从 XML 响应中解组子元素
Unmarshalling SubElements from XML Response
我正在尝试解组 XML 响应。结构如下
<employee>
<name> David </name>
<age> 33 </age>
<dateofjoin>
<year>2012</year>
<month>10</month>
<day>01</day>
</dateofjoin>
</employee>
以下是我的员工Class
public class Employee
{
private Dateofjoin dateofjoin;
private String name;
private String age;
public Dateofjoin getDateofjoin ()
{
return dateofjoin;
}
public void setDateofjoin (Dateofjoin dateofjoin)
{
this.dateofjoin = dateofjoin;
}
public String getName ()
{
return name;
}
public void setName (String name)
{
this.name = name;
}
public String getAge ()
{
return age;
}
public void setAge (String age)
{
this.age = age;
}
}
以下是我的加入日期class
public class Dateofjoin
{
private String month;
private String year;
private String day;
public String getMonth ()
{
return month;
}
public void setMonth (String month)
{
this.month = month;
}
public String getYear ()
{
return year;
}
public void setYear (String year)
{
this.year = year;
}
public String getDay ()
{
return day;
}
public void setDay (String day)
{
this.day = day;
}
}
我正在尝试通过以下方式一次解组这些响应
但我能够得到员工姓名、年龄。对于加入元素的日期,我如何在解组过程中获取元素值。我不想进行解组 again.I 可以看出这不是一种好的编程方式
while(xsr1.hasNext()) {
if(xsr1.isStartElement() && xsr1.getLocalName().equals("employee")) {
JAXBContext jc = JAXBContext.newInstance(Employee.class);
Unmarshaller unmarshaller = jc.createUnmarshaller();
Employee jb = unmarshaller.unmarshal(xsr1,Employee.class).getValue();
System.out.println("Employee" + jb.getName()+"Age.getAge());
}
xsr1.next();
}
非常感谢任何帮助。也希望看到一些解释,以便我可以根据问题清楚地理解
dataofjoin 应该已经包含在未编组的员工对象中
我正在尝试解组 XML 响应。结构如下
<employee>
<name> David </name>
<age> 33 </age>
<dateofjoin>
<year>2012</year>
<month>10</month>
<day>01</day>
</dateofjoin>
</employee>
以下是我的员工Class
public class Employee
{
private Dateofjoin dateofjoin;
private String name;
private String age;
public Dateofjoin getDateofjoin ()
{
return dateofjoin;
}
public void setDateofjoin (Dateofjoin dateofjoin)
{
this.dateofjoin = dateofjoin;
}
public String getName ()
{
return name;
}
public void setName (String name)
{
this.name = name;
}
public String getAge ()
{
return age;
}
public void setAge (String age)
{
this.age = age;
}
}
以下是我的加入日期class
public class Dateofjoin
{
private String month;
private String year;
private String day;
public String getMonth ()
{
return month;
}
public void setMonth (String month)
{
this.month = month;
}
public String getYear ()
{
return year;
}
public void setYear (String year)
{
this.year = year;
}
public String getDay ()
{
return day;
}
public void setDay (String day)
{
this.day = day;
}
}
我正在尝试通过以下方式一次解组这些响应 但我能够得到员工姓名、年龄。对于加入元素的日期,我如何在解组过程中获取元素值。我不想进行解组 again.I 可以看出这不是一种好的编程方式
while(xsr1.hasNext()) {
if(xsr1.isStartElement() && xsr1.getLocalName().equals("employee")) {
JAXBContext jc = JAXBContext.newInstance(Employee.class);
Unmarshaller unmarshaller = jc.createUnmarshaller();
Employee jb = unmarshaller.unmarshal(xsr1,Employee.class).getValue();
System.out.println("Employee" + jb.getName()+"Age.getAge());
}
xsr1.next();
}
非常感谢任何帮助。也希望看到一些解释,以便我可以根据问题清楚地理解
dataofjoin 应该已经包含在未编组的员工对象中