zsh: command not found: laravel

Question

zsh 5.7.1 (x86_64-apple-darwin19.0)

Here is my $PATH: in my.bash_profile

export PATH="$PATH:$HOME/.composer/vendor/bin"
export PATH="$PATH:/usr/local/opt/mysql@5.7/bin"
export PATH="$PATH:$HOME/.npm-packages/bin"
export PATH="$PATH:/usr/local/opt/node@8/bin"
export PATH="$PATH:/usr/local/bin"
export PATH="$PATH:/usr/local/sbin"
export PATH="$PATH:/usr/local/opt/php@7.2/bin"
export PATH="$PATH:/usr/local/opt/php@7.2/sbin"

I'm not sure why the laravel command isn't being found because it used to work just fine. I have the $HOME/.composer/vendor/bin added to the path as shown in the laravel documentation https://laravel.com/docs/7.x

Any ideas why this might not be working?

Answer 1

I solved my problem by reinstalling composer and Laravel as well as adding composer to my path within the ~/.zshrc file. After that, running composer commands started working again. This problem was caused by my swap from bash to zsh.

Answer 2

Bash and zsh are different shells and therefore use different config files.

.bash_profile is for bash only and .zshrc is for zsh only. You can change between the shells by typing their respective names as commands ( bash will get you into bash from zsh and vice versa).