文件保存到数据库

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将 XML 文件保存到数据库/Spring Boot

本文介绍了将 XML 文件保存到数据库/Spring Boot的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我是一名初级开发人员，无法完成我的项目.我需要将从 XML 文件接收的数据保存到数据库中.我写了代码，但是我无法将接收到的数据保存到数据库中:

I am a beginner developer and cannot finish my project. I need the data received from the XML file to be saved to the database. I wrote the code, but I can't save the received data to the database:

import org.springframework.stereotype.Service;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;

import javax.xml.XMLConstants;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.net.URL;

@Service
public class UploadURLServiceImpl implements UploadURLService {

    private final FileDTORepository fileDTORepository;

    @Autowired
    public UploadURLServiceImpl(FileDTORepository fileDTORepository) {
        this.fileDTORepository = fileDTORepository;
    }

    public void store(MultipartFile file) throws IOException {

        String fileName = StringUtils.cleanPath(file.getOriginalFilename());
        SnEntryDTO snEntryDTO = new SnEntryDTO(fileName, file.getContentType(), file.getContentType(), file.getBytes());

        fileDTORepository.save(snEntryDTO);
    }

    @Override
    public boolean uploadData(String url) {
        try (BufferedInputStream in = new BufferedInputStream(new URL(url).openStream());
             FileOutputStream fileOutputStream = new FileOutputStream(new File("sn.xml"))) {
            byte dataBuffer[] = new byte[1024];
            int bytesRead;
            while ((bytesRead = in.read(dataBuffer, 0, 1024)) != -1) {
                fileOutputStream.write(dataBuffer, 0, bytesRead);
            }
        } catch (IOException e) {
            System.out.println(e);
            return false;
        }
        return true;
    }

    @Override
    public void parseSdnFile(String fileName) {

        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();

        try {

            dbf.setFeature(XMLConstants.FEATURE_SECURE_PROCESSING, true);

            DocumentBuilder db = dbf.newDocumentBuilder();

            Document doc = db.parse(new File(fileName));

            doc.getDocumentElement().normalize();

            System.out.println("Root Element :" + doc.getDocumentElement().getNodeName());
            System.out.println("------");

            // get <staff>
            NodeList list = doc.getElementsByTagName("snEntry");

            for (int temp = 0; temp < list.getLength(); temp++) {
                Node node = list.item(temp);
                if (node.getNodeType() == Node.ELEMENT_NODE) {
                    Element element = (Element) node;
                    String id = element.getElementsByTagName("uid").item(0).getTextContent();
                    String lastName = element.getElementsByTagName("lastName").item(0).getTextContent();
                    String snType = element.getElementsByTagName("snType").item(0).getTextContent();
                    String programList = element.getElementsByTagName("program").item(0).getTextContent();
                    System.out.println(id + " " + lastName + " snType " + snType + " programList " + programList);

                }
            }

        } catch (ParserConfigurationException | SAXException | IOException e) {
            e.printStackTrace();
        }
    }
}

请告诉我.据我了解，我还需要创建一个对象吗?

Tell me please. As I understand it, I still need to create an object?

这部分代码，我需要把它塑造成一个对象吗?

this part of the code, I need to shape it as an object?

            for (int temp = 0; temp < list.getLength(); temp++) {
                Node node = list.item(temp);
                if (node.getNodeType() == Node.ELEMENT_NODE) {
                    Element element = (Element) node;
                    String id = element.getElementsByTagName("uid").item(0).getTextContent();
                    String lastName = element.getElementsByTagName("lastName").item(0).getTextContent();
                    String snType = element.getElementsByTagName("snType").item(0).getTextContent();
                    String programList = element.getElementsByTagName("program").item(0).getTextContent();
                    System.out.println(id + " " + lastName + " snType " + snType + " programList " + programList);

推荐答案

您的问题有多种解决方案.最常见的方法是将您的 xml 序列化为实体并将其持久化到数据库中.如我所见，您正在使用 spring-package 中的注释 @Service.这使得可以使用 Spring-Data JPA 或普通的 Hibernate.我将为您提供 Postgresql 和 Spring-Data 的示例.

Your question has several solutions.The most common way is serealize your xml into entity and persist it to database. As I see, you are use annotation @Service from spring-package. This makes available to use Spring-Data JPA or plain Hibernate. I'll provide you example with Postgresql and Spring-Data.

  1. 将 db 驱动程序和 spring-data-jpa 启动程序添加到您的项目

  1. Add db driver and spring-data-jpa starter to your project

 <dependency>
     <groupId>org.springframework.boot</groupId>
     <artifactId>spring-boot-starter-data-jpa</artifactId>
 </dependency>
 <dependency>
     <groupId>org.postgresql</groupId>
     <artifactId>postgresql</artifactId>
     <scope>runtime</scope>
 </dependency>

  • 将您的数据库凭据提供到文件 application.properties

     spring.datasource.driver-class-name=org.postgresql.Driver
 spring.datasource.url=jdbc:postgresql://localhost:5432/mydatabase
 spring.datasource.username=postgres
 spring.datasource.password=12345
 spring.jpa.show-sql=true
 spring.jpa.hibernate.ddl-auto=update
 spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.PostgreSQLDialect

    • 小心 spring.jpa.hibernate.ddl-auto 将休眠模式设置为更改数据库.为避免数据丢失，如果您使用不害怕丢失数据的本地/测试数据库，请使用 create、create-drop.如果您使用数据丢失至关重要的阶段或生产数据库，请使用更新或验证.

    Be careful spring.jpa.hibernate.ddl-auto sets hibernate mode to changing database. To avoid data lose use create, create-drop if you work with your local/test database where you are not afraid of losing data. Use update or validate if you working with stage or production database where data losing will be critical.

    1. 创建将持久化到数据库的实体类:

    1. Create Entity class that will be persisted to database:

     @Entity
 @Table(name = "my_entity")
 public class MyEntity {
     @Id
     @Column(name = "id",  length = 16, unique = true, nullable = false)
     private UUID uid = randomUuid();

     @Column(name = "last_name")
     private String lastName;

     @Column(name = "sn_type")
     private String snType;

     @Column(name = "program")
     private String program;

     // Constructors
     // Getters and setters
}

  • 创建简单的存储库

  • Create simple Repository

     @Repository
 public interface MyRepository extends CrudRepository<MyEntity, UUID> {
 }

  • 在代码中使用创建的存储库

  • Use created repository in your code

      @Autowired
  private MyRepository myRepo; //Inject your repo in service


  @Override
  public void parseSdnFile(String fileName) {

      //...

      for (int temp = 0; temp < list.getLength(); temp++) {
             Node node = list.item(temp);
             if (node.getNodeType() == Node.ELEMENT_NODE) {
                 Element element = (Element) node;
                 String id = element.getElementsByTagName("uid").item(0).getTextContent();
                 String lastName = element.getElementsByTagName("lastName").item(0).getTextContent();
                 String snType = element.getElementsByTagName("snType").item(0).getTextContent();
                 String programList = element.getElementsByTagName("program").item(0).getTextContent();

                 MyEntity entity = new MyEntity(id, lastName, snType, program); //serealize your data into entity
                 myRepo.save(entity) //saving to database
                 System.out.println(id + " " + lastName + " snType " + snType + " programList " + programList);

    • 这篇关于将 XML 文件保存到数据库/Spring Boot的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

    09-05 23:32
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