the*_*e_V 3 .net c# xml serialization
这是一个代码示例:
public class Person
{
public string FirstName { get; set; }
public string LastName { get; set; }
}
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...
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);
}
}
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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>
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这是一个问题.如何摆脱根元素并渲染人像这样:
<?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>
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谢谢!
这是XML你想要的格式错误,不可能通过它获得它XmlSerializer,但你可以改变ArrayOfPersno元素名称以换取其他内容:
例:
XmlSerializer xs = new XmlSerializer(typeof(Person[]),
new XmlRootAttribute("Persons"));
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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>
...
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