Tomcat server absolute file access in war webapp
Solution 1
To read files:
ServletContext application = ...;
InputStream in = null;
try {
in = application.getResourceAtStream("/WEB-INF/web.xml"); // example
// read your file
} finally {
if(null != in) try { in.close(); }
catch (IOException ioe) { /* log this */ }
}
To write files:
ServletContext application = ...;
File tmpdir = (File)application.getAttribute("javax.servlet.context.tempdir");
if(null == tmpdir)
throw new IllegalStateException("Container does not provide a temp dir"); // Or handle otherwise
File targetFile = new File(tmpDir, "my-temp-filename.txt");
BufferedWriter out = null;
try {
out = new BufferedWriter(new FileWriter(targetFile));
// write to output stream
} finally {
if(null != out) try { out.close(); }
catch (IOException ioe) { /* log this */ }
}
If you don't want to use the tmpdir provided by the servlet container, then you should use someplace that is entirely outside of the servlet context's purvue, like /path/to/temporary/files
or something like that. You definitely don't want to use the container's temporary directory for anything other than truly temporary files which are okay to delete on re-deployment, etc.
Solution 2
It's a WAR. You don't read/write files inside it.
Reading is trivial: put the files on the classpath, and then read as a resource.
You shouldn't be writing inside a web app anyway since, even if it wasn't a WAR, things inside the context could disappear during a redeploy, or it might only be on one server if you're clustered, etc. Instead file writes should live somewhere configurable.
Admin
Updated on February 13, 2020Comments
-
Admin over 4 years
I have a Spring webapp whose
.war
file has been uploaded to a Tomcat server. Most of the basic functions are working as intended - page views and form submission.My problem now is that my webapp needs to read and write files and I am clueless as to how I can achieve this (the file I/O returns
java.lang.NullPointerException
).I used the following code to get the absolute path of a given file suggested by Titi Wangsa Bin Damhore to know the path relative to the server:
HttpSession session = request.getSession(); ServletContext sc = session.getServletContext(); String file = sc.getRealPath("src/test.arff"); logger.info("File path: " + file);
Here is the output path:
/home/username/tomcat/webapps/appname/src/test.arff
But when I checked the file directory via WinSCP, the file's actual path is:
/home/username/tomcat/webapps/appname/WEB-INF/classes/test.arff
Here are my questions:
- How do I transform these paths into something like
C:/Users/Workspace/appname/src/test.arff
(the original path in my local machine that works perfectly)? It's servers areApache Tomcat 6.0.35
andApache Tomcat 6.0.35
. - Why is the code returning a different path as opposed to the actual path?
- If file I/O is not applicable, what alternatives can I use?
PS I just need to access two files (< 1MB each) so I don't think I may need to use a database to contain them as suggested by minus in this thread.
File I/O
Below is the code I use for accessing the file I need.
BufferedWriter writer; try { URI uri = new URI("/test.arff"); writer = new BufferedWriter(new FileWriter( calcModelService.getAbsolutePath() + uri)); writer.write(data.toString()); writer.flush(); writer.close(); } catch (IOException e) { e.printStackTrace(); } catch (URISyntaxException e) { e.printStackTrace(); }
- How do I transform these paths into something like