JAXB Java生成XML,为什么是小写?
发布时间:2020-09-06 07:02:46 所属栏目:Java 来源:互联网
导读:当我运行这个代码: import javax.xml.bind.JAXBContext;import javax.xml.bind.Marshaller;import javax.xml.bind.annotation.XmlAccessType;import javax.xml.bind.annotation.XmlAccessorType;import javax.xml.bind.a
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当我运行这个代码: import javax.xml.bind.JAXBContext;
import javax.xml.bind.Marshaller;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlRootElement;
public class JavaToXMLDemo {
public static void main(String[] args) throws Exception {
JAXBContext context = JAXBContext.newInstance(Employee.class);
Marshaller m = context.createMarshaller();
m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT,true);
Employee object = new Employee();
object.setCode("CA");
object.setName("Cath");
object.setSalary(300);
m.marshal(object,System.out);
}
}
@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
class Employee {
private String code;
private String name;
private int salary;
public String getCode() {
return code;
}
public void setCode(String code) {
this.code = code;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getSalary() {
return salary;
}
public void setSalary(int population) {
this.salary = population;
}
}
我得到 <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<employee>
<code>CA</code>
<name>Cath</name>
<salary>300</salary>
</employee>
这是正确的,所以我的问题是为什么把员工改为员工? 这是我实际想要的: <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Employee>
<code>CA</code>
<name>Cath</name>
<salary>300</salary>
</Employee>
谢谢! 解决方法您所看到的行为是标准的 JAXB (JSR-222) XML名称到Java名称转换算法的结果.您可以使用@XmlRootElement注释来指定名称: @XmlRootElement(name="Employee")
@XmlAccessorType(XmlAccessType.FIELD)
class Employee {
...
}
我是EclipseLink JAXB (MOXy)领先,我们有一个扩展,它允许你覆盖,你可能会感兴趣的默认名称转换算法: > http://blog.bdoughan.com/2011/05/overriding-jaxbs-name-mangling.html (编辑:哈尔滨站长网) 【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! |
