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:
下面是我要写入外部文件的内容
// new user is created
Bank bank = new Bank();
System.out.println("Enter your full name below (e.g. 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);
这一切都会创建一个Bank用户,它会被投入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。
答案 0 :(得分:3)
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_;
}
//下面的课实际写了
public void writeXmlFile(ArrayList<details> 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");
}
}
答案 1 :(得分:1)
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 data provider。
答案 2 :(得分:1)
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。
new GradientPaint(0, 0, Color.RED, 0, 1000, Color.BLUE)
同样,反向解码。