Question

我正在使用JAXB来解析基于GWT的应用程序中的XML文件。 XML看起来像这样（一个简化的例子）：

<addressbook>

    <company name="abc">
        <contact>
            <name>...</name>
            <address>...</address>
        </contact>

        <contact>
            <name>...</name>
            <address>...</address>
        </contact>

        <contact>
            <name>...</name>
            <address>...</address>
        </contact>
        ... 
        ... 
    </company>

    <company name="def">
        <contact>
            <name>...</name>
            <address>...</address>
        </contact>
        ...
        ...
    </company>

    ...
    ...

</addressbook>

我已经定义了如下所示的类：

@XmlRootElement(name="addressbook")
public class Addressbook implements Serializable {

    private ArrayList<Company> companyList = new ArrayList<Company>();

    public Addressbook() {            
    }

    @XmlElement(name = "company")
    public ArrayList<Company> getCompanyList() {
        return companyList;
    }


}

=============================

@XmlRootElement(name="company")
public class Company implements Serializable {

    private String name;

    private ArrayList<Contact> contactList = new ArrayList<Contact>();

    public Company() {      
    }

    @XmlAttribute
    public String getName() {
        return name;
    }

    @XmlElement(name = "contact")
    public ArrayList<Contact> getContactList() {
        return contactList;
    }

    ...
    ...
}

=============================

@XmlRootElement(name="contact")
public class Contact implements Serializable
{
    private String name;
    private String address;

    public Contact() {
    }

    @XmlElement
    public String getName ()
    {
        return name;
    }

    @XmlElement
    public String getAddress ()
    {
        return address;
    }

    ...
    ...
}

这是代码：

try {
    JAXBContext jc = JAXBContext.newInstance(Addressbook.class);
    Unmarshaller um = jc.createUnmarshaller();
    addressbook = (Addressbook) um.unmarshal(new FileReader("ds/addressbook.xml"));        
} catch (JAXBException e) {
    e.printStackTrace();
}

我需要根据公司名称获取联系人列表。例如，获取公司“abc”的所有联系人。我可以解析整个XML文件，然后手动过滤记录。但是如果输入文件很大，那么仅解析我需要的内容可能更有效。那么是否可以预先指定一个标准并仅解析特定记录？

感谢。

Answer 1

您可以使用EclipseLink JAXB (MOXy中的@XmlPath扩展名来处理这种情况（我是MOXy技术主管）：

@XmlRootElement(name="addressbook")
public class Addressbook implements Serializable {

    private ArrayList<Company> companyList = new ArrayList<Company>();

    public Addressbook() {            
    }

    @XmlPath("company[@name='abc']")
    public ArrayList<Company> getCompanyList() {
        return companyList;
    }


}

更多信息：

http://bdoughan.blogspot.com/2011/03/map-to-element-based-on-attribute-value.html

更新 - 使用StreamFilter

下面的示例演示了如何在此用例中使用StreamFilter：

import java.io.FileInputStream;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.Marshaller;
import javax.xml.bind.Unmarshaller;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamReader;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(Addressbook.class);

        XMLInputFactory xif = XMLInputFactory.newFactory();
        FileInputStream xmlStream = new FileInputStream("input.xml");
        XMLStreamReader xsr = xif.createXMLStreamReader(xmlStream);
        xsr = xif.createFilteredReader(xsr, new CompanyFilter());

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        Addressbook addressbook = (Addressbook) unmarshaller.unmarshal(xsr);

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(addressbook, System.out);
    }
}

StreamFilter的实现如下：

import javax.xml.stream.StreamFilter;
import javax.xml.stream.XMLStreamReader;

public class CompanyFilter implements StreamFilter {

    private boolean accept = true;

    public boolean accept(XMLStreamReader reader) {
        if(reader.isStartElement() && "company".equals(reader.getLocalName())) {
            accept = "abc".equals(reader.getAttributeValue(null, "name"));
        } else if(reader.isEndElement()) {
            boolean returnValue = accept;
            accept = true;
            return returnValue;
        }
        return accept;
    }

}

Answer 2

你可以

将XSLT转换应用于XML文件，或
将文件解组为DOM，并使用XPath选择所需的节点

将结果对象传递给unmarshal方法之前

创建一个由公司名称键入的内存Map可能更简单：

public class SearchableAddressBook {

    public final Map<String, Company> companyMap = new HashMap<String,Company>();

    public SearchableAddressBook(List<Company> companyList) {
        for (Company company: companyList) {
            companyMap.add(company.getName(), company));
        }

}

如果你真的想要过度设计它，可以创建一个内存数据库。

JAXB过滤了解析

2 个答案: