 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
Java Forum / First Aid / September 2005finding container directory of an executing class?
Hi, How do I find the container directory for a particular file when it gets executed? Ex: c:\Tomcat\shared\lib\xyz.jar (xyz.jar contains abc.class) I want to find out in which directory abc.class lies. If I try to find the current directory from within the class using any of the following, System.getProperty("user.dir"); String currentPath = new File(".").getPath(); String currentPath = new File(".").getCanonicalPath(); String currentPath = new File(".").getAbsolutePath(); I get "c:\win32" as my directory, probably because its the "current" working directory for execution. 1) How can I get the path returned to be: c:\Tomcat\shared\lib\ ? 2) Are these any Tomcat APIs that might help me get the full path of where this file lies? Thx for any help. Rohit object.getClass().getResource("yourclassname").getPath(); ------------ FeedFeeds : A new way to read news and blogs! http://www.feedfeeds.com >How do I find the container directory for a particular file when it >gets executed? If you mean class files: see http://mindprod.com/jgloss/classpath.html#WHERE if you mean exe files you exec, see http://mindprod.com/projects/which.html  SignatureCanadian Mind Products, Roedy Green. http://mindprod.com Again taking new Java programming contracts. Hi thx for the replies. Roedy, I tried out your code with a simple example. c:\Level1\Level2\Test.java String where = MyClass.class.getResource( "MyClass" ).getPath(); String where = Test.class.getResource( "Test" ).getPath(); This throws a null pointer exception. What am I doing wrong? > >How do I find the container directory for a particular file when it > >gets executed? [quoted text clipped - 7 lines] > Canadian Mind Products, Roedy Green. > http://mindprod.com Again taking new Java programming contracts. Hi, > Hi thx for the replies. > [quoted text clipped - 6 lines] > > This throws a null pointer exception. In which line? > What am I doing wrong? Are there *really files called "MyClass" and "Test" in this directory? Or are there only the files "MyClass.class" or "Text.class"? Ciao, Ingo My directory contains the files Test.java and Test.class. At Runtime, I get a NullPointerException pointing to the line: String currentDir = Test.class.getResource("Test").getPath(); Aftr compilation I also tried removing the .java from the directory and executed the .class. Still get the same exception. This is my file: ---------------------------------------------------------------- public class Test{ public static String getCommonDataPath() { String currentDir = Test.class.getResource("Test").getPath(); return currentDir; } public static void main(String [] args) { System.out.println("currentDir : " + getCommonDataPath()); } } ------------------------------------------------------------------------- > Hi, > [quoted text clipped - 18 lines] > Ciao, > Ingo Hi, > My directory contains the files Test.java and Test.class. At Runtime, I > get a NullPointerException pointing to the line: > > String currentDir = Test.class.getResource("Test").getPath(); As I expected. What exactly do you expect? There is *no* resource/file called "Test"! Hint (is this really neccessary to say explicitely?): Try String currentDir = Test.class.getResource("Test.class").getPath(); or perhaps: String currentDir = Test.class.getResource(".").getPath(); or perhaps: String currentDir = Test.class.getResource("/").getPath(); Ciao, Ingo Thanks Ingo. That fixed my problem. Finally have it working!! Never thought of explicitly naming the file! I also tried this out with one of my files packaged in a .jar, nested in the Tomcat directory structure. Curiously, the path I get is prefixed by "file:" i.e. something like "file:\C:\Program Files\.....\Tomcat\webapps\commonfiles" is it because this is getting executed from c:\win32 rather than the actual location? This doesn't seem to be causing any problems - windows seems to understand it well. but if i were to deploy on Unix, that would mean trouble. when I tried the earlier example with "Test.java" and ran from the container directory for Test.java, I got the path without the "file:\" prefix?.. > Hi, > [quoted text clipped - 15 lines] > Ciao, > Ingo > Thanks Ingo. That fixed my problem. Finally have it working!! Never > thought of explicitly naming the file! [quoted text clipped - 7 lines] > is it because this is getting executed from c:\win32 rather than the > actual location? ? > This doesn't seem to be causing any problems - windows seems to > understand it well. but if i were to deploy on Unix, that would mean > trouble. No trouble at all. So long as the resource is found, it will return a valid URL that points to the resource. You might note that your resources in .jar archives will have a '!' after the jar name to indicate the resource was found inside a jar, rather than an oddly named directory. >String where = Test.class.getResource( "Test" ).getPath(); > >This throws a null pointer exception. What am I doing wrong? Oops. as a resource, I should be looking for "MyClass.class" not "MyClass" I will fix that. SignatureCanadian Mind Products, Roedy Green. http://mindprod.com Again taking new Java programming contracts. > I want to find out in which directory abc.class lies. Another method is to use <classname>.class.getProtectionDomain().getCodeSource().getLocation(). Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
|
Reply to this Message
|
 Sign In
 Join
 My Latest Posts My Monitored Threads My Blog My Photo Gallery My Profile My Homepage Start New Thread Enable EMail Alerts Rate this Thread
|
©2008 Advenet LLC Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.