Question

i want to open a file and return its content. Although it is in the same directory like the class that wants to open the file, the file can’t be found. Would be cool if you could help me solving the problem.

Here is the code:

@GET @Produces("text/html") @Path("/{partNO}/") @Consumes("text/html")
public String getPartNoResponseHTML(@PathParam("partNO") String parID) throws WebApplicationException {
PartNoTemplate partNo = getPartNoResponse(parID);

String result = "";

try {
  result = readFile(PART_NO_TEMPLATE_FILE);
} catch (FileNotFoundException e) {
  e.printStackTrace(System.out);
  return e.getMessage() + e.toString();
  // throw new WebApplicationException(Response.Status.NOT_FOUND);
} finally {
  result = result.replace("{partNO}", parID);
  result = result.replace("{inputFormat}", partNo.getFormat().toString());
}

return result;
}

I guess it can’t find the file, because its running on tomcat. I’m also using Jersey and JAX-RS. Thank you for your help,

Maxi

Answer 1

Try to get the path of the file from ServletContext.

ServletContext context = //Get the servlet context

In JAX-RS to get servlet context use this:

@javax.ws.rs.core.Context 
ServletContext context;

Then get the file from your web application:

File file = new File(context.getRealPath("/someFolder/myFile.txt"));