Question

我想序列化包含此内容的xml文件：

<?xml version="1.0" encoding="utf-8" ?>
<patients>
  <patient>
    <firstname>Patience_name_1</firstname>
    <lastname>Patience_surname_1</lastname>
    <age>20</age>
    <gender>Male</gender>
    <exams>
      <exam>
        <id>1</id>
        <date>02/27/2016</date>
        <comment>Exam completed for patience1</comment>
      </exam>
    </exams>
  </patient>
  <patient>
    <firstname>Patience_name_2</firstname>
    <lastname>Patience_surname_2</lastname>
    <age>22</age>
    <gender>Male</gender>
    <exams>
      <exam>
        <id>2</id>
        <date>02/27/2016</date>
        <comment>Exam completed fro patience 2</comment>
      </exam>
    </exams>
  </patient>
  <patient>
    <firstname>Patience_name_3</firstname>
    <lastname>Patience_surname_3</lastname>
    <age>23</age>
    <gender>Female</gender>
    <exams>
      <exam>
        <id>3</id>
        <date>02/26/2016</date>
        <comment>Exam completed for patience 3</comment>
      </exam>
    </exams>
  </patient>
</patients>

这是我的序列化代码..我突出了与之相关的点异常＆＃34;没想到。＆＃34;错误消息

    try
    {
        System.Xml.Serialization.XmlSerializer reader =
                new System.Xml.Serialization.XmlSerializer(typeof(List<patient>));
        System.IO.StreamReader file = new System.IO.StreamReader("data.xml");
        overview = (List<patient>)reader.Deserialize(file);// throws exception here
        file.Close();
    }
    catch (Exception ex)
    {
        Console.WriteLine(ex.InnerException.Message);
    }

我的模特课：

namespace WpfApplication1
{

    [XmlRoot("patients"), XmlType("patient")]
    public class patient
    {
        public patient()
        {
            this.Items = new ObservableCollection<patient>();
        }
        [XmlElement(ElementName = "firstname")]    
        public string name { get; set; }
        [XmlElement(ElementName = "lastname")] 
        public string surname { get; set; }
        [XmlElement(ElementName = "age")] 
        public int age { get; set; }
        [XmlElement(ElementName = "gender")] 
        public string gender { get; set; }
        [XmlElement(ElementName = "exams")] 
        public List<exam> exams { get; set; }

        [XmlElement(ElementName = "patients")] 
        public ObservableCollection<patient> Items { get; set; }
    }
    [XmlRoot("exams"), XmlType("exam")]
    public class exam
    {
        [XmlElement(ElementName = "id")] 
        public int id { get; set; }
        [XmlElement(ElementName = "date")] 
        public DateTime date { get; set; }
        [XmlElement(ElementName = "comment")] 
        public string comment { get; set; }
    }
}

我在网上搜索得很多，一切都很好，但我错过了一些东西..

Answer 1

将XmlRootAttribute添加到XmlSerializer：

var rootAttribute = new XmlRootAttribute();
rootAttribute.ElementName = "patients";
rootAttribute.IsNullable = true;
XmlSerializer reader =
        new XmlSerializer(typeof(List<patient>), rootAttribute);

此外，此行不需要：

[XmlRoot("patients"), XmlType("patient")]

XML序列化程序类型错误

1 个答案: