Mat*_*nit 5 java list jaxb azure
我知道这是一个初学者的问题,但是我现在一直在撞墙挡了两个小时,试图解决这个问题.
我从REST服务(Windows Azure Management API)返回XML,如下所示:
<HostedServices
xmlns="http://schemas.microsoft.com/windowsazure"
xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<HostedService>
<Url>https://management.core.windows.net/XXXXX</Url>
<ServiceName>foo</ServiceName>
</HostedService>
<HostedService>
<Url>https://management.core.windows.net/XXXXX</Url>
<ServiceName>bar</ServiceName>
</HostedService>
</HostedServices>
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当我尝试使用JAXB解组它时,服务列表始终为空.
我想尽可能避免编写XSD(微软不提供).这是JAXB代码:
JAXBContext context = JAXBContext.newInstance(HostedServices.class, HostedService.class);
Unmarshaller unmarshaller = context.createUnmarshaller();
HostedServices hostedServices = (HostedServices)unmarshaller.unmarshal(new StringReader(responseXML));
// This is always 0:
System.out.println(hostedServices.getHostedServices().size());
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以下是Java类:
@XmlRootElement(name="HostedServices", namespace="http://schemas.microsoft.com/windowsazure")
public class HostedServices
{
private List<HostedService> m_hostedServices = new ArrayList<HostedService>();
@XmlElement(name="HostedService")
public List<HostedService> getHostedServices()
{
return m_hostedServices;
}
public void setHostedServices(List<HostedService> services)
{
m_hostedServices = services;
}
}
@XmlType
public class HostedService
{
private String m_url;
private String m_name;
@XmlElement(name="Url")
public String getUrl()
{
return m_url;
}
public void setUrl(String url)
{
m_url = url;
}
@XmlElement(name="ServiceName")
public String getServiceName()
{
return m_name;
}
public void setServiceName(String name)
{
m_name = name;
}
}
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任何帮助将是真诚的感谢.
@XmlRootElement的namespace不会传播到其子.您应该明确指定命名空间:
...
@XmlElement(name="HostedService", namespace="http://schemas.microsoft.com/windowsazure")
...
@XmlElement(name="Url", namespace="http://schemas.microsoft.com/windowsazure")
...
@XmlElement(name="ServiceName", namespace="http://schemas.microsoft.com/windowsazure")
...
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