C# 将 XML 转换为通用列表
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Converting a XML to Generic List
提问by user2067567
I am trying to convert XML to List
我正在尝试将 XML 转换为列表
<School>
<Student>
<Id>2</Id>
<Name>dummy</Name>
<Section>12</Section>
</Student>
<Student>
<Id>3</Id>
<Name>dummy</Name>
<Section>11</Section>
</Student>
</School>
I tried few things using LINQ and am not so clear on proceeding.
我使用 LINQ 尝试了几件事，但对继续操作不太清楚。
dox.Descendants("Student").Select(d=>d.Value).ToList();
Am getting count 2 but values are like 2dummy12 3dummy11
正在计数 2 但值就像 2dummy12 3dummy11
Is it possible to convert the above XML to a generic List of type Student which has Id,Name and Section Properties ?
是否可以将上述 XML 转换为具有 Id、Name 和 Section Properties 的 Student 类型的通用列表？
What is the best way I can implement this ?
我可以实现的最佳方法是什么？
采纳答案by Anirudha
You can create an anonymous type
您可以创建一个匿名类型
var studentLst=dox.Descendants("Student").Select(d=>
new{
id=d.Element("Id").Value,
Name=d.Element("Name").Value,
Section=d.Element("Section").Value
}).ToList();
This creates a list of anonymous type..
这将创建一个匿名类型列表..
If you want to create a list of Student type
如果要创建 Student 类型的列表
class Student{public int id;public string name,string section}
List<Student> studentLst=dox.Descendants("Student").Select(d=>
new Student{
id=d.Element("Id").Value,
name=d.Element("Name").Value,
section=d.Element("Section").Value
}).ToList();
回答by Azhar Khorasany
var students = from student in dox.Descendants("Student")
select new
{
id=d.Element("Id").Value,
Name=d.Element("Name").Value,
Section=d.Element("Section").Value
}).ToList();
or you can create a class call Student with id, name and section as properties and do:
或者您可以创建一个类调用 Student，将 id、name 和 section 作为属性，然后执行以下操作：
var students = from student in dox.Descendants("Student")
select new Student
{
id=d.Element("Id").Value,
Name=d.Element("Name").Value,
Section=d.Element("Section").Value
}).ToList();
回答by Adil Mammadov
I see that you have accepted an answer. But I just want to show another way which I like. First you will need classes as below:
我看到你已经接受了一个答案。但我只想展示我喜欢的另一种方式。首先，您将需要以下课程：
public class Student
{
[XmlElement("Id")]
public int StudentID { get; set; }
[XmlElement("Name")]
public string StudentName { get; set; }
[XmlElement("Section")]
public int Section { get; set; }
}
[XmlRoot("School")]
public class School
{
[XmlElement("Student", typeof(Student))]
public List<Student> StudentList { get; set; }
}
Then you can deserialize this xml:
然后你可以反序列化这个xml：
string path = //path to xml file
using (StreamReader reader = new StreamReader(path))
{
XmlSerializer serializer = new XmlSerializer(typeof(School));
School school = (School)serializer.Deserialize(reader);
}
Hope it will be helpful.
希望它会有所帮助。
