Search code examples
gopathzsh

exec: "firefox": executable file not found in $PATH

I am reading a go book to learn go writing command line tools. In one of the examples I need to open firefox browser with an html file. The command run is:

browserPath,_ := exec.LookPath("firefox")

// Open the file on the browser
if err := exec.Command(browserPath, "index.html").Start(); err != nil {
    return err
}

But get the error:

exec: "firefox": executable file not found in $PATH

My $PATH is:

$HOME/bin:/usr/local/bin:$PATH

I am running on mac with zsh. I looked at similar problems but can't solve it yet, anyone see what I am missing?

Solution

  • First of all thank you for reopening the question, I understand it might be trivial to many, but it can help someone else to see this answer.

    The issue was (as @xarantolus commented) that my PATH did not contain the route to my /Applications folder. Since I use zsh and mac, I did the following steps to fix it:

    First find where firefox executable was:

    $ type -a firefox

    Printed route:

    firefox is /Applications/Firefox.app/Contents/MacOS/firefox

    Now open zshrc file:

    $ vim ~/.zshrc

    Inside the file, my $PATH was $HOME/bin:/usr/local/bin:$PATH, and I added /Applications so the line ended up like this:

    export PATH=$HOME/bin:/usr/local/bin:/Applications/:$PATH

    Notice that :$PATH will be at end also after adding the new path

    Then run command to reload .zshrc

    source ~/.zshrc

    If you dont use zsh, use the file .bashrc instead of .zshrc

    Go could now see the firefox executable and it opened it as expected.