Question

我在完成这项看似简单的任务时遇到了困难。我想加载XML文件同样容易加载艺术资产：

        content  = new ContentManager(Services);
        content.RootDirectory = "Content";
        Texture2d background = content.Load<Texture2D>("images\\ice");

我不知道该怎么做。这个tutorial似乎有帮助，但如何获得StorageDevice实例？

我现在有一些工作，但感觉非常hacky：

public IDictionary<string, string> Get(string typeName)
        {
            IDictionary<String, String> result = new Dictionary<String, String>();
            xmlReader.Read(); // get past the XML declaration

            string element = null;
            string text = null;

            while (xmlReader.Read())
            {

                switch (xmlReader.NodeType)
                {
                    case XmlNodeType.Element:
                        element = xmlReader.Name;
                        break;
                    case XmlNodeType.Text:
                        text = xmlReader.Value;
                        break;
                }

                if (text != null && element != null)
                {
                    result[element] = text;
                    text = null;
                    element = null;
                }

            }
            return result;
        }

我将其应用于以下XML文件：

<?xml version="1.0" encoding="utf-8" ?>
<zombies>
  <zombie>
    <health>100</health>
    <positionX>23</positionX>
    <positionY>12</positionY>
    <speed>2</speed>
  </zombie>
</zombies>

它能够通过这个单元测试：

    internal virtual IPersistentState CreateIPersistentState(string fullpath)
    {
        IPersistentState target = new ReadWriteXML(File.Open(fullpath, FileMode.Open));
        return target;
    }

    /// <summary>
    ///A test for Get with one zombie.
    ///</summary>
    //[TestMethod()]
    public void SimpleGetTest()
    {
        string fullPath = "C:\\pathTo\\Data\\SavedZombies.xml";
        IPersistentState target = CreateIPersistentState(fullPath);
        string typeName = "zombie"; 

        IDictionary<string, string> expected = new Dictionary<string, string>();
        expected["health"] = "100";
        expected["positionX"] = "23";
        expected["positionY"] = "12";
        expected["speed"] = "2";

        IDictionary<string, string> actual = target.Get(typeName);

        foreach (KeyValuePair<string, string> entry in expected)
        {
            Assert.AreEqual(entry.Value, expected[entry.Key]);
        }
    }

当前方法的缺点：文件加载效果不佳，并且将值与值匹配似乎比不必要的更省力。此外，我怀疑这种方法会因XML中的多个条目而崩溃。

我无法想象这是最佳实施方式。

更新：根据@Peter Lillevold的建议，我改变了一下：

    public IDictionary<string, string> Get(string typeName)
    {
        IDictionary<String, String> result = new Dictionary<String, String>();

        IEnumerable<XElement> zombieValues = root.Element(@typeName).Elements();

        //result["health"] = zombie.Element("health").ToString();

        IDictionary<string, XElement> nameToElement = zombieValues.ToDictionary(element => element.Name.ToString());

        foreach (KeyValuePair<string, XElement> entry in nameToElement)
        {
            result[entry.Key] = entry.Value.FirstNode.ToString();
        }

        return result;
    }

    public ReadWriteXML(string uri)
    {
        root = XElement.Load(uri);
    }

    internal virtual IPersistentState CreateIPersistentState(string fullpath)
    {
        return new ReadWriteXML(fullpath);
    }

    /// <summary>
    ///A test for Get with one zombie.
    ///</summary>
    [TestMethod()]
    public void SimpleGetTest()
    {
        IPersistentState target = CreateIPersistentState("../../../path/Data/SavedZombies.xml");
        string typeName = "zombie"; 

        IDictionary<string, string> expected = new Dictionary<string, string>();
        expected["health"] = "100";
        expected["positionX"] = "23";
        expected["positionY"] = "12";
        expected["speed"] = "2";

        IDictionary<string, string> actual = target.Get(typeName);

        foreach (KeyValuePair<string, string> entry in expected)
        {
            Assert.AreEqual(entry.Value, actual[entry.Key]);
        }
    }

加载仍然很糟糕，不知怎的，我无法让单行ToDictionary与这两个lambdas一起工作。我不得不求助于那个foreach循环。我在那里做错了什么？

Answer 1

还有新的闪亮XElement（运动Linq to XML）。此示例将加载xml文件，查找僵尸并将值转储到字典中：

var doc = XElement.Load("filename");
var zombieValues = doc.Element("zombie").Elements();
var zombieDictionary = 
    zombieValues.ToDictionary(
        element => element.Name.ToString(), 
        element => element.Value);

如果您更愿意明确选择每个值（并使用强制转换为自动转换为正确的值类型），您可以这样做：

var zombie = doc.Element("zombie");
var health = (int)zombie.Element("health");
var positionX = (int)zombie.Element("positionX");
var positionY = (int)zombie.Element("positionY");
var speed = (int)zombie.Element("speed");

更新：修正一些拼写错误并清理一下，您的Get方法应如下所示：

public IDictionary<string, string> Get(string typeName)
{
    var zombie = root.Element(typeName);
    return zombie.Elements()
          .ToDictionary(
                  element => element.Name.ToString(),
                  element => element.Value);
}

Answer 2

System.Xml.XmlDocument doc = new System.Xml.XmlDocument();
doc.LoadXml(xmlString);

string health = doc["zombies"]["zombie"]["health"].InnerText;
// etc..

// or looping

foreach( XmlNode node in doc["zombies"].ChildNodes )
{
    string health = node["health"].InnerText;
    // etc...
}

或者这在XNA中不起作用？

XNA：加载和读取XML文件的最佳方式？

2 个答案: