java 将java中对象的数据写入xml文件
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Writing data from objects in java into a xml file
提问by Gopal Samant
I have a piece of code as follows :
我有一段代码如下：
Person person1 = new Person();
person1.setName("abc");
person1.setAge(23);
Person person2 = new Person();
person2.setName("xyz");
person2.setAge(32);
And it needs to re represented in a xml file as :
它需要在 xml 文件中重新表示为：
<Person>
<person1>
<name>abc</name>
<age>23</age>
</person1>
<person2>
<name>abc</name>
<age>23</age>
</person2>
</Person>
How do I do it?
我该怎么做？
回答by user1635014
I guess you are looking for java-xml binding. You can JAXB binding and marshall. Please check the link http://www.mkyong.com/java/jaxb-hello-world-example/for sample.
我猜您正在寻找 java-xml 绑定。您可以通过 JAXB 绑定和编组。请查看链接 http://www.mkyong.com/java/jaxb-hello-world-example/以获取示例。
回答by Gopal Samant
checking out at the code in your link i implemented it for my code. The code is as follows. I can write xml data only for one object and if I loop it just writes the last object. So when I tried to used the file in append mode it stopped functioning :
查看您链接中的代码，我为我的代码实现了它。代码如下。我只能为一个对象编写 xml 数据，如果我循环它只写最后一个对象。因此，当我尝试在追加模式下使用该文件时，它停止运行：
try {
FileOutputStream file = new FileOutputStream("file.xml", true);
JAXBContext jaxbContext = JAXBContext.newInstance(NeuronNode.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
// output pretty printed
jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
for(int i=0; i<neuronNodes.length; i++)
{
jaxbMarshaller.marshal(neuronNodes[i], file);
jaxbMarshaller.marshal(neuronNodes[i], System.out);
}
} catch (JAXBException e | IOException e) {
e.printStackTrace();
}
回答by pidabrow
You could also use Serialization and XML-Parser (DOM for example). If you create an XML document then you can use XPath for querying.
您还可以使用序列化和 XML-Parser（例如 DOM）。如果创建 XML 文档，则可以使用 XPath 进行查询。
I'd also consider on using http://simple.sourceforge.net/
回答by Atul Sharma
Have sorted out this problem using Hymanson Library and XMLMapper method. Here is the code for the same.
已经使用Hymanson Library 和 XMLMapper 方法解决了这个问题。这是相同的代码。
**MainClass.java:-**
import java.io.IOException;
import com.fasterxml.Hymanson.databind.ObjectMapper;
import com.fasterxml.Hymanson.databind.SerializationFeature;
import com.fasterxml.Hymanson.dataformat.xml.XmlMapper;
public class MainClass {
public static void main(String[] args) throws IOException {
String xmlString = " <Person>\r\n" + " <person1>\r\n" + "
<name>abc</name>\r\n"
+ " <age>23</age>\r\n" + " </person1>\r\n" + "
<person2>\r\n"
+ " <name>abc</name>\r\n" + " <age>23</age>\r\n" + "
</person2>\r\n"
+ " </Person>";
ObjectMapper xmlMapper = new XmlMapper();
xmlMapper.enable(SerializationFeature.INDENT_OUTPUT);
Person1 person1 = new Person1();
person1.setName("abc");
person1.setAge("23");
Person2 person2 = new Person2();
person2.setName("xyz");
person2.setAge("32");
Person person = new Person();
person.setPerson1(person1);
person.setPerson2(person2);
String XMLStr = xmlMapper.writeValueAsString(person);
System.out.println(XMLStr);
} }
**Person.java:-**
public class Person {
Person1 Person1Object;
Person2 Person2Object;
// Getter Methods
public Person1 getPerson1() {
return Person1Object;
}
public Person2 getPerson2() {
return Person2Object;
}
// Setter Methods
public void setPerson1(Person1 person1Object) {
this.Person1Object = person1Object;
}
public void setPerson2(Person2 person2Object) {
this.Person2Object = person2Object;
} }
**Person1.java:-**
public class Person1 {
private String name;
private String age;
// Getter Methods
public String getName() {
return name;
}
public String getAge() {
return age;
}
// Setter Methods
public void setName(String name) {
this.name = name;
}
public void setAge(String age) {
this.age = age;
} }
**Person2.java:-**
public class Person2 {
private String name;
private String age;
// Getter Methods
public String getName() {
return name;
}
public String getAge() {
return age;
}
// Setter Methods
public void setName(String name) {
this.name = name;
}
public void setAge(String age) {
this.age = age;
} }
**OUTPUT:-**
<Person>
<person1>
<name>abc</name>
<age>23</age>
</person1>
<person2>
<name>xyz</name>
<age>32</age>
</person2>
</Person>
