How to get application context path in spring-ws?

Question

I am using Spring -WS to create a web service. In my project, I created a class Helperto read a sample response and request an XML file, which is located in my folder /src/main/resource.

When I do unit testing of my webservice application "locally", I use System.getProperty("user.dir")to get the application context folder. Below is the method I created in the class Helperto help me get the file that interests me in my resources folder.

 public static File getFileFromResources(String filename) {
  System.out.println("Getting file from resource folder");
  File request = null;
  String curDir = System.getProperty("user.dir");
  String contextpath = "C:\\src\\main\\resources\\";
  request = new File(curDir + contextpath + filename);
  return request;
 }

However, after "publishing" the compiled WAR file to a folder ../webappsin the Apache Tomcat directory, I realize that it System.getProperty("user.dir")no longer returns the path to the application context. Instead, it returns the Apache Tomcat root directory as shown

C:\Program Files\Apache Software Foundation\Tomcat 6.0\src\main\resources\SampleClientFile

- -. - Spring, , :

request.getSession().getServletContext().getContextPath()

- Spring, . Spring -WS, - . webservice.

-

C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps\clientWebService\WEB-INF\classes

- ?

Answer 1

RequestContextHolder:

HttpServletRequest request = ((ServletRequestAttributes) RequestContextHolder.currentRequestAttributes()).getRequest();

, , , .

Answer 2

, , . , .

, , . ServletContext javax.servlet.ServletContext # getResourceAsStream, . ( ).

, org.springframework.web.context.ServletContextAware bean.