如何在java中解组这种类型的xml

Question

我是Rest API的新手。我有一个xml输出需要解封。以下是xml输出：

<dsml>
    <entries>
        <entry dn="uid=7686,c=in,ou=pages,o=example.com">
            <att name="uid">
                <value>7568766</value>
            </att>
            <att name="email">
                <value>[e-mail]</value>
            </att>
            <att name="callname">
                <value>John</value>
            </att>
        </entry>
        <entry dn="uid=7689,c=in,ou=pages,o=example.com">
            <att name="uid">
                <value>7678766</value>
            </att>
            <att name="callname">
                <value>Mike</value>
            </att>
        </entry>
        <entry dn="uid=7690,c=in,ou=pages,o=example.com">
            <att name="uid">
                <value>75858766</value>
            </att>
            <att name="email">
                <value>[e-mail]</value>
            </att>
            <att name="callname">
                <value>rahul</value>
            </att>
        </entry>
    </entries>
</dsml>

实际的xml输出总共有37个条目。下面是模型类：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "user")
public class User implements Serializable{
    private static final long serialVersionUID = 1L;

    @XmlElement(name = "uid")
    private int uid;

    @XmlElement(name = "callname")
    private String callname;

    @XmlElement(name = "email")
    private String email;

    public int getId() {
        return uid;
    }
    public void setId(int uid) {
        this.uid = uid;
    }
    public String getFirstName() {
        return callname;
    }
    public void setFirstName(String callname) {
        this.callname = callname;
    }
    public String getmail() {
        return email;
    }
    public void setmail(String email) {
        this.email = email;
    }
}

下面是Rest Api类代码：

public class UsingRestAPI {
    public static void main(String[] args) 
    {
        try
        {
            URL url = new URL("www.example.com");
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setRequestMethod("GET");
            conn.setRequestProperty("Accept", "application/xml");

            if (conn.getResponseCode() != 200) 
            {
                throw new RuntimeException("Failed : HTTP error code : " + conn.getResponseCode());
            }

            BufferedReader br = new BufferedReader(new InputStreamReader((conn.getInputStream())));
            String apioutput="",temp="";
            while ((apioutput = br.readLine()) != null) {
                temp += apioutput;
                System.out.println(apioutput);
            }


            JAXBContext jaxbContext = JAXBContext.newInstance(User.class);
            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
            User user = (User) jaxbUnmarshaller.unmarshal(new StringReader(temp));

            System.out.println(user.getId());
            System.out.println(user.getFirstName());
            System.out.println(user.getmail());

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

在执行此代码时，我收到错误

javax.xml.bind.UnMarshalException-带有链接异常：[org.xml.sax.saxParseException；LineNumber：1；ColumnNumber：110；外部DTD：无法读取外部DTD“dsml.DTD”，因为accessExternalDTD属性设置的限制不允许“HTTP”访问。]

我尝试将属性ACCESS_EXTERNAL_DTD更改为true，但它给出了另一个错误：

javax.xml.bind.PropertyException: name: http://javax.xml.XMLConstants/property/accessExternalDTD value: true
    at javax.xml.bind.helpers.AbstractUnmarshallerImpl.setProperty(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.setProperty(Unknown Source)
    at UsingRestAPI.main(UsingRestAPI.java:60)

Answer 1

为什么不将Xml保存在文件中，然后将其传递到unmarshal中

新版本_:

User user = (User) jaxbUnmarshaller.unmarshal(new File("c:/temp/user.xml") );//File path

它不会给出任何错误。