更新C＃中的特定XML元素

Question

我正在用C＃開發一個游戲，它將在XML文件中存儲多個用戶的數據。 我在弄清楚如何僅為當前播放器（例如Jack）更新XML數據時遇到了麻煩：

    <?xml version="1.0" encoding="utf-8"?>
<PlayerStats>
  <Name>Jack</Name>
  <WinCount>15</WinCount>
  <PlayCount>37</PlayCount>

  <Name>John</Name>
  <WinCount>12</WinCount>
  <PlayCount>27</PlayCount>
</PlayerStats>

XML文件中的元素應與C＃（Jack）中的字符串變量“ strPlayerName”匹配。 然后，僅應更新Jack的WinCount和PlayCount編號。

如何將元素與strPlayerName字符串變量匹配，然后僅針對此播放器更新XML文檔中的和數字？ 謝謝，

Answer 1

正如Matthew Watson建議的那樣，一個好的解決方案是使用XML和序列化。

在您的項目中創建一個xml，並確保將“構建操作”和“始終復制”或“如果復制到輸出目錄較新”，則將其屬性設置為none。

這是xml文件的示例：

<?xml version="1.0" encoding="utf-8" ?>
<ArrayOfPlayer xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Player>
    <Name>Jack</Name>
    <WinCount>15</WinCount>
    <PlayCount>37</PlayCount>
  </Player>
  <Player>
    <Name>John</Name>
    <WinCount>12</WinCount>
    <PlayCount>27</PlayCount>
  </Player>
</ArrayOfPlayer>

現在，我們將使用此XML將其反序列化為播放器列表。 我在下面有一個用於序列化的幫助程序類。 您將讀取XML文件的內容，並將其傳遞給Deserialize方法，如圖所示。 當您想要保存播放器列表時，將該列表傳遞給Serializer並保存回您的文件。

序列化器幫助程序類：

public static class Serializer
{
    public static string SerializeObject(object objectToSerialize)
    {
        XmlSerializer x = new XmlSerializer(objectToSerialize.GetType());

        StringWriter writer = new StringWriter();
        x.Serialize(writer, objectToSerialize);

        return writer.ToString();
    }

    public static T DeserializeObject<T>(string serializedObject)
    {
        XmlSerializer xs = new XmlSerializer(typeof(T));
        StringReader reader = new StringReader(serializedObject);
        return (T)xs.Deserialize(reader);
    } 
}

使用該類進行反序列化：

//Change this as needed to read your XML file. 
string playersXML = File.ReadAllText(@"./Players.xml");
List<Player> players = Serializer.DeserializeObject<List<Player>>(playersXML);

使用該類來序列化並保存列表：

string newPlayersXML = Serializer.SerializeObject(players);
//Change this as needed to point to the XML location
File.WriteAllText(@"./Players.xml", newPlayersXML);

最后是Player類：

public class Player
{
    public string Name { get; set; }
    public int WinCount { get; set; }
    public int PlayCount { get; set; }
}

您將在代碼中根據需要使用Player類和列表。

Answer 2

假設您的XML文件如下所示：

<PlayerStats>
  <Player>
    <Name>Jack</Name>
    <WinCount>15</WinCount>
    <PlayCount>37</PlayCount>
  </Player>
  <Player>
    <Name>John</Name>
    <WinCount>12</WinCount>
    <PlayCount>27</PlayCount>
  </Player>
</PlayerStats>

這是使用XmlNode API更新John的統計信息的方法：

string name = "John";

XmlNode player = doc.SelectNodes("/PlayerStats/Player")
                    .OfType<XmlNode>()
                    .FirstOrDefault(n => n["Name"].InnerText == name);

if (player != null)
{
    player["WinCount"].InnerText = "21";
    player["PlayCount"].InnerText = "22";
}

或使用LINQ to XML：

var player2 = xe.Descendants("Player")
                .FirstOrDefault(n => (string)n.Element("Name") == name);

if (player != null)
{
    player2.Element("WinCount").SetValue(21);
    player2.Element("PlayCount").SetValue(22);
}

盡管正如其他人所說，對於這樣的任務，序列化，反序列化可能是解決之道。

Answer 3

更改您的XML結構，如下所示：

<?xml version="1.0" encoding="utf-8" ?>
<PlayerStats>
    <Player>
        <Name>Jack</Name>
        <WinCount>15</WinCount>
        <PlayCount>37</PlayCount>
    </Player>

    <Player>
        <Name>John</Name>
        <WinCount>12</WinCount>
        <PlayCount>27</PlayCount>
    </Player>
</PlayerStats>

創建一些類來保存您的XML數據：

[XmlRoot("PlayerStats")]
public class PlayerStats
{
    [XmlElement(ElementName = "Player")]
    public List<Player> Players { get; set; }
}

public class Player
{
    public string Name { get; set; }
    public int WinCount { get; set; }
    public int PlayCount { get; set; }
}

然后，您可以執行以下操作來讀取，更新和重寫文件。

PlayerStats stats;
using (var fileStream = new System.IO.FileStream("Sample.XML", System.IO.FileMode.Open))
{
    XmlSerializer xmlSerializer = new XmlSerializer(typeof(PlayerStats));
    stats = (PlayerStats)xmlSerializer.Deserialize(fileStream);
}

var player = stats.Players.Where(p => p.Name == "Jack").FirstOrDefault();
if (player != null)
{
    // Update the record
    player.WinCount = player.WinCount + 1;
    player.PlayCount = player.PlayCount + 1;

    // Save back to file
    using (var fileStream = new System.IO.FileStream("Sample.XML", System.IO.FileMode.Create))
    {
        XmlSerializer xmlSerializer = new XmlSerializer(typeof(PlayerStats));
        xmlSerializer.Serialize(fileStream, stats);
    }
}