Path of files in Tomcat webapplication

Question

I want to load a few parameters from a xml file on my Tomcat server. let´s say thats the folder structure:

 Tomcat
 |-webapps
   |-MyWebApp.war
   |-MyWebApp //(Source files)
      |-META-INF
      |-WEB-INF
         |-configFile.xml

I tried several different methods like

File inputfile = new File("/MyWebApp/WEB-INF/configFile.xml");
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(inputfile);

which just delivered a filemissing exception...

So the question is:

HOW do I have to set the path in my Java code to refer to files on my Tomcat ?

score 1 · Accepted Answer · answered Feb 04 '16 at 10:25

1

You should avoid to use files in a web application.

Here is the way to go: You need to obtain a ServletContext (e.g. being inside a Servlet).

Then use

ServletContext context = ...
InputStream in = context.getResourceAsStream("/WEB-INF/configFile.xml");

to open an InputStream to your XML file and pass it to the DocumentBuilder.

Document doc = dBuilder.parse(in);

answered Feb 04 '16 at 10:25

wero

32,544
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How can I get the ServletContext ? I just have a simple REST service class which calls my (service)class to open the file and get the prams. – JohnDizzle Feb 04 '16 at 10:36
@JohnDizzle inject the context into your service class, e.g. for JAX-RS see http://stackoverflow.com/questions/1814517/get-servletcontext-on-tomcat-from-jax-rs-jersey – wero Feb 04 '16 at 10:40

score -1 · Answer 2 · answered Feb 04 '16 at 10:29

-1

Try this

File inputfile = new File("/WEB-INF/configFile.xml");

OR

File inputfile = new File("../WEB-INF/configFile.xml");

answered Feb 04 '16 at 10:29

Piyush Ghediya

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Path of files in Tomcat webapplication

2 Answers2