[英]In ActionScript 3, how to decode from xml to ActionScript class?
In ActionScript 3, how to decode from xml to ActionScript class ? 在ActionScript 3中,如何从xml解码为ActionScript类?
I could encode from ActionScript class to xml by using XmlEncoder. 我可以使用XmlEncoder从ActionScript类编码为xml。
The xml schema I used at that time is this. 我当时使用的xml模式就是这个。
[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>
This schema is created by Ant(schemagen) task using POJO(User.java) with no annotations. 此架构是由Ant(schemagen)任务使用POJO(User.java)创建的,没有注释。
But I could not decode from xml to ActionScript class by using this schema and XmlDecoder. 但是我无法使用此架构和XmlDecoder从xml解码为ActionScript类。 (In correct, I can't be done casting from Object type to User type.) (正确的说,我无法完成从对象类型到用户类型的转换。)
I want not to put any annotations like @XmlRootElement or @XmlType in Java class. 我不想在Java类中放置任何@XmlRootElement或@XmlType之类的注释。
However, I need a schema file for client side of ActionScript to marshal and unmarshall. 但是,我需要一个模式文件供ActionScript的客户端编组和解组。
Please tell me any solutions or examples... 请告诉我任何解决方案或示例...
The following class: 以下课程:
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;
}
}
Can be used to unmarshal the following XML: 可用于解组以下XML:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<id>123</id>
<password>foo</password>
<userDate>2011-01-07T09:15:00</userDate>
</root>
Using the following code without requiring any annotations on the User class: 使用以下代码,而无需在User类上添加任何注释:
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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