java.net.MalformedURLException:解析xml时没有协议异常

java.net.MalformedURLException:解析xml时没有协议异常,java,xml-parsing,Java,Xml Parsing,我正在执行以下代码: DocumentBuilder docBuilder = docFactory.newDocumentBuilder(); doc = docBuilder.parse(sourcepath); node = doc.getElementsByTagName("NAME").item(0); node.setTextContent("Bob"); node

我正在执行以下代码:

        DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
        doc = docBuilder.parse(sourcepath);                  
        node = doc.getElementsByTagName("NAME").item(0);
        node.setTextContent("Bob");

        node = doc.getElementsByTagName("ADDRESS1").item(0);
        node.setTextContent("US");


        node = doc.getElementsByTagName("EMAIL").item(0);
        node.setTextContent("abc@defg.com");




        dateFormat = new SimpleDateFormat("dd/MM/yy HH:mm");
        Date date = new Date();

        node = doc.getElementsByTagName("DATESUBMITTED").item(0);
        node.setTextContent(dateFormat.format(date));


        // write the content into xml file
        TransformerFactory transformerFactory = TransformerFactory.newInstance();
        Transformer transformer = transformerFactory.newTransformer();
        DOMSource source = new DOMSource(doc);
        System.out.println(rail_feedback_dest);
        StreamResult result = new StreamResult(new File(destinationPath));          transformer.transform(source, result);
但当我在变量“sourcepath”和“destinationPath”中给出path时,它给出的是错误:java.net.MalformedURLException:无协议:
但是,在提供完整路径而不是在变量中提供路径时,代码工作得很好……

那么sourcepath是什么样子的呢
MalformedURLException
可能意味着它获取了错误的源路径。源路径和目标路径的类型是什么?URL、字符串或文件?“C:/Users/user123/Desktop/summit/Dummy Xmls/Feedback.xml”,当您不提供完整路径,但在变量中提供时,它是什么样子?你能打印它/断点它,看看sourcepath的值是多少吗?好的。。知道了。。引用正在产生问题。。。谢谢