I've recently searched how I could get the application's directory in Java. I've finally found the answer but I've needed surprisingly long because searching for such a generic term isn't easy. I think it would be a good idea to compile a list of how to achieve this in multiple languages.
Feel free to up/downvote if you (don't) like the idea and please contribute if you like it.
Clarification:
There's a fine distinction between the directory that contains the executable file and the current working directory (given by pwd under Unix). I was originally interested in the former but feel free to post methods for determining the latter as well (clarifying which one you mean).
In Java the calls
System.getProperty("user.dir")
and
new java.io.File(".").getAbsolutePath();
return the current working directory.
The call to
getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
returns the path to the JAR file containing the current class, or the CLASSPATH element (path) that yielded the current class if you're running directly from the filesystem.
Example:
Your application is located at
C:\MyJar.jar
Open the shell (cmd.exe) and cd to C:\test\subdirectory.
Start the application using the command java -jar C:\MyJar.jar.
The first two calls return 'C:\test\subdirectory'; the third call returns 'C:\MyJar.jar'.
When running from a filesystem rather than a JAR file, the result will be the path to the root of the generated class files, for instance
c:\eclipse\workspaces\YourProject\bin\
The path does not include the package directories for the generated class files.
A complete example to get the application directory without .jar file name, or the corresponding path to the class files if running directly from the filesystem (e.g. when debugging):
String applicationDir = getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
if (applicationDir.endsWith(".jar"))
{
applicationDir = new File(applicationDir).getParent();
}
// else we already have the correct answer
In .NET (C#, VB, …), you can query the current Assembly instance for its Location. However, this has the executable's file name appended. The following code sanitizes the path (using System.IO and using System.Reflection):
Directory.GetParent(Assembly.GetExecutingAssembly().Location)
Alternatively, you can use the information provided by AppDomain to search for referenced assemblies:
System.AppDomain.CurrentDomain.BaseDirectory
VB allows another shortcut via the My namespace:
My.Application.Info.DirectoryPath
In Windows, use the WinAPI function GetModuleFileName(). Pass in NULL for the module handle to get the path for the current module.
Python
path = os.path.dirname(__file__)
That gets the path of the current module.
Objective-C Cocoa (Mac OS X, I don't know for iPhone specificities):
NSString * applicationPath = [[NSBundle mainBundle] bundlePath];
In Java, there are two ways to find the application's path. One is to employ System.getProperty:
System.getProperty("user.dir");
Another possibility is the use of java.io.File:
new java.io.File("").getAbsolutePath();
Yet another possibilty uses reflection:
getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
In VB6, you can get the application path using the App.Path property.
Note that this will not have a trailing \ EXCEPT when the application is in the root of the drive.
In the IDE:
?App.Path
C:\Program Files\Microsoft Visual Studio\VB98
In .Net you can use
System.IO.Directory.GetCurrentDirectory
to get the current working directory of the application, and in VB.NET specifically you can use
My.Application.Info.DirectoryPath
to get the directory of the exe.
Delphi
In Windows applications:
Unit Forms;
path := ExtractFilePath(Application.ExeName);
In console applications:
Independent of language, the first command line parameter is the fully qualified executable name:
Unit System;
path := ExtractFilePath(ParamStr(0));
Libc
In *nix type environment (also Cygwin in Windows):
#include <unistd.h>
char *getcwd(char *buf, size_t size);
char *getwd(char *buf); //deprecated
char *get_current_dir_name(void);
See man page
Unix
In unix one can find the path to the executable that was started using the environment variables. It is not necessarily an absolute path, so you would need to combine the current working directory (in the shell: pwd) and/or PATH variable with the value of the 0'th element of the environment.
The value is limited in unix though, as the executable can for example be called through a symbolic link, and only the initial link is used for the environment variable. In general applications on unix are not very robust if they use this for any interesting thing (such as loading resources). On unix, it is common to use hard-coded locations for things, for example a configuration file in /etc where the resource locations are specified.
In bash, the 'pwd' command returns the current working directory.
In PHP :
<?php
echo __DIR__; //same as dirname(__FILE__). will return the directory of the running script
echo $_SERVER["DOCUMENT_ROOT"]; // will return the document root directory under which the current script is executing, as defined in the server's configuration file.
echo getcwd(); //will return the current working directory (it may differ from the current script location).
?>
in Android its
getApplicationInfo().dataDir;
to get SD card, I use
Environment.getExternalStorageDirectory();
Environment.getExternalStoragePublicDirectory(String type);
where the latter is used to store a specific type of file (Audio / Movies etc). You have constants for these strings in Environment class.
Basically, for anything to with app use ApplicationInfo class and for anything to do with data in SD card / External Directory using Environment class.
Docs :
ApplicationInfo ,
Environment
In Tcl
Path of current script:
set path [info script]
Tcl shell path:
set path [info nameofexecutable]
If you need the directory of any of these, do:
set dir [file dirname $path]
Get current (working) directory:
set dir [pwd]
Java:
On all systems (Windows, Linux, Mac OS X) works for me only this:
public static File getApplicationDir()
{
URL url = ClassLoader.getSystemClassLoader().getResource(".");
File applicationDir = null;
try {
applicationDir = new File(url.toURI());
} catch(URISyntaxException e) {
applicationDir = new File(url.getPath());
}
return applicationDir;
}
in Ruby, the following snippet returns the path of the current source file:
path = File.dirname(__FILE__)
In CFML there are two functions for accessing the path of a script:
getBaseTemplatePath()
getCurrentTemplatePath()
Calling getBaseTemplatePath returns the path of the 'base' script - i.e. the one that was requested by the web server.
Calling getCurrentTemplatePath returns the path of the current script - i.e. the one that is currently executing.
Both paths are absolute and contain the full directory+filename of the script.
To determine just the directory, use the function getDirectoryFromPath( ... ) on the results.
So, to determine the directory location of an application, you could do:
<cfset Application.Paths.Root = getDirectoryFromPath( getCurrentTemplatePath() ) />
Inside of the onApplicationStart event for your Application.cfc
To determine the path where the app server running your CFML engine is at, you can access shell commands with cfexecute, so (bearing in mind above discussions on pwd/etc) you can do:
Unix:
<cfexecute name="pwd"/>
for Windows, create a pwd.bat containing text #cd, then:
<cfexecute name="C:\docume~1\myuser\pwd.bat"/>
(Use the variable attribute of cfexecute to store the value instead of outputting to screen.)
In cmd (the Microsoft command line shell)
You can get the name of the script with %* (may be relative to pwd)
This gets directory of script:
set oldpwd=%cd%
cd %0\..
set app_dir=%pwd%
cd %oldpwd%
If you find any bugs, which you will. Then please fix or comment.
I released https://github.com/gpakosz/whereami which solves the problem in C and gives you:
the path to the current executable
the path to the current module (differs from path to executable when calling from a shared library).
It uses GetModuleFileNameW on Windows, parses /proc/self/maps on Linux and Android and uses _NSGetExecutablePath or dladdr on Mac and iOS.
Note to answer "20 above regarding Mac OSX only: If a JAR executable is transformed to an "app" via the OSX JAR BUNDLER, then the getClass().getProtectionDomain().getCodeSource().getLocation(); will NOT return the current directory of the app, but will add the internal directory structure of the app to the response. This internal structure of an app is /theCurrentFolderWhereTheAppReside/Contents/Resources/Java/yourfile
Perhaps this is a little bug in Java. Anyway, one must use method one or two to get the correct answer, and both will deliver the correct answer even if the app is started e.g. via a shortcut located in a different folder or on the desktop.
carl
SoundPimp.com
Related
Im making a basic .bat file that runs the application. This is to get auto startup feature.
However for some reason System.getProperty("user.dir") does not always get the correct path to the program.
Basically i am saving this to the .bat file:
protected final String fileSeparator=System.getProperty("file.separator");
out.println("#echo off");
out.println("start " + System.getProperty("user.dir") + fileSeparator +"App.jar");
out.println("exit");
On Windows server it returns the correct path but on Vista it does not.
Any ideas how i could get this to work on all versions of Windows?
You can do this by providing user.dir when you start your Java program like this
java somepackage.Main -Duser.dir=C:/Users/myUser
user.dir = User working directory
[from documentation ]
It is difficult to take decision based on value of this variable. Depending on the program starting the "java", it could have different values. For example, a bat file could have different working directory from invoked from different command-windows.
You could use %~dp0 to get location of the batch script and then put other paths relative to this.
Another option is to use tools like launch4j, which allows an easy way to control program directory : How to get the path to the executable when using launch4j?
I am writing a Java utility that executes a batch file to generate a PDF using the DITA toolkit and Apache FOP. It finishes by using pdftk to watermark the front page. If I execute the batch file in Windows using Start>CMD, this line in the batch file works:
pdftk "%DITA_OUTPUT%book.pdf" multibackground C:\doc_build_system\watermark.pdf output "%DITA_OUTPUT%external.pdf" compress verbose
When I execute the batch file by Runtime.exec() the same line fails.
The cause of the failure is that the PATH variable is incomplete when executed through Java. It should have an entry like:
C:\Program Files (x86)\PDFtk Server\bin\
...but it does not. I tried to force execution through CMD by invoking runtime.exec("cmd /c batchfile.bat") (rather than just directly invoking the batchfile) but this also had no effect.
Any ideas?
You can try to set the path manualy before your start your java in cmd:
start cmd.exe. Then type:
SET PATH=%PATH%;C:\Program Files (x86)\PDFtk Server\bin
java MyProgram
If that is working you have to check if edited the right PATH variable. In Windows you can have different PATH environment variables for each user, plus there is one system-wide PATH variable (see screenshot) that will always be applied and combined with the user variables.
e.g. if you did set the path for your user and then use administrator for elevated rights to execute java, the PATH won't be set properly.
Make sure to use the system variable.
Also make sure to restart windows after you did edit the variable, because open applications and consoles will usually only fetch environment variables once at start up.
Just take pathman from the Windows Server 2003 Resource Kit:
USAGE:
/as path[;path[;path ...]]
Adds the semicolon-separated paths to the system path.
/au path[;path[;path ...]]
Adds the semicolon-separated paths to the user path.
/rs path[;path[;path ...]]
Removes the semicolon-separated paths from the system path.
/ru path[;path[;path ...]]
Removes the semicolon-separated paths from the user path.
I have problems compiling Java programs from command line interfaces (both command prompt and NetBeans terminal).
I added the jdk directory to the PATH system variable but not all commands are recognizable, some commands (in the jdk1.6.0_37\bin folder) are recognized (like: java, javaw, packger)and others I'd have to type the full directory name for it to work (like: jar, javac, javah, javap) (this applies for both cmd and NB).
I don't know why this should be, all of these files are .exe , all are Java Platform SE binary, the only difference I can see is the icon of the files, the ones that work have the Java logo (Coffee Mug) and the rest have the standard .exe logo. Any help would be appreciated. Thanks
You have to add the jdk1.6.0_37\bin directory to the PATH variable. That is where all the commands that you want to run are in. After you add the directory to the PATH variable, open a new command prompt and run those commands again.
I'm searched many answers that suggest me to type in cmd:
set path = "%path%;c:program files\java\jdk1.7.0\bin"
but this is WRONG!
the right solution is that you leave "set" and just type
path = %path%;c:program files\java\jdk1.7.0\bin
P/s: of course you have to replace "jdk1.7.0" folder by your current java version folder
I am trying to run the command:
cmd.exe /X /C "svn --non-interactive info <file name>
by using commandline.class from the API (org.codehaus.plexus.util.cli) which is called by
SVNInfoCommand.class(org.apache.maven.scm.provider.svn.svnexe.command.info)
It works fine when I pass a filename which exists in C:\ drive.
But it returns the following error when I pass a filename which exists in shared location or network drive. Error: "UNC paths are not supported. Defaulting to Windows directory. svn: '.' is not a working copy"
Current working directory is being set using the method setWorkingDirectory( String path ), but I suspect working directory is not being set when we use shared locations(UNC Paths).
EX: \Test_Location\Test_File
This command works fine when I pass the “absolute path” of filename located in shared drive.
(\Test_Location\Test_File)
Please let me know why we get this error when we use UNC paths even after setting the working directory.
By default, the Windows command processor (cmd.exe) doesn't support UNC paths.
You can alter this behavior by adding a dedicated registry key, as described in Microsoft KB 156276.
Alternatively, you can map your UNC path to a standard drive letter, and then use that.
From Java, I'm extracting an executable into a location specified using File.createTempFile(). When I try to run my executable, my program hangs when it tries to read the first line of output.
I have discovered that if I try to run the same extracted executable from another program, it works if I specify the directory as C:\Documents and Settings\username\Local Settings\Temp\prog.exe. But if I specify the directory as C:\DOCUME~1\USERNA~1\LOCALS~1\Temp\prog.exe I get the hang.
Is there a way to unmangle the tilde filename in my program so I can specify a directory name that will work?
(And since I always like addressing the language and API design issues, is there any reason why Java File.createTempFile() and java.io.tmpdir have to evaluate to mangled filenames?)
You can use getCanonicalPath() to get the expanded path. E.g.:
try
{
File file = File.createTempFile("abc", null);
System.out.println(file.getPath());
System.out.println(file.getCanonicalPath());
}
catch (IOException e) {}
... produces ...
C:\DOCUME~1\USERNA~1\LOCALS~1\Temp\abc49634.tmp
C:\Documents and Settings\username\Local Settings\Temp\abc49634.tmp
I tested this on XP, but assume it would work similarly on other Windows operating systems.
See #raviaw's answer to your second question.
Wow, I never saw that. The fact is that the environment variable %TEMP% returns a mangled name (this is from my computer):
TEMP=C:\DOCUME~1\raviw\LOCALS~1\Temp
TMP=C:\DOCUME~1\raviw\LOCALS~1\Temp
Assuming that a newly create java VM uses the environment variable to get the temporary folder location, it is not VM's fault that the directories are coming mangled.
And even if you try to use System.getenv() to get the temporary folder, you will still have the same problem.
I would make sure that:
The problem is not caused by the fact that you have a directory called "prog.exe" (based on your question, I am assuming this);
If the file is "prog.exe", if it was not in use by any other program (an antivirus, maybe);
Checking if your computer is sane (this would be a very critical bug for any application that is not a web application and that need temporary files).