Question

这是一个代码示例：

public class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

...

static void Main() 
{
    Person[] persons = new Person[] 
    {
        new Person{ FirstName = "John", LastName = "Smith"},
        new Person{ FirstName = "Mark", LastName = "Jones"},
        new Person{ FirstName= "Alex", LastName="Hackman"}
    };

    XmlSerializer xs = new XmlSerializer(typeof(Person[]), "");

    using (FileStream stream = File.Create("persons-" + Guid.NewGuid().ToString().Substring(0, 4) + ".xml")) 
    {
        xs.Serialize(stream, persons);
    }
}

这是输出：

<?xml version="1.0"?>
<ArrayOfPerson xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Person>
    <FirstName>John</FirstName>
    <LastName>Smith</LastName>
  </Person>
  <Person>
    <FirstName>Mark</FirstName>
    <LastName>Jones</LastName>
  </Person>
  <Person>
    <FirstName>Alex</FirstName>
    <LastName>Hackman</LastName>
  </Person>
</ArrayOfPerson>

这是一个问题。如何摆脱根元素并呈现这样的人：

<?xml version="1.0"?>
<Person>
  <FirstName>John</FirstName>
  <LastName>Smith</LastName>
</Person>
<Person>
  <FirstName>Mark</FirstName>
  <LastName>Jones</LastName>
</Person>
<Person>
  <FirstName>Alex</FirstName>
  <LastName>Hackman</LastName>
</Person>

谢谢！

Answer 1

这是您想要的格式错误XML，无法通过XmlSerializer获取，但您可以将ArrayOfPersno元素名称更改为其他内容：

示例：

XmlSerializer xs = new XmlSerializer(typeof(Person[]),
                                     new XmlRootAttribute("Persons"));

会给你：

<?xml version="1.0"?>
<Persons xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Person>
    <FirstName>John</FirstName>
    <LastName>Smith</LastName>
  </Person>
  ...

Answer 2

IMO你应该使用顶级对象，即

[XmlRoot("whatever")]
public class Foo {
    [XmlElement("Person")]
    public List<Person> People {get;set;}        
}

哪个应序列化为具有多个“Person”子元素的“whatever”元素。

在序列化数组</arrayofclassname>时去掉<arrayofclassname>根元素

2 个答案: