JAXB Java 生成 XML，为什么要小写？
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JAXB Java generating XML, Why lowercase?
提问by user1293962
When I run this code:
当我运行此代码时：
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;
}
}
I get
我得到
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<employee>
<code>CA</code>
<name>Cath</name>
<salary>300</salary>
</employee>
Which is correct, so my question is why does it change the Employee to employee? Is it possible to make it print with uppercase E, instead of employee?
哪个是正确的，所以我的问题是为什么它将员工更改为员工？是否可以使用大写 E 而不是员工来打印？
This is what I actually wanted to have:
这就是我真正想要的：
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Employee>
<code>CA</code>
<name>Cath</name>
<salary>300</salary>
</Employee>
Thanks!
谢谢！
回答by bdoughan
The behaviour you are seeing is the result of the standard JAXB (JSR-222)XML name to Java name conversion algorithm.
您看到的行为是标准JAXB (JSR-222)XML 名称到 Java 名称转换算法的结果。
You can use the @XmlRootElementannotation to specify a name:
您可以使用@XmlRootElement注释来指定名称：
@XmlRootElement(name="Employee")
@XmlAccessorType(XmlAccessType.FIELD)
class Employee {
...
}
I'm the EclipseLink JAXB (MOXy)lead, and we have an extension that allows you to override the default name conversion algorithm that you may be interested in:
我是EclipseLink JAXB (MOXy) 的负责人，我们有一个扩展，允许您覆盖您可能感兴趣的默认名称转换算法：
回答by FAtBalloon
For specific elements...
对于特定元素...
@XmlElement( name = "Code")
private String code;
For the object....
对于对象....
@XmlRootElement(name="Employee")
public class Employee{ ...
回答by Rid
My solution after put @XmlElement(name="Xxxxx") to fields and used XStream.aliasField(). This is more generic because it uses annotations and scans other class calls in the same package.
将 @XmlElement(name="Xxxxx") 放入字段并使用 XStream.aliasField() 后我的解决方案。这更通用，因为它使用注释并扫描同一包中的其他类调用。
import java.lang.reflect.Field;
import java.util.Map;
import java.util.TreeMap;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlAttribute;
import com.thoughtworks.xstream.XStream;
import my.MyClassGeneratedFromXsdToJaxB;
public class TestChangeFirstLetterXml {
public static void main(String[] args) throws ClassNotFoundException {
MyClassGeneratedFromXsdToJaxB myClassGeneratedFromXsdToJaxB=new MyClassGeneratedFromXsdToJaxB();
XStream xstream = new XStream();
xstream.autodetectAnnotations(true);
xstream = makeAliasAnnotatedFields(xstream, MyClassGeneratedFromXsdToJaxB.class, "FirstTagOrRoot");
//System.out.println(xstream.toXML(myClassGeneratedFromXsdToJaxB));
}
public static XStream makeAliasAnnotatedFields(XStream xstream, Class myclass, String firstTag)
throws ClassNotFoundException {
xstream.alias(firstTag, myclass);
Map<String, Object[]> aliaslist = getListAlias(myclass);
for (String key : aliaslist.keySet()) {
Object[] aliasvalue = new Object[3];
aliasvalue = aliaslist.get(key);
String xmlTag = new String((String) aliasvalue[0]);
Class<?> classJaxb = (Class<?>) aliasvalue[1];
String tagToRename = new String((String) aliasvalue[2]);
xstream.aliasField(xmlTag, classJaxb, tagToRename);
System.out.println("AliasField " + xmlTag + " " + classJaxb.getName() + " " + tagToRename);
}
return xstream;
}
public static Map<String, Object[]> getListAlias(Class<?> classToCheck)
throws ClassNotFoundException {
/* Read recursive fields of the class */
Field[] fs = classToCheck.getDeclaredFields();
String annotationsPackage = classToCheck.getPackage().getName();
String classSimpleName = new String(classToCheck.getSimpleName());
/* it is necessary avoid loop */
Map<String, Object[]> aliasStart = new TreeMap<String, Object[]>();
/* */
for (int i = 0; i < fs.length; i++) {
String nameField = fs[i].getName();
String classFieldName = new String(fs[i].getType().getName());
String nameXmlXsd = new String("");
String idkey = new String(annotationsPackage + ".");
if (fs[i].isAnnotationPresent(javax.xml.bind.annotation.XmlElement.class)) {
XmlElement atrib = fs[i].getAnnotation(XmlElement.class);
nameXmlXsd = new String(atrib.name());
idkey = new String(idkey + classSimpleName + ".Element." + nameField);
} else if (fs[i].isAnnotationPresent(javax.xml.bind.annotation.XmlAttribute.class)) {
XmlAttribute atrib = fs[i].getAnnotation(XmlAttribute.class);
nameXmlXsd = new String(atrib.name());
idkey = new String(idkey + classSimpleName + ".Type." + nameField);
}
if (aliasStart.containsKey(idkey)) /* avoid loop */
continue;
if (nameXmlXsd.equals("Signature")) // My particular condition
continue;
if (!nameXmlXsd.equals(classFieldName)) {
// xstrem.aliasField(a,b,c)
Object[] alias = new Object[3];
alias[0] = new String(nameXmlXsd);
alias[1] = classToCheck;
alias[2] = new String(nameField);
aliasStart.put(idkey, alias);
}
if (classFieldName.indexOf(annotationsPackage) > -1) {
Class<?> c = Class.forName(classFieldName);
Map<String, Object[]> aliaslist = getListAlias(c);
for (String key : aliaslist.keySet()) {
Object[] aliasvalue = new Object[3];
aliasvalue = aliaslist.get(key);
aliasStart.put(key, aliasvalue);
}
}
}
return aliasStart;
}
}
回答by cody.tv.weber
An alternative answer, if JAXB is not a MUST, then you can actually use org.json jar to convert the object to a JSONObject, then from there, you can use the XML object to convert the JSONObject to an XML. You will need a few tweaks before it can be a standalone XML.
另一个答案，如果 JAXB 不是必须的，那么您实际上可以使用 org.json jar 将对象转换为 JSONObject，然后从那里，您可以使用 XML 对象将 JSONObject 转换为 XML。在它成为独立的 XML 之前，您需要进行一些调整。
A code snippet example:
代码片段示例：
public static String getXMLString(Object o){
JSONObject json = new JSONObject(o);
String result =
"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" +
XML.toString(json, o.getClass().getSimpleName());
return result;
}
