我正在用C#构建一个游戏,它将多个用户的数据存储在XML文件中。我无法确定如何仅为当前播放器(例如Jack)更新XML数据:
<?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,只应更新Jack的WinCount和PlayCount数字。
如何将元素与strPlayerName字符串变量匹配,然后仅为此播放器更新XML文档中的和数字?谢谢,
答案 0 :(得分:1)
正如Matthew Watson所推荐的,一个好的解决方案是使用XML和序列化。
在项目中创建一个xml,并确保其属性设置为none,用于Build Action和Copy Always,如果更新则复制到输出目录。
以下是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。我有一个帮助类,用于下面的序列化。您将读取XML文件内容并将其传递给Deserialize方法,如图所示。如果要保存播放器列表,请将列表传递给Serializer并保存回文件。
Serializer helper class:
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);
}
}
使用类反序列化:
//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; }
}
您可以在代码中根据需要使用您的Player类和列表。
答案 1 :(得分:0)
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>
以下是使用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);
}
虽然正如其他人所说,对于这样的任务,序列化,反序列化可能是要走的路。
答案 2 :(得分:0)
将XML结构更改为:
<?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阅读,更新并重写该文件。
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);
}
}