讓一個班級跟蹤相關的字符串

Question

有趣的是，隨着下面的類我知道我的輸出應該如何，但我無法弄清楚如何爲它保存數據。

請參閱下面的代碼：

public class QuickFailureReportText 
{ 
    public string[] device { get; set; } 
    public string[] group { get; set; } 
    public string[] pin { get; set; } 


    public override string ToString() 
    { 
     TextWriter tw; 
     StringBuilder sb = new StringBuilder(); 
     tw = new StringWriter(sb); 

     tw.WriteLine("Quick Failure Report"); 

     foreach (string dev in device) 
     { 
      tw.WriteLine("Failures in " + dev); 

      foreach (string grp in group) 
      { 
       tw.Write("Group " + grp + " : "); 

       foreach(string p in pin) 
       { 
        tw.Write(p + ", "); 
       } 
       tw.WriteLine(); //new line 
      } 
      tw.WriteLine(); //new line 
     } 

     return tw.ToString(); 
    } 
} 

所以我想做的事，就是我希望能夠somohe涉及三個不同的字符串「設備，組針」 somhow一個PIN屬於GROUP和一個GROUP屬於DEVICE。這怎麼可能？

如果我不夠清楚，請讓我知道。

---

UPDATE 好吧，我有，我可以沒有問題從中讀取數據的XML文件。 xml文件看起來是這樣的：

<?xml version="1.0" encoding="utf-8"?> 
<DEVICES> 

    <device> 
    <name>device 1</name> 
    <groups> 
     <group> 
     <group_name>group 1</group_name> 
     <pins> 
     <pin result="fail">A1</pin> 
     <pin result="pass">A2</pin> 
     </pins> 
     </group> 
     <group> 
     <group_name>group 2</group_name> 
     <pins> 
     <pin result="fail">B1</pin> 
     <pin result="pass">B2</pin> 
     </pins> 
     </group>  
    </groups> 
    </device> 

</DEVICES> 

所以我想從這個XML（可有很多設備的）收集數據，並使用我上面寫的類，過濾器失敗的引腳。

Answer 1

我已經寫了一些代碼，你可以用它來讀取XML文件中的變量設備所需的信息，存儲

public class Device 
{ 
    public string Name; 
    public Dictionary<string, Group> Groups = new Dictionary<string, Group>(); 
} 

public class Group 
{ 
    public string Name; 
    public List<string> Pins = new List<string>(); 
} 

public class QuickFailureReportText 
{ 
    public Dictionary<string, Device> devices = new Dictionary<string, Device>(); 

    public void AddLog(string deviceName, string groupName, string pin) 
    { 
     if (!devices.ContainsKey(deviceName)) 
      devices.Add(deviceName, new Device() 
       { Name = deviceName, Groups = new Dictionary<string, Group>() }); 

     if (!devices[deviceName].Groups.ContainsKey(groupName)) 
      devices[deviceName].Groups.Add(groupName, new Group() 
       { Name = groupName, Pins = new List<string>() }); 

     devices[deviceName].Groups[groupName].Pins.Add(pin); 
    } 


    public override string ToString() 
    { 
     TextWriter tw; 
     StringBuilder sb = new StringBuilder(); 
     tw = new StringWriter(sb); 
     tw.WriteLine("Quick Failure Report"); 

     XDocument xDoc = XDocument.Load(@"devices.xml"); 
     foreach (XElement device in xDoc.XPathSelectElements("DEVICES/device")) 
     { 
      foreach (XElement group in device.XPathSelectElements("groups/group")) 
      { 
       foreach (XElement pin in group.XPathSelectElements("pins/pin")) 
       { 
        if (pin.Attribute("result").Value == "fail") 
        { 
         AddLog(device.XPathSelectElement("name").Value, 
         group.XPathSelectElement("group_name").Value, pin.Value); 
        } 
       } 
      } 
     } 

     foreach (var device in devices.Values) 
     { 
      tw.WriteLine("Failures in " + device.Name); 

      foreach (var grp in device.Groups.Values) 
      { 
       tw.Write("Group " + grp.Name + " : "); 

       foreach (string p in grp.Pins) 
       { 
        tw.Write(p + ", "); 
       } 
       tw.WriteLine(); //new line 
      } 
      tw.WriteLine(); //new line 
     } 


     return tw.ToString(); 
    } 
} 

class Program 
{ 
    static void Main(string[] args) 
    { 
     string s = new QuickFailureReportText().ToString(); 
    } 
} 

下面是「S」字符串爲您的示例文件的價值：

Quick Failure Report 
Failures in device 1 
Group group 1 : A1, 
Group group 2 : B1, 

Answer 2

我會實現對象數據模型，用三類：

DEVICE HAS GROUP

GROUP HAS PIN

---

UPDATE

第1類：DEVICE，與成員字段list_of_groups（可以使用不同的名稱）

2類：GROUP，與構件字段list_of_pins

3類：PIN，與構件字段result（布爾值）

Answer 3

類似的東西？

public class Device 
{ 
    public string Name; 
    public List<Group> Groups = new List<Group>(); 
} 

public class Group 
{ 
    public string Name; 
    public List<Pin> Pins = new List<Pin>(); 
} 

public class Pin 
{ 
    public string Name; 
    public string Result; 
} 

Answer 4

我認爲這是更好，如果你用這種方式改變的xml：

<?xml version="1.0" encoding="utf-8"?> 
<DEVICES> 

    <device> 
    <name>device 1</name> 
    <groups> 
     <group> 
     <group_name>group 1</group_name> 
     <pins> 
     <pin result="fail">A1</pin> 
     <pin result="pass">A2</pin> 
     </pins> 
     </group> 
     <group> 
     <group_name>group 2</group_name> 
     <pins> 
     <pin result="fail">B1</pin> 
     <pin result="pass">B2</pin> 
     </pins> 
     </group>  
    </groups> 
    </device> 

</DEVICES> 

所以，你可以定義對象Device塔包含Group列表包含對象Pin的列表。