如何在java中解组这种类型的xml

Question

我是 Rest API 的新手。我有一个需要解组的 xml 输出。下面是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>new@gmail.com</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>old@gmail.com</value>
            </att>
            <att name="callname">
                <value>rahul</value>
            </att>
        </entry>
    </entries>
</dsml>

实际的 xml 输出共有 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();
        }
    }
}

执行此代码时，出现错误

javax.xml.bind.UnmarshalException - 有关联的例外： [org.xml.sax.SAXParseException;行号：1；列号：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。]

我已尝试将属性 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，然后将其传递给 unmarshal

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。

访问-：http://howtodoinjava.com/jaxb/jaxb-exmaple-marshalling-and-unmarshalling-list-or-set-of-objects/

Answer 2

（已编辑答案：已编辑 UserClass 和 DsmlFactory getEntities() 方法）

我认为这是因为您的 Xml 结构和您的类结构不同。我永远不会使用具有相同名称但值类型不同的元素（例如您的 att 和 value 元素）。为什么？这对 Marshaller 来说不是问题，而是 Unmarshaller 的问题。因为，据我所知，它无法根据父级的 name 属性定义子元素值的类型。即使你定义了泛型类型的静态内部类（例如interface Attribute<V> 和static class Uid<Integer>、static class Email<String> 等等，你只会让你的工作更加复杂。

但是如果你写了@XmlElement(name="att") Set<Attribute> attributes，其中属性计数对于 Marshaller 和 Unmarshaller 并不重要，那就是另外一回事了。

不过为了简单起见，我真的推荐<email>{value}</email>、<uid>{value}</uid>等结构。

我已经制作了一个简单的 Xml Marshaller/Unmarshaller 来测试（没有 REST），这很有效：

我的Dslm Xml 根类：

@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class Dsml {

    @XmlElement
    private Entities entities;

    public Dsml() {}
    public Dsml(Entities entities) {
        this.entities = entities;
    }

    public Entities getEntities() {
        return entities;
    }

    public void print() {
        entities.print();
    }
}

---

我的Entities 班级：

@XmlAccessorType(XmlAccessType.FIELD)
public class Entities {
    @XmlElement(name = "entity")
    private Set<User> users = new LinkedHashSet<>();

    public Entities() {}

    public Entities(Set<User> users) {
        this.users = users;
    }

    public Set<User> getUsers() {
        return Collections.unmodifiableSet(users);
    }

    public void print() {
        if (users != null && !users.isEmpty()) {
            users.stream().forEach(User::print);
        } else {
            System.out.println("No entities found");
        }
    }
}

---

我的User 班级：

@XmlAccessorType(XmlAccessType.FIELD)
public class User {

    @XmlAttribute
    private String dn;

    @XmlElement
    private int uid;

    @XmlElement
    private String email;

    @XmlElement
    private String callname;


    public User() {}

    public User(int uid, String dn, String email, String callname) {
        setUid(uid);
        setEmail(email);
        setCallname(callname);
        setDn(dn);
    }

    public int getUid() {
        return uid;
    }

    public void setUid(int uid) {
        this.uid = uid;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getCallname() {
        return callname;
    }

    public void setCallname(String callname) {
        this.callname = callname;
    }

    public String getDn() {
        return dn;
    }

    public void setDn(String dn) {
        this.dn = dn;
    }

    public void print() {
        System.out.println("Dn: " + dn);
        System.out.println("Uid: " + uid);
        System.out.println("Email: " + email);
        System.out.println("Callname: " + callname);
    }
}

---

我的DsmlFactory Marshaller/Unmarshaller 测试类（确保您有D:\ 分区，并且您对它有Read|ReadWrite 权限或在方法saveXml() 和readAndPrint() 中键入任何其他路径：

public final class DsmlFactory {
    private DsmlFactory() {}
    public static Entities getEntities() {
        Set<User> users = new LinkedHashSet<>();
        users.add(new User(1000, "dn1", "email1", "callname1"));
        users.add(new User(2000, "dn2", "email2", "callname2"));
        users.add(new User(3000, "dn3", "email3", "callname3"));
        users.add(new User(4000, "dn4", "email4", "callname4"));
        users.add(new User(5000, "dn5", "email5", "callname5"));
        return new Entities(users);
    }

    public static void saveXmlTest() {
        try {
            File file = new File("D:/test.xml");
            if (!file.exists()) {
                try {
                    file.createNewFile();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }

            JAXBContext jaxbContext = JAXBContext.newInstance(Dsml.class);
            Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

            jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            Dsml dsml = new Dsml(getEntities());
            jaxbMarshaller.marshal(dsml, file);
            jaxbMarshaller.marshal(dsml, System.out);

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

    public static void readAndPrintXml() {
        JAXBContext jaxbContext;
        try {
            jaxbContext = JAXBContext.newInstance(Dsml.class);
            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();

            Dsml dsml = (Dsml) jaxbUnmarshaller.unmarshal( new File("D:/test.xml") );
            dsml.print();
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }

    public static void testIt() {
        saveXmlTest();
        readAndPrintXml();
    }
}

---

还有我的mainTestDrive：

public class Main {
    public static void main(String[] args) {
        DsmlFactory.testIt();
    }
}

Answer 3

我建议您尝试以下方法：

第一步：创建一个名为Attribute的类，如下：

属性.java

import java.io.Serializable;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlElement;

public class Attribute implements Serializable {
private static final long serialVersionUID = 1L;
private String name;
private String value;
public Attribute() {
 }
public Attribute(String name, String value) {
    this.name = name;
    this.value = value;
 }
@XmlAttribute
public String getName() {
    return name;
 }
public void setName(String name) {
    this.name = name;
 }
@XmlElement
public String getValue() {
    return value;
 }
public void setValue(String value) {
    this.value = value;
 }
}

第 2 步：请修改您的名为 Student 的类，如下所示：

学生.java

import java.io.Serializable;
import java.util.List;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlElement;

public class Student implements Serializable {
private static final long serialVersionUID = 1L;
private List<Attribute> attributes;
private String distinguishedName;
public Student() {
 }
public Student(String distinguishedName, List<Attribute> attributes) {
    this.distinguishedName = distinguishedName;
    this.attributes = attributes;
 }
@XmlAttribute(name = "dn")
public String getDistinguishedName() {
    return distinguishedName;
 }
public void setDistinguishedName(String distinguishedName) {
    this.distinguishedName = distinguishedName;
 }
@XmlElement(name = "att", type = Attribute.class)
public List<Attribute> getAttributes() {
    return attributes;
 }
public void setAttributes(List<Attribute> attributes) {
    this.attributes = attributes;
 }

}

第 3 步：请创建一个名为 DSML 的类，如下所示：

DSML.java

import java.io.Serializable;
import java.util.List;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlElementWrapper;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name = "dsml")
public class DSML implements Serializable {
private static final long serialVersionUID = 1L;
private List<Student> entries;

@XmlElementWrapper(name = "entries")
@XmlElement(name = "entry", type = Student.class)
public List<Student> getEntries() {
    return entries;
 }
public void setEntries(List<Student> entries) {
    this.entries = entries;
 }
}

第 4 步：现在是时候将 XML 转换为 Java 对象了，如下所示：

UsingRestAPI.java

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