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Asked: 2026-05-11T18:11:15+00:00

I have the following code which should put programs startable in Bash. if [

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I have the following code which should put programs startable in Bash.

if [ "`uname`" = "Darwin" ]; then
   compctl -f -x 'p[2]' -s "`/bin/ls -d1 /Applications/*/*.app
/Application:/*.app | sed 's|^.*/\([^/]*\)\.app.*|\\1|;s/ /\\\\ /g'`"                 
-- open
   alias run='open -a' 
fi

However, it does not work in my Zsh at all. I can open no programs with it.

Another bug which it has is that it opens all programs.
I want have only programs startable which permissions is 700.

I know that you can search these programs rather well by

 find -perm 700 -type f *.app

However, I could not get my find -command work exactly for each program file.
This suggests me that there may be a better way to make programs startable in terminal.

How can you make programs startable in Zsh in Mac/Ubuntu?

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1 Answer

  1. Editorial Team

    How do you feel about aliases instead of completion? The aliases still tab complete. Splitting across lines to make it easier to read…

    `print -l /Applications/**/MacOS/*(*f:700:) | 
grep -v "Contents.*Contents" | 
sed -e "s#\(.*/\)\([^./]*\)\(\.app\)\(.*\)#alias \2=\"\1\2\3\4\"#g"`

    The first prints each match on its own line, the second removes sub-packages of an app, and the last generates the alias command. Remove the backticks to see and confirm the commands that are generated.

    [Edit:] This runs the executable directly – if you prefer to use the ‘open blah.app’ method, you could change the final portion of the sed command to:

    #alias \2=\"open \1\2\3\"#
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