How to get application context path in spring-ws?
- by Dhaliwal
I am using Spring-WS to create a webservice. In my project, I have created a Helper class to reads sample response and request xml file which are located in my /src/main/resource folder.
When I am unit-testing my webservice application 'locally', I use the System.getProperty("user.dir") to get the application context folder. The following is a method that I created in the Helper class to help me retrieve the file that I am interested in my resource 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 the ../webapps folder to the Apache Tomcat directory, I realise that System.getProperty("user.dir") no longer returns my application context path. Instead, it is returning the Apache Tomcat root directory as shown
C:\Program Files\Apache Software Foundation\Tomcat 6.0\src\main\resources\SampleClientFile
I cant seem to find any information about getting the root folder of my webservice. I have seen examples on Spring web application where I can retrieve the context path by using the following :
request.getSession().getServletContext().getContextPath()
But in this case, I am using a Spring web application where there is a servlet context in the request. But the Spring-WS, my entry point is an endpoint. How can I get the context path of my webservice application.
I am expecting a context path of something like
C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps\clientWebService\WEB-INF\classes
Could someone suggest a way to achieve this?
