生成XML文件
发布时间:2020-12-16 07:40:13 所属栏目:百科 来源:网络整理
导读:public void testSave() throws Exception{ListPerson persons=new ArrayListPerson();persons.add(new Person(1,"aa",23));persons.add(new Person(2,"bb",24));File file=new File(getContext().getFilesDir(),"person.xml");OutputStream output=new File
public void testSave() throws Exception{ List<Person> persons=new ArrayList<Person>(); persons.add(new Person(1,"aa",23)); persons.add(new Person(2,"bb",24)); File file=new File(getContext().getFilesDir(),"person.xml"); OutputStream output=new FileOutputStream(file); XmlSerializer ser=Xml.newSerializer(); ser.setOutput(output,"UTF-8"); ser.startDocument("UTF-8",true); ser.startTag(null,"persons"); for(Person person : persons){ ser.startTag(null,"person"); ser.attribute(null,"id",String.valueOf( person.getId())); ser.startTag(null,"name"); ser.text(person.getName()); ser.endTag(null,"name"); ser.startTag(null,String.valueOf("age")); ser.text(String.valueOf(person.getAge())); ser.endTag(null,String.valueOf("age")); ser.endTag(null,"person"); } ser.endTag(null,"persons"); ser.endDocument(); output.flush(); output.close(); } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |