如何将ArrayList写入XML文件？

我正在尝试将ArrayList存储到XML文件中，以便稍后可以检索该信息，然后将其显示回控制台。

有人能告诉我最有效的方法吗？

编辑：

inheritance人我试图写入外部文件

 // new user is created Bank bank = new Bank(); System.out.println("Enter your full name below (eg John M. Smith): "); String name = scanner.nextLine(); System.out.println("Create a username: "); String userName = scanner.nextLine(); System.out.println("Enter your starting deposit amount: "); int balance = scanner.nextInt(); System.out.print(dash); System.out.print("Generating your information...\n"); System.out.print(dash); int pin = bank.PIN(); String accountNum = bank.accountNum(); User user = new User(name, userName, pin, accountNum, balance); //new user gets added to the array list Bank.users.add(user); System.out.println(user); 

这一切都创建了一个Bank用户，它被抛入ArrayList ，然后我想存储他们的信息，以便我可以稍后再回来重新显示它。

Answer 1

 public static void main(String[] args) { // TODO Auto-generated method stub WriteFile ob = new WriteFile(); ArrayList list = new ArrayList(); list.add(new details("A", 20, 1)); list.add(new details("B", 30, 2)); ob.writeXmlFile(list); } 

//根据您的需要修改下面的课程

 class details { String name; public String getName() { return name; } public void setName(String name) { this.name = name; } public int getAge() { return age; } public void setAge(int age) { this.age = age; } public int getId() { return id; } public void setId(int id) { this.id = id; } int age; int id; public details() { } public details(String name_, int age_, int id_) { name = name_; age = age_; id = id_; } 

//下面的课实际上写了

 public void writeXmlFile(ArrayList
 list) { try { DocumentBuilderFactory dFact = DocumentBuilderFactory.newInstance(); DocumentBuilder build = dFact.newDocumentBuilder(); Document doc = build.newDocument(); Element root = doc.createElement("Studentinfo"); doc.appendChild(root); Element Details = doc.createElement("Details"); root.appendChild(Details); for (details dtl : list) { Element name = doc.createElement("Name"); name.appendChild(doc.createTextNode(String.valueOf(dtl .getName()))); Details.appendChild(name); Element id = doc.createElement("ID"); id.appendChild(doc.createTextNode(String.valueOf(dtl.getId()))); Details.appendChild(id); Element mmi = doc.createElement("Age"); mmi.appendChild(doc.createTextNode(String.valueOf(dtl.getAge()))); Details.appendChild(mmi); } // Save the document to the disk file TransformerFactory tranFactory = TransformerFactory.newInstance(); Transformer aTransformer = tranFactory.newTransformer(); // format the XML nicely aTransformer.setOutputProperty(OutputKeys.ENCODING, "ISO-8859-1"); aTransformer.setOutputProperty( "{http://xml.apache.org/xslt}indent-amount", "4"); aTransformer.setOutputProperty(OutputKeys.INDENT, "yes"); DOMSource source = new DOMSource(doc); try { // location and name of XML file you can change as per need FileWriter fos = new FileWriter("./ros.xml"); StreamResult result = new StreamResult(fos); aTransformer.transform(source, result); } catch (IOException e) { e.printStackTrace(); } } catch (TransformerException ex) { System.out.println("Error outputting document"); } catch (ParserConfigurationException ex) { System.out.println("Error building document"); } } 

Answer 2

我这样做的方法是使用XStream或Jackson API（首选）将java对象序列化为XML或JSON文件。

例如，请参阅我编写的用于TestNG或JUnit参数化测试的XStream数据提供程序 。

Answer 3

如何使用XMLEncoder / XMLDecoder？

http://docs.oracle.com/javase/6/docs/api/java/beans/XMLEncoder.html

从javadoc复制和释义。

 ArrayList arr = new ArrayList(); // populate your array XMLEncoder e = new XMLEncoder(BufferedOutputStream( new FileOutputStream("Test.xml"))); e.writeObject(arr); e.close(); 

同样，反向解码。