From command line, how to know which Firefox version is installed in windows/linux?

Question

I need to know which Firefox version is installed on my system from command line of Windows or Linux.

Answer 1

7 years later...

On newer version (I tried on 51.0.1) @Nömmik's answer doesn't work on Windows.

But you can try this workaround:

firefox -v | find "Mozilla"

Answer 2

On linux (debian, lenny), you can use -v:

firefox -v
Mozilla Firefox 3.0.19, Copyright (c) 1998 - 2010 mozilla.org

(that was on a random server just now)

Answer 3

According to this link, it seems that the -v argument (and more) is broken on Windows. If you follow the bug link on the site, it seems there is a workaround. I quote the comment:

Workaround (works with Firefox, Thunderbird, and, I suppose, other programs too):

<program-name> -h | more

The bug happens because, without redirection, the program releases its stdout before handling the -help parameter. With redirection, stdout is not released and you can see the output.

So for example for Firefox:

C:\Program Files (x86)\Mozilla Firefox>firefox -v | more
Mozilla Firefox 3.6.13, Copyright (c) 1998 - 2010 mozilla.org

It works for me at least. Without | more I get nothing printed. On Linux it works with or without the piping.

Answer 4

On my Windows 8 box I was aware that Firefox Quantum was installed and the firefox binary is placed within:

C:\Program Files\Mozilla Firefox

So when I changed the directory to C:\Program Files\Mozilla Firefox and issued the command firefox -v nothing was shown:

But when I issued the command firefox -v|more it worked perfecto:

Answer 5

In 2018, FF 60:

firefox --version

output:

Mozilla Firefox 60.0.1

Answer 6

If this does not work , try using ./firefox -v