阿帕奇常见的JXPath XML解析 - XML URL为null

Question

这里是我已经习惯了学习的JXPath XML解析的示例代码，

import java.net.URL; 
import java.util.Iterator; 

import org.apache.commons.jxpath.Container; 
import org.apache.commons.jxpath.JXPathContext; 
import org.apache.commons.jxpath.xml.DocumentContainer; 

public class DocumentContainerTest { 

    /** 
    * @param args 
    */ 
    public static void main(String[] args) { 

     //Get the URL the of XML document 
     URL url = DocumentContainerTest.class.getClassLoader().getResource("student_class.xml"); 

     //Construct document container from the URL to XML 
     Container container = new DocumentContainer(url); 

     JXPathContext context = JXPathContext.newContext(container); 

     Iterator<?> subjects = context.iterate("/studentClass/subjects_list/subject"); 
     while (subjects.hasNext()) { 
      System.out.println(subjects.next()); 
     } 

     Iterator<?> stdNames = context.iterate("/studentClass/student_list/student/firstName"); 
     while (stdNames.hasNext()) { 
      System.out.println(stdNames.next()); 
     } 

     System.out.println(context.getValue("/studentClass/student_list/student[@id='1']/firstName")); 
    } 

} 

，这里是我的XML文件中使用

<?xml version="1.0" encoding="UTF-8" standalone="yes"?> 
<studentClass> 
    <name>MPC</name> 
    <subjects_list> 
     <subject>Maths</subject> 
     <subject>Physics</subject> 
     <subject>Chemistry</subject> 
    </subjects_list> 
    <student_list> 
     <student id="1"> 
      <age>2</age> 
      <dob>2011-09-25T16:41:56.250+05:30</dob> 
      <firstName>Sriram</firstName> 
      <hobby>Painting</hobby> 
      <lastName>Kasireddi</lastName> 
     </student> 
     <student id="2"> 
      <age>26</age> 
      <dob>2011-09-25T16:41:56.250+05:30</dob> 
      <firstName>Sudhakar</firstName> 
      <hobby>Coding</hobby> 
      <lastName>Kasireddi</lastName> 
     </student> 
    </student_list> 
</studentClass> 

我得到下面的错误，

Exception in thread "main" org.apache.commons.jxpath.JXPathException: XML URL is null 
    at org.apache.commons.jxpath.xml.DocumentContainer.<init>(DocumentContainer.java:106) 
    at org.apache.commons.jxpath.xml.DocumentContainer.<init>(DocumentContainer.java:92) 
    at org.apache.commons.jxpath.XMLDocumentContainer.<init>(XMLDocumentContainer.java:58) 
    at jxpath_ex1.DocumentContainerTest.main(DocumentContainerTest.java:26) 
Java Result: 1 

我已经添加的lib文件，公地的JXPath-1.3.jar，公地的BeanUtils-1.3.jar，Apache的公地loggi ng.jar。

Answer 1

定义url已经试过了，像下面这样，它的工作原理！

URL url = DocumentContainerTest.class.getResource("student_class.xml");