How to find the working folder of a servlet based application in order to load resources

You could use ServletContext#getRealPath() to convert a relative web content path to an absolute disk file system path.

String relativeWebPath = "/WEB-INF/static/file1.ext";
String absoluteDiskPath = getServletContext().getRealPath(relativeWebPath);
File file = new File(absoluteDiskPath);
InputStream input = new FileInputStream(file);
// ...

However, if your sole intent is to get an InputStream out of it, better use ServletContext#getResourceAsStream() instead because getRealPath() may return null whenever the WAR is not expanded into local disk file system but instead into memory and/or a virtual disk:

String relativeWebPath = "/WEB-INF/static/file1.ext";
InputStream input = getServletContext().getResourceAsStream(relativeWebPath);
// ...

This is much more robust than the java.io.File approach. Moreover, using getRealPath() is considered bad practice.

See also:

• getResourceAsStream() vs FileInputStream
• What does servletcontext.getRealPath("/") mean and when should I use it
• Where to place and how to read configuration resource files in servlet based application?
• Recommended way to save uploaded files in a servlet application