Xml序列化c＃

Question

無法理解我做錯了什么，結果集是空的。
我的代碼：

class Class1
    {

        public static object DeSerialize()
        {
            object resultObject;

            XmlSerializer serializer = new XmlSerializer(typeof(PointsContainer));
           using (TextReader textReader = new StreamReader(@"d:\point.xml"))
            {
                resultObject = serializer.Deserialize(textReader);
            }

            return resultObject;


        }
    }

    [Serializable]
    [XmlRoot("Points")]
    public class PointsContainer
    {
        [XmlElement("Point")]       
        private List<Point> items = new List<Point>();

        public List<Point> Items
        {
            get { return items; }
            set { items = value; }
        }


    }


    [Serializable]   
    public class Point
    {      
        [XmlAttribute]
        public bool x { get; set; }

        [XmlAttribute]
        public bool y { get; set; }
    }

XML：

<Points>  
   <Point x="1" y="5"/>
   <Point x="21" y="3"/>
   <Point x="3" y="7"/>
</Points>

Answer 1

將[XmlElement]屬性移動到屬性。
XmlSerializer忽略私有成員。

Answer 2

正如SLaks所說

你的Point對象也顯示兩個字段為bool但xml文件中的值至少是ints（21,3,5,7等）

Answer 3

bool變量可以是true或false，其整數值為1和0.因此，您的XML具有無效數據和/或您的類屬性的類型錯誤。

Answer 4

[XmlElement("Point")]
public List<Point> Items
{
  get { return items; }
  set { items = value; }
}

在你的觀點類中，x和y都不應該是bool。

Answer 5

解：

namespace XmlStackProblem
{
    class Class1
    {

        public static void Main()
        {
            Points resultObject;

            XmlSerializer serializer = new XmlSerializer(typeof(Points));
            using (TextReader textReader = new StreamReader(@"d:\points.xml"))
            {
                resultObject = serializer.Deserialize(textReader) as Points;
            }
        }
    }

    [Serializable]
    [XmlRoot(IsNullable = false)]
    public class Points
    {
        [XmlElementAttribute("Point")]
        public List<Point> Point
        {
            get; set;
        }
    }

    [Serializable]
    [XmlType(AnonymousType = true)]
    public class Point
    {
        [XmlAttribute]
        public int x
        {
            get;
            set;
        }

        [XmlAttribute]
        public int y { get; set; }
    }
}

Answer 6

您可以使用返回對象類型的DeSerialize函數，如下例所示：

public T DeSerializeFromString<T>(string data)
        {
            T result;
            StringReader rdr = null;
            try
            {
                rdr = new StringReader(data);
                XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
                result = (T)xmlSerializer.Deserialize(rdr);
            }
            catch (Exception)
            {
                throw;
            }
            finally
            {
                rdr.Close();
            }
            return result;
        }