Question

我有一个名为Student的对象

SELECT i.*,
       d.Date,
       a.in_time,
       a.out_time,
       SEC_TO_TIME(SUM(TIME_TO_SEC(a.out_time))-(TIME_TO_SEC(a.in_time))) AS duration,
       c.cardtype,
       a.wo,
       v.activity,
       a.quty,
       a.wastage,
       a.mcusage,
       a.actual_wastage
FROM employee_details i
INNER JOIN actual_alldetails a ON i.emp_code=a.emp_code
INNER JOIN attendance_date d ON d.date_id=a.date_id
INNER JOIN card_type c ON c.card_id=a.card_id
INNER JOIN activities v ON v.activity_id=a.activity_id
WHERE d.Date='2016-01-30'
ORDER BY v.activity;

我的xml看起来像这样

class Student{
  int a;
  String name;
  List<Student> childStudents;
}

现在我想要的是读取XML并将值填充到学生对象，在第一个学生对象中，根据XML只有值“a”和“name”，它只有没有子学生列表但在第二个元素中在xml中也有“a，name”和子学生，它与父对象学生“a，名字和学生列表”相同，所以我想要的是按照xml值用这样的第一个studnet填充对象

<root>
   <student>
      <a>1</a>
      <name>ABC</name>
   </student>
   <student>
      <a>2</a>
      <name>XYZ</name>
      <student>
           <a>3</a>
           <name>PQR</name>
       </student>
   </student>
</root>

和第二个像这样的对象

Student
[ a =1 
  name = "ABC"
  listofStudents = null;
]

Any One做了同样的请链接，提取方案不一样只是我在问题中提到的样本简化方式才能理解

我试过这样的事情

Student
[ a =2 
  name = "XYZ"
  listofStudents = (size)1;
            Student[
               a = 3
               name = "PQR"
               listofStudents = null;
            ]
]

我试过的第二次尝试解决方案是

public void A(){
    nodeList = doc.getElementsByTagName("student");
    for (int msgIndex = 0; msgIndex < nodeList.getLength(); msgIndex++) {
            Node message = nodeList.item(msgIndex);
            if (message.getNodeType() == Node.ELEMENT_NODE) {

                Element eElement = (Element) message;
                String msgName = eElement.getAttribute("a");
                String msgCateg = eElement.getAttribute("name");


                Student msg = new Student();
                msg.setMsgName(msgName);
                msg.setMsgA(msgCateg);            

            }
        }
}

和XML Read

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

/**
 * Created by sadeeshal on 3/30/2016.
 */
public class Book {

    String author;
    double price;
    String pubdate;
    List<Book> listOfBooks = new ArrayList<Book>();

    public void addBookToList(Book b) {
        getListOfBooks().add(b);
    }

    public String getAuthor() {
        return author;
    }

    public void setAuthor(String author) {
        this.author = author;
    }

    public double getPrice() {
        return price;
    }

    public void setPrice(double price) {
        this.price = price;
    }

    public String getPubdate() {
        return pubdate;
    }

    public void setPubdate(String pubdate) {
        this.pubdate = pubdate;
    }

    public List<Book> getListOfBooks() {
        return listOfBooks;
    }

    public void setListOfBooks(List<Book> listOfBooks) {
        this.listOfBooks = listOfBooks;
    }
}

XML文件

import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import java.io.File;
import java.util.ArrayList;

/**
 * Created by sadeeshal on 3/30/2016.
 */
public class XML {

    public static void main(String[] args) {
        XML xml = new XML();
        xml.readXML();
    }

    public void readXML(){

        try {
            File file = new File("C:/Users/sadeeshal/Downloads/A.xml");
            DocumentBuilder dBuilder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
            Document doc = dBuilder.parse(file);
            doc.getDocumentElement().normalize();

            NodeList nodeList = doc.getElementsByTagName("book");
            Book bookParent = new Book();
            read(nodeList, bookParent);

        } catch (Exception e) {
            System.out.print(e.getMessage());
        }
    }

    public void read( NodeList nList,Book bookParent){
        for (int temp = 0; temp < nList.getLength(); temp++) {
            Node nNode = nList.item(temp);
            if (nNode.getNodeType() == Node.ELEMENT_NODE) {
                Element eElement = (Element) nNode;
                String name = eElement.getElementsByTagName("author").item(0).getTextContent();
                String pric = eElement.getElementsByTagName("price").item(0).getTextContent();
                String pub  =  eElement .getElementsByTagName("pubdate").item(0).getTextContent();

                Book book = new Book();
                book.setAuthor(name);
                book.setPrice(Double.parseDouble(pric));
                book.setPubdate(pub);

                if(nNode.hasChildNodes()){
                   read(((Element) nNode).getElementsByTagName("book"),book);
                   bookParent.addBookToList(book);
                }
            }
        }
    }
}

并按如下方式加载

但实际上需要发生的事情并不像img carsonChild应该只有listofbooks数组下的单个对象而不是父母

Answer 1

在ReadXML中使用这个级别可以修复。（第二个解决方案）所以自动修复第一个

XPathFactory xpathFactory = XPathFactory.newInstance();
            XPath xpath = xpathFactory.newXPath();
            NodeList links = (NodeList) xpath.evaluate("books/book", doc, XPathConstants.NODESET);

查看对象创建的附件

与child和parent具有相同元素的Xml文件读取并设置为object

1 个答案: