更新C＃中的特定XML元素

Question

我正在用C＃开发一个游戏，它将在XML文件中存储多个用户的数据。 我在弄清楚如何仅为当前播放器（例如Jack）更新XML数据时遇到了麻烦：

    <?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文件中的元素应与C＃（Jack）中的字符串变量“ strPlayerName”匹配。 然后，仅应更新Jack的WinCount和PlayCount编号。

如何将元素与strPlayerName字符串变量匹配，然后仅针对此播放器更新XML文档中的和数字？ 谢谢，

Answer 1

正如Matthew Watson建议的那样，一个好的解决方案是使用XML和序列化。

在您的项目中创建一个xml，并确保将“构建操作”和“始终复制”或“如果复制到输出目录较新”，则将其属性设置为none。

这是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>

现在，我们将使用此XML将其反序列化为播放器列表。 我在下面有一个用于序列化的帮助程序类。 您将读取XML文件的内容，并将其传递给Deserialize方法，如图所示。 当您想要保存播放器列表时，将该列表传递给Serializer并保存回您的文件。

序列化器帮助程序类：

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);
    } 
}

使用该类进行反序列化：

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

使用该类来序列化并保存列表：

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

最后是Player类：

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

您将在代码中根据需要使用Player类和列表。

Answer 2

假设您的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>

这是使用XmlNode API更新John的统计信息的方法：

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);
}

尽管正如其他人所说，对于这样的任务，序列化，反序列化可能是解决之道。

Answer 3

更改您的XML结构，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>

创建一些类来保存您的XML数据：

[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，您可以执行以下操作来读取，更新和重写文件。

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);
    }
}