On Linux System.getenv("HOME") return absolute path /home/user, but on Windows return Users\user.
On Windows the home drive is specified separately in HOMEDRIVE variable. Concatenating it with HOMEPATH gives you the absolute path:
String home = System.getenv("HOMEDRIVE")+System.getenv("HOMEPATH");
You need to use System.getProperty("user.home") if you want something that works the same on multiple operating systems.
System.getenv is operating-system or context dependent - there is no guarantee whatsoever that System.getenv("HOME") returns anything in particular on a given operating system, it's just luck that what you get on Unix is anything similar to what you get on Windows.
From the Javadoc for System.getenv:
An environment variable is a system-dependent external named value.
For System.getProperty, there is a list of properties that you can get in a system-independent way:
https://docs.oracle.com/javase/7/docs/api/java/lang/System.html#getProperties()
One of them is user.home: "User's home directory"
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
In a java method i am loading a path name into a string variable, not knowing whether it is a Unix or a Windows path name.
How can i programmatically check whether it is a Unix or a Windows path?
if path.contains(":") execute a Windows-related function
else execute a Unix-related function
won't work, because Unix path names can contain a ":" as far as i know. Same with "/" for Unix , because Windows path names can contain a "/" .
So how could i do this best?
EDIT: I am loading directory names for remote machines, so i doubt i can use Use System.getProperty("os.name").
i was thinking about using Regex, but since i have no idea about regex in Java i have not considered that yet.
IMHO, using path names to determine OS platform is not the best idea.
Use System.getProperty("os.name")
and have a look here:
How do I programmatically determine operating system in Java?
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?
Please I am facing the following issue:
Throughout my Java program, i am accessing some files which it seems they are being accessed in a different way under windows compared to Linux. For example, if i wanted to access the following file within the same folder as the project i would write the following:
Under Linux: File Operations_File = new File("Data/Operations.txt");
Under Windows: File Operations_File = new File("Data\\Operations.txt");
I will be needing a standard methodology that works under all operating systems (or at least those two). As coding two versions of my code is not elegant at all.
My Two operating system that I am operating on are: Linux Mint 9 and Windows XP. I used NetBeans 6.9.1 throughout all the project.
Your help is greatly appreciated!
File.separator is exactly for this.
File f = new File("Data" + File.separator + "Operations.txt");
Don't get confused with File.pathSeparator, that is used to separate paths from each other. For example:
/usr/local/lib:/usr/lib:/var/lib
In the above example, : is the path separator (windows uses ; for path separators).
You can also create a File representing the directory and another File representing something in that directory like this:
File dataDir = new File("Data");
File operationsFile = new File(dataDir, "Operations.txt");
You could also skip the File for the directory and just do this as well:
File operationsFile = new File("Data", "Operations.txt");
Under Windows, printing out operationsFile gives Data\Operations.txt as expected.