getting the contextpath to read and write text files

[Vanitha Sugumaran]

Hi,

First I have posted my query here in tomcat section, but I think this would be the right forum to ask

https://coderanch.com/t/84787/Tomcat/reading-text-files

My problem:
I have my java class files in tomcat

webapps/applicationName/JavaFiles/classes/JavaHelper.class

Text File folder is in

webapps/webResources/textFiles/*.txt

I need to get my helper java class to read and write the text file in 'textFiles' folder?

so from Tomcat section I got the response to write a servlet and get the context path. I don't have requests for the servlets from a webpage,

how can a java class request a servlet?

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
 
public class SomeServlet extends HttpServlet {
 
     private String myContextPath ="";
  public void doGet(HttpServletRequest request,
                    HttpServletResponse response)
      throws ServletException, IOException {
      
    myContextPath = request.getContextPath();
  }
 
  public String getMyContextPath(){
return myContextPath;
 
 }
}
  

1)Can this be possible?

2)Is there any other way to do this without servlets?

Thanks.

[Sripathi Krishnamurthy]

If reading a text file from a servlet is your concern, this is for you

http://www.jguru.com/faq/view.jsp?EID=11569

Cheers

[Vanitha Sugumaran]

I need to read and write from a java class which is inside Tomcat's webapps folder. (I have folder structure in my first post)

I wrote teh servlet to get the context path, but not sure how to use it.. because I need to call this servlet from a class file not a web page.

thanks,

[Eugene Lucash]

ServletContext.getRealPath(String)

[Vanitha Sugumaran]

Hi,

I tried this..

 
try {
File myFile = new File("..//myTest//xml//defaultXmlFiles",fileName);
 
Writer output = null;
 
 
      //use buffering
      output = new BufferedWriter( new FileWriter(myFile) );
      output.write( aContents );
      //flush and close both "output" and its underlying FileWriter
      if (output != null) output.close();
     
 
 
} catch (MalformedURLException e2) {
 
e2.printStackTrace();
} catch (IOException e1) {
 
e1.printStackTrace();
}
 

my Class file is in :

tomcatHOme\webapps\myServices\WEB-INF\classes\fileWriterTest\

I have
\Tomcat 5.5\webapps\myServices\WEB-INF\myTest\xml\defaultXmlFiles\

when i run the FileWrtierClass from command line it works fine.

but when I call this class from another application. it throws file not found error.

it should be something simple I am missing here..

[Ben Souther]

Trying to find a file to write to dynamically is not a best practice.

While some app servers will unpack a war file into a directory structure, others will not. getRealPath(String) will return null if your app is being run from a war file (as opposed to an exploded directory structure).

You're better off configuring an absolute path as a servlet init param or a context init param.

[Bhiku Mhatre]

Hi Vani,
File myFile = new File("..//myTest//xml//defaultXmlFiles",fileName); is not the right way to do it. You are mapping a relative path. But in this case not relative to root context of the web app but relative to tomcat root directory.
You have following solutions.
1) Provide absolute path..(Hardcoding ..Bad idea)
2)Calibrate the path WRT tomcat root. (Tomcat specific Hardcoding .. Bad idea.)
3) Use ServletContext.getRealPath("/webResources/textFiles/*.txt") (Should work in all web containers.)

Regards,
Milind.

[Ben Souther]

3) Use ServletContext.getRealPath("/webResources/textFiles/*.txt") (Should work in all web containers.)

This will not work in all containers.
The servlet spec requires that apps be able to be run from a war file. It doesn't require that appservers unpack the war file into a directory structure on file system. When apps are run from a war file (not unpacked), getRealPath will return null.

The most dependable solution you will come up with is to use an absolute path, configured as either a servlet init param or a context init param.

Also: if this path points to a directory in your webapp and you are running the app in a packed war file, you will also run into a problem.

Another option is to use the 'tempdir' provided to each app by the container.

[Vanitha Sugumaran]

Thanks for your replies.

I don't have war files.

How would I give init param or context param to a servlet?

<servlet>
        <servlet-name>MyServlet</servlet-name>
        <display-name>MyServlet</display-name>
<init-param>
    <param-name>path</param-name>
    <param-value>//myTestFile/xml/defaultXmlFiles</param-value>
  </init-param>
 
        <description>Simple text servlet</description>
        <servlet-class>image_gallery.MyServlet</servlet-class>
</servlet>

How would I call this servlet from java class instead of webpage?

thanks again,

[Ben Souther]

Take a look at "Servlet Init Params" at http://simple.souther.us.

[Vanitha Sugumaran]

how could this init param help to get the context path?
[code]
<servlet>
<servlet-name>MyServlet</servlet-name>
<display-name>MyServlet</display-name>

<init-param>
<param-name>path</param-name>
<param-value>//myTestFile/xml/defaultXmlFiles</param-value>
</init-param>
<description>Simple text servlet</description>
<servlet-class>image_gallery.MyServlet</servlet-class>
</servlet>
[code]

[Ben Souther]

My point was that you shouldn't write apps that depend on being able to find a particular directory based on the location of the context because, depending, on how your app is run, the context may not have a physical location on the file system.