Get("XML_Output_Path"); 
       connection.Open(); 
       SqlDataAdapter adapter = new SqlDataAdapter(sql, connection); 
       adapter.Fill(ds, "Requirement"); 
       var sb = new StringBuilder(); 
       XmlWriter xmlWrite = XmlWriter.Create(sb); 
       ds.WriteXml(xmlWrite); 
       XML = ds.GetXml(); 
      } 

我從下面的代碼中獲得XML，這是正確的。

<?xml version="1.0" encoding="utf-8"?> 
<Requirements> 
    <Requirement> 
    <ID>Req97</ID> 
    <VagueWord>or</VagueWord> 
    <Suggestion>Keep each requirement in a single sentence.</Suggestion> 
    <Explanation>Suggests that you are combining requirements. Requirements that contain conjunctions/disjunctions (AND/OR) are dangerous and can lead to downstream problems in defining scope of the requirement.</Explanation> 
    <VaguePhrase>Marketing or Servicing</VaguePhrase> 
    <ContentText>If a user is identified as Marketing or Servicing, then the Campaign Management (CM) hyperlink should be displayed.</ContentText> 
    <VagueTypeText>Not Standard</VagueTypeText> 
    </Requirement> 
    <Requirement> 
    <ID>Req97</ID> 
    <VagueWord>should</VagueWord> 
    <Suggestion>Use 'shall/must/will' for requirements,</Suggestion> 
    <Explanation>Is often ambiguous, or inappropriate. Some readers will interpret these as optional or advisory, others as required.</Explanation> 
    <ContentText>If a user is identified as Marketing or Servicing, then the Campaign Management (CM) hyperlink should be displayed.</ContentText> 
    <VagueTypeText>Not Standard</VagueTypeText> 
    </Requirement> 
    <Requirement> 
    <ID>Req98</ID> 
    <VagueWord>Unless</VagueWord> 
    <Suggestion>Specify each conditions explicitly. One condition per requirement.</Suggestion> 
    <Explanation>Is an escape clause. Requirements with escape clauses are not testable. The word implies additional condition to the requirement.</Explanation> 
    <ContentText>Unless Sleep, Latency, Noise, or apply conditions are present, the data transmissions will contain the code for Normal Operation.</ContentText> 
    <VagueTypeText>Not Standard</VagueTypeText> 
    </Requirement> 
</Requirements> 

但現在我需要轉換XML，檢查需求節點內的ID元素。如果它在Requirement節點下面重複，則將其重命名爲Requirement節點內的所有其他元素，以便追加id = 1，數字依此類推。 以上XML的預期輸出如下所示。

<?xml version="1.0" encoding="utf-8"?> 
<Requirements> 
    <Requirement> 
    <ID "id=1">Req97</ID> 
    <VagueWord "id=1">or</VagueWord> 
    <Suggestion "id=1">Keep each requirement in a single sentence.</Suggestion> 
    <Explanation "id=1">Suggests that you are combining requirements. Requirements that contain conjunctions/disjunctions (AND/OR) are dangerous and can lead to downstream problems in defining scope of the requirement.</Explanation> 
    <VaguePhrase "id=1">Marketing or Servicing</VaguePhrase> 
    <ContentText "id=1">If a user is identified as Marketing or Servicing, then the Campaign Management (CM) hyperlink should be displayed.</ContentText> 
    <VagueTypeText "id=1">Not Standard</VagueTypeText> 
    </Requirement> 
    <Requirement> 
    <ID "id=2">Req97</ID> 
    <VagueWord "id=2">should</VagueWord> 
    <Suggestion "id=2">Use 'shall/must/will' for requirements,</Suggestion> 
    <Explanation "id=2">Is often ambiguous, or inappropriate. Some readers will interpret these as optional or advisory, others as required.</Explanation> 
    <ContentText "id=2">If a user is identified as Marketing or Servicing, then the Campaign Management (CM) hyperlink should be displayed.</ContentText> 
    <VagueTypeText "id=2">Not Standard</VagueTypeText> 
    </Requirement> 
    <Requirement> 
    <ID>Req98</ID> 
    <VagueWord>Unless</VagueWord> 
    <Suggestion>Specify each conditions explicitly. One condition per requirement.</Suggestion> 
    <Explanation>Is an escape clause. Requirements with escape clauses are not testable. The word implies additional condition to the requirement.</Explanation> 
    <ContentText>Unless Sleep, Latency, Noise, or apply conditions are present, the data transmissions will contain the code for Normal Operation.</ContentText> 
    <VagueTypeText>Not Standard</VagueTypeText> 
    </Requirement> 
</Requirements> 

Answer 1

您可以反序列化（創建XML對象），修改您的ID作爲對象，然後序列化對象返回XML或者你可以直接編輯XML。

這是如何直接編輯XML的一個很好的例子：How to modify existing XML file with XmlDocument and XmlNode in C#。

Answer 2

var doc = XElement.Load("test.xml"); 

var dict = doc.Elements("Requirement") 
    .Elements("ID") 
    .GroupBy(id => id.Value) 
    .ToDictionary(id => id.Key, id => id.Count() > 1 ? 1 : 0); 

foreach (var req in doc.Elements("Requirement")) 
{ 
    var id = req.Element("ID").Value; 

    if (dict[id] > 0) 
    { 
     foreach (var elem in req.Elements()) 
     { 
      elem.Add(new XAttribute("id", dict[id])); 
     } 
     dict[id]++; 
    } 
} 

現在doc含有添加了id屬性在需要XML。

Answer 3

我已經使用下面的解決方案，不知道它的優化

XmlElement root = returnXmlFile.DocumentElement; 
     XmlNodeList nodes = root.ChildNodes; 
     for (int i = 0; i < nodes.Count; i++) 
     { 
      string OuterIDvalue = nodes[i].ChildNodes[0].InnerText.ToString(); 
      const int idVal = 1; 
      int counter = 1; 
      if (nodes[i].FirstChild.Attributes.Count == 0) 
      { 
       for (int ctrInner = 0; ctrInner < nodes[i].ChildNodes.Count; ctrInner++) 
       { 
        XmlAttribute xKey = returnXmlFile.CreateAttribute("id"); 
        xKey.Value = idVal.ToString(); 
        nodes[i].ChildNodes[ctrInner].Attributes.Append(xKey); 
       } 
      }  
       for (int j = i + 1; j < nodes.Count; j++) 
       { 
        string innerIDvalue = nodes[j].ChildNodes[0].InnerText.ToString(); 
        if (OuterIDvalue == innerIDvalue && nodes[j].FirstChild.Attributes.Count == 0) 
        { 
         counter++; 
         for (int ctr = 0; ctr < nodes[j].ChildNodes.Count; ctr++) 
         { 
          XmlAttribute xKey = returnXmlFile.CreateAttribute("id"); 
          xKey.Value = counter.ToString(); 
          nodes[j].ChildNodes[ctr].Attributes.Append(xKey); 
         } 
        } 
       } 
     } 
     string xmlnew = returnXmlFile.InnerXml; 
     returnXmlFile.Save(FileName); 
     return xmlnew; 
    }