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

Erel Segal-Halevi picture Erel Segal-Halevi · Jul 5, 2011 · Viewed 35.8k times · Source

I write a Java servlet that I want to install on many instances of Tomcat on different servers. The servlet uses some static files that are packed with the war file under WEB-INF. This is the directory structure in a typical installation:

- tomcat
-- webapps
--- myapp
---- index.html
---- WEB-INF
----- web.xml
----- classes
------ src
------- .....
----- MY_STATIC_FOLDER
------ file1
------ file2
------ file3

How can I know the absolute path of MY_STATIC_FOLDER, so that I can read the static files?

I cannot rely on the "current folder" (what I get in a new File(".")) because it depends on where the Tomcat server was started from, which is different in every installation!

Answer

BalusC picture BalusC · Jul 5, 2011

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.

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