Question

<?xml version="1.0" encoding="UTF-8"?>
<objects xmlns="XXXXXXX">
<cuobject type-id="EmailSubscription" object-id="jjjil@gmail.com">
    <object-attribute attribute-id="exported">true</object-attribute>
    <object-attribute attribute-id="firstName">jjj</object-attribute>
    <object-attribute attribute-id="lastName">jjj</object-attribute>
    <object-attribute attribute-id="subscribed">true</object-attribute>
</cuobject>

<cuobject type-id="EmailSubscription" object-id="gggj@gmail.com">
    <object-attribute attribute-id="exported">true</object-attribute>
    <object-attribute attribute-id="firstName">ghh</object-attribute>
    <object-attribute attribute-id="lastName">fhh</object-attribute>
    <object-attribute attribute-id="subscribed">true</object-attribute>
</cuobject>

<cuobject type-id="EmailSubscription" object-id="mmm@gmail.com">
    <object-attribute attribute-id="exported">true</object-attribute>
    <object-attribute attribute-id="firstName">mmm</object-attribute>
    <object-attribute attribute-id="lastName">mmm</object-attribute>
    <object-attribute attribute-id="subscribed">true</object-attribute>
</cuobject>
</objects>

如何在java中读取此xml文件（需要使用exported，firstName，lastName，...）？我们怎么做？我这样做是为了

NodeList nList = doc.getElementsByTagName("cuobject");

for (int temp = 0; temp < nList.getLength(); temp++) 
{               

    Node nNode = nList.item(temp);       

    System.out.println("\nCurrent Element :" + nNode.getNodeName());         

    if (nNode.getNodeType() == Node.ELEMENT_NODE) 
    {        

        Element eElement = (Element) nNode;

        String email = eElement.getAttribute("object-id");

        String firstName =  eElement.

        getElementsByTagName("object-attribute").item(1).getTextContent();

        String lastName = eElement.

        getElementsByTagName("object-attribute").item(2).getTextContent();

        String subscribed = eElement.

        getElementsByTagName("object-attribute").item(3).getTextContent();

    }
}

Answer 1

您可以使用XPath查询和Java支持查询Xml文档。

您可以参考How to read XML using XPath in Java了解更多信息

Answer 2

如果您对JAXB没有任何问题，请转到此处

将xml转换为XSD
您可以在线转换xml到XSD，例如（http://www.freeformatter.com/xsd-generator.html）
将XSD转换为Java类

如果您使用的是eclipse juno，则可以轻松地将其模式转换为Java类。请查看此处的详细步骤（http://theopentutorials.com/examples/java/jaxb/generate-java-class-from-xml-schema-in-eclipse-ide/）

使用XML作为对象（与OOPS一致，非常适合您的客户详细信息XML）来访问值

公共类MyTestClass {

/**
 * @param args
 */
public static void main(String[] args) {
    File file = new File("C:\\kar\\file1.xml");
    try{
    JAXBContext jaxbContext = JAXBContext.newInstance(Objects.class);
    Unmarshaller jaxbUnMarshaller = jaxbContext.createUnmarshaller();


    Objects listcustomers = (Objects) jaxbUnMarshaller.unmarshal(file);
    List<Cuobject> cusmters=listcustomers.getCuobject();
    Iterator<Cuobject> it = cusmters.iterator();
    while(it.hasNext()){
        Cuobject customer = it.next();
        System.out.println("Customer ID: "+ customer.typeId + " Customer email" + customer.objectId);
        List<ObjectAttribute> details =customer.objectAttribute;
        Iterator<ObjectAttribute> it1 = details.iterator();
        System.out.println(" -------Details------------");
        while(it1.hasNext()){
            ObjectAttribute detail = it1.next();

            System.out.println(detail.attributeId + " =" + detail.value);
        }
    }
    }catch(Exception ex){
        System.out.println(ex.getMessage());
    }
}

}

输出就是这样的

客户ID：EmailSubscription客户emailjjjil@gmail.com - - - -细节 - - - - - - exported = true firstName = jjj lastName = jjj subscribed = true

Customer ID: EmailSubscription Customer emailgggj@gmail.com
 -------Details------------
exported =true
firstName =ghh
lastName =fhh
subscribed =true
Customer ID: EmailSubscription Customer emailmmm@gmail.com
 -------Details------------
exported =true
firstName =mmm
lastName =mmm
subscribed =true

如何在java中读取这个xml文件（需要使用exported，firstName，lastName，...）？

2 个答案: