Get the application's path

Question

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:

1. Your application is located at

    C:\MyJar.jar
   

2. Open the shell (cmd.exe) and cd to C:\test\subdirectory.

3. Start the application using the command java -jar C:\MyJar.jar.

4. 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();