java - 在 Java 中读取多级 XML 文件

Question

直到最近，我的 XML 文件的标签结构还相当简单。但是现在我多了一层带有标签的标签，解析 XML 变得更加复杂。

这是我的新 XML 文件的示例(我更改了标签名称以使其更易于理解):

<SchoolRoster>
   <Student>
      <name>John</name>
      <age>14</age>
      <course>
         <math>A</math>
         <english>B</english>
      </course>
      <course>
         <government>A+</government>
      </course>
   </Student>
   <Student>
      <name>Tom</name>
      <age>13</age>
      <course>
         <gym>A</gym>
         <geography>incomplete</geography>
      </course>
   </Student>
</SchoolRoster>

上面的 XML 的重要特征是我可以有多个“类(class)”属性，并且在其中我可以有任意命名的标签作为它们的 child 。可以有任意数量的这些 child ，我想将其读入“名称”、“值”的 HashMap。

public static TreeMap getAllSchoolRosterInformation(String fileName) {
    TreeMap SchoolRoster = new TreeMap();

    try {
        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        DocumentBuilder db = dbf.newDocumentBuilder();
        File file = new File(fileName);
        if (file.exists()) {
            Document doc = db.parse(file);
            Element docEle = doc.getDocumentElement();
            NodeList studentList = docEle.getElementsByTagName("Student");

            if (studentList != null && studentList.getLength() > 0) {
                for (int i = 0; i < studentList.getLength(); i++) {

                    Student aStudent = new Student();
                    Node node = studentList.item(i);

                    if (node.getNodeType() == Node.ELEMENT_NODE) {
                        Element e = (Element) node;
                        NodeList nodeList = e.getElementsByTagName("name");
                        aStudent.setName(nodeList.item(0).getChildNodes().item(0).getNodeValue());

                        nodeList = e.getElementsByTagName("age");
                        aStudent.setAge(Integer.parseInt(nodeList.item(0).getChildNodes().item(0).getNodeValue()));

                        nodeList = e.getElementsByTagName("course");
                        if (nodeList != null && nodeList.getLength() > 0) {
                            Course[] courses = new Course[nodeList.getLength()];
                            for (int j = 0; j < nodeList.getLength(); j++) {

                                Course singleCourse = new Course();
                                HashMap classGrades = new HashMap();
                                NodeList CourseNodeList = nodeList.item(j).getChildNodes();

                                for (int k = 0; k < CourseNodeList.getLength(); k++) {
                                    if (CourseNodeList.item(k).getNodeType() == Node.ELEMENT_NODE && CourseNodeList != null) {
                                        classGrades.put(CourseNodeList.item(k).getNodeName(), CourseNodeList.item(k).getNodeValue());
                                    }
                                }
                                singleCourse.setRewards(classGrades);
                                Courses[j] = singleCourse;
                            }
                            aStudent.setCourses(Courses);
                        }
                    }
                    SchoolRoster.put(aStudent.getName(), aStudent);
                }
            }
        } else {
            System.exit(1);
        }
    } catch (Exception e) {
        System.out.println(e);
    }
    return SchoolRoster;
}

我遇到的问题是，不是让学生在“数学”中获得“A”，而是在“数学”中获得空值。 (如果这篇文章太长，我可以尝试找到一些缩短它的方法。)

Answer 1

如果这是我的项目，我会避免尝试手动剖析 HTML 中的数据，而是让 Java 使用 JAXB 为我做这件事。我越使用这个工具，就越喜欢它。我敦促您考虑尝试这样做，因为如果您这样做，将 XML 更改为 Java 对象所需要做的就是在 Java 类中添加适当的注释，然后解码 XML。使用的代码会简单得多，因此出错的可能性也会大大降低。

例如，以下代码非常容易且干净地将信息编码到 XML 中:

import java.util.ArrayList;
import java.util.List;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlType;

@XmlRootElement
public class SchoolRoster {
   @XmlElement(name = "student")
   private List<Student> students = new ArrayList<Student>();

   public SchoolRoster() {
   }

   public List<Student> getStudents() {
      return students;
   }

   public void addStudent(Student student) {
      students.add(student);
   }

   public static void main(String[] args) {
      Student john = new Student("John", 14);
      john.addCourse(new Course("math", "A"));
      john.addCourse(new Course("english", "B"));

      Student tom = new Student("Tom", 13);
      tom.addCourse(new Course("gym", "A"));
      tom.addCourse(new Course("geography", "incomplete"));

      SchoolRoster roster = new SchoolRoster();
      roster.addStudent(tom);
      roster.addStudent(john);

      try {
         JAXBContext context = JAXBContext.newInstance(SchoolRoster.class);
         Marshaller marshaller = context.createMarshaller();
         marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

         String pathname = "MySchoolRoster.xml";
         File rosterFile = new File(pathname );
         marshaller.marshal(roster, rosterFile);
         marshaller.marshal(roster, System.out);

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

@XmlRootElement
@XmlType(propOrder = { "name", "age", "courses" })
class Student {
  // TODO: completion left as an exercise for the original poster
}

@XmlRootElement
@XmlType(propOrder = { "name", "grade" })
class Course {
  // TODO: completion left as an exercise for the original poster
}

这导致了以下 XML:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<schoolRoster>
    <student>
        <name>Tom</name>
        <age>13</age>
        <courses>
            <course>
                <name>gym</name>
                <grade>A</grade>
            </course>
            <course>
                <name>geography</name>
                <grade>incomplete</grade>
            </course>
        </courses>
    </student>
    <student>
        <name>John</name>
        <age>14</age>
        <courses>
            <course>
                <name>math</name>
                <grade>A</grade>
            </course>
            <course>
                <name>english</name>
                <grade>B</grade>
            </course>
        </courses>
    </student>
</schoolRoster>

将其解码到充满数据的 SchoolRoster 类中只需几行代码。

private static void unmarshallTest() {
  try {
     JAXBContext context = JAXBContext.newInstance(SchoolRoster.class);
     Unmarshaller unmarshaller = context.createUnmarshaller();

     String pathname = "MySchoolRoster.xml"; // whatever the file name should be
     File rosterFile = new File(pathname );
     SchoolRoster roster = (SchoolRoster) unmarshaller.unmarshal(rosterFile);
     System.out.println(roster);
  } catch (JAXBException e) {
     e.printStackTrace();
  }
}

在我的类中添加了 toString() 方法之后，结果是:

SchoolRoster 
  [students=
    [Student [name=Tom, age=13, courses=[Course [name=gym, grade=A], Course [name=geography, grade=incomplete]]], 
    Student [name=John, age=14, courses=[Course [name=math, grade=A], Course [name=english, grade=B]]]]]