在ActionScript 3，如何去code从XML到ActionScript类？ [英] In ActionScript 3, how to decode from xml to ActionScript class?

问题描述

在ActionScript 3，如何去code从XML到ActionScript类？

In ActionScript 3, how to decode from xml to ActionScript class ?

我可以连接code从ActionScript类XML使用XmlEn codeR。

I could encode from ActionScript class to xml by using XmlEncoder.

我当时使用XML架构是这样的。

The xml schema I used at that time is this.

[schema1.xsd]

[schema1.xsd]

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<xs:schema version="1.0" xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xs:complexType name="user">
    <xs:sequence>
      <xs:element name="id" type="xs:string" minOccurs="0"/>
      <xs:element name="password" type="xs:string" minOccurs="0"/>
      <xs:element name="userDate" type="xs:dateTime" minOccurs="0"/>
    </xs:sequence>
  </xs:complexType>
</xs:schema>

此模式是通过使用POJO（User.java）不带注释的蚂蚁（schemagen）任务创建。

This schema is created by Ant(schemagen) task using POJO(User.java) with no annotations.

但我不能去$ C $从XML到ActionScript类使用此架构和XmlDe codeR℃。 （在正确的，我不能这样做从Object类型用户类型进行转换。）

But I could not decode from xml to ActionScript class by using this schema and XmlDecoder. (In correct, I can't be done casting from Object type to User type.)

我要不要放像@XmlRootElement或@XmlType任何注解的Java类。

I want not to put any annotations like @XmlRootElement or @XmlType in Java class.

不过，我需要的ActionScript客户端编组和解组模式文件。

However, I need a schema file for client side of ActionScript to marshal and unmarshall.

请告诉我，任何解决方案或实例...

Please tell me any solutions or examples...

推荐答案

下面的类：

import java.util.Date;

public class User {

    private String id;
    private String password;
    private Date userDate;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public Date getUserDate() {
        return userDate;
    }

    public void setUserDate(Date userDate) {
        this.userDate = userDate;
    }

}

可用于解组下面的XML：

Can be used to unmarshal the following XML:

<?xml version="1.0" encoding="UTF-8"?>
<root>
   <id>123</id>
   <password>foo</password>
   <userDate>2011-01-07T09:15:00</userDate>
</root>

使用下面的code，而不需要任何注释上的用户等级：

Using the following code without requiring any annotations on the User class:

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBElement;
import javax.xml.bind.Unmarshaller;
import javax.xml.transform.stream.StreamSource;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(User.class);

        StreamSource source = new StreamSource("input.xml");
        Unmarshaller unmarshaller = jc.createUnmarshaller();
        JAXBElement<User> root = unmarshaller.unmarshal(source, User.class);

        User user = root.getValue();
        System.out.println(user.getId());
        System.out.println(user.getPassword());
        System.out.println(user.getUserDate());
    }
}

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