FireFox.exe path as per the operating system

Question

How could we identify the installation path of firefox.exe using java code.

    accumulator = (accumulator) && (runCommand("C:\\Program Files (x86)\\Mozilla Firefox\\firefox.exe", "-P PAX-PROFILE-1 %s", "google"));
    accumulator = (accumulator) && (runCommand("C:\\Program Files (x86)\\Mozilla Firefox\\firefox.exe", "-P PAX-PROFILE-2 %s", "google"));
    accumulator = (accumulator) && (runCommand("C:\\Program Files (x86)\\Mozilla Firefox\\firefox.exe", "-P PAX-PROFILE-3 %s", "google"));

Here I am passing windows complete firefox.exe path.

Right now i am executing these command in Java, using

      Process p = Runtime.getRuntime().exec(parts);

Is there any way through which we can identify the location of

firefox.exe

as per the operating system.

Answer 1

Since you tagged the question as 'unix', on a Unix / Linux / MacOSX system, you can do this from the command line using the which command; eg

$ which firefox
/usr/bin/firefox

So, to do the same in Java, you can use a Process to run that command, and read the output into a string. It is also possible to do the same thing by reading the $PATH environment variable, splitting it, and checking each of the directories on the path to see if they contain a "firefox" executable. (You could probably do the same thing on Windows).

BTW, executables on Unix / Linux / MacOSX do not have a ".exe" file suffix. The ".exe" suffix is a Windows-ism.

Answer 2

This is not java-specific. On unix systems (and DOS shells) the general contract of locating an executable is as follows:

1. get the PATH environment variable
2. split it by the path separator character
3. consider each split result as a directory
4. check whether it contains an executable file with the name you're looking for.

At least that's what's used on the CLI. Desktop environments may use separate application registries.