【发布时间】:2014-04-01 10:02:22
【问题描述】:
我正在使用 JAXB 将给定的输入 Xml 文件解组为 Java 对象 然后将其整理回 Xml 字符串。 我的 Xml 文件如下所示:
<bpmn2:definitions xmlns:bpmn2="http://www.omg.org/spec/BPMN/20100524/MODEL" id="_Definitions_1">
<bpmn2:process id="_500441" name="process">
</bpmn2:process>
</bpmn2:definitions>
Definitions.class:
@XmlRootElement(namespace = "http://www.omg.org/spec/BPMN/20100524/MODEL")
public class Definitions {
@XmlAttribute
private String id;
@XmlElement(name = "bpmn2:process")
private Process process;
@XmlElement(name = "bpmndi:BPMNDiagram")
private Diagram diagram;
public Definitions() {
}
public Definitions(String id, Process process, Diagram diagram) {
this.id = id;
this.process = process;
this.diagram = diagram;
}
public Process getProcess() {
return process;
}
public Diagram getDiagram() {
return diagram;
}
public String getId() {
return id;
}
}
Process.class:
@XmlAccessorType(XmlAccessType.FIELD)
public class Process {
@XmlAttribute
private String id;
public Process() {
}
public Process(String id) {
this.id = id;
}
public String getId() {
return id;
}
}
Model.class:
public class Model {
@XmlElement
private Process process;
public Model() {
}
public Model(String processId, Process p) {
this.id = processId;
this.process = p;
}
}
主要方法:
public static void main(String[] args) throws IOException, JSONException, JAXBException {
BpmnToJsonImport bj = new BpmnToJsonImport();
InputStream is = BpmnToJsonImport.class.getResourceAsStream("myXml.txt");
String Str = IOUtils.toString(is);
StringReader sr = new StringReader(Str);
JAXBContext context = JAXBContext.newInstance(Definitions.class, Model.class);
Unmarshaller unmarshaller = context.createUnmarshaller();
Definitions d = (Definitions) unmarshaller.unmarshal(sr);
Model model = new Model(d.getProcess().getId(), d.getProcess());
StringWriter sw = new StringWriter();
Marshaller marshaller = context.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.setProperty(Marshaller.JAXB_ENCODING, "UTF-8");
marshaller.marshal(model, sw);
String str = sw.toString();
System.out.println(str);
}
当它尝试使用 d.getProcess.getId 检索进程 ID 时,我得到了 java.lang.NullPointerException
【问题讨论】:
-
尝试为您的 XML 文件使用完全限定的路径。例如:C:\MyFolder\myXml.txt 可能是找不到 myXml.txt。如果找到了,可能是你的“Process.class”没有“name”字段?
-
问题不在于文件,因为当我这样做时:System.out.println(d.getId()) 它给了我 bpmn2:definitions 的 id。我不需要名称字段
-
这里有很多问题。 Model.java 无法编译,而且您还没有提供 Diagram 类。我能够通过从 definitions.class 中删除“bpmn2”命名空间限定符来克服您的错误,因此我会说您在某处有注释错误。
标签: java xml jaxb marshalling unmarshalling