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I googled many solutions but did not find one that fully worked. My own rough working is:

[ $(which curl) -eq "" ] || sudo apt install curl

This doesn't quite work, saying "unary operator expected". I've tried = or ==, so I'm not getting how to do a test like this. Is it possible that the empty output from which curl is a $null and I have to test for that?

Alternatively, is there a way to just run sudo apt install curl and then 2>&null at the end to suppress if it's already installed?

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  • The trivial fix is to quote the command substitution. Sure you can run sudo apt -y install curl >/dev/null 2>&1 but that is going to fail or hang if the user does not have their sudo password cached. Commented Nov 26, 2020 at 13:14
  • @YorSubs : Technically speaking, you don't test whether an executable curl exists, you just test whether it exists and is in your PATH. Also, you don't know whether (whatever is hidden behind the name curl) is that curl in that version you need. One alternative would be to run curl --version and parse its output. Commented Nov 26, 2020 at 13:17
  • I see your point. Would testing for /usr/bin/curl be more rigorous (I don't need completely bulletproof, and it would be very odd for someone to put a file there that was not a downloaded version of curl). My only question then is: do different distros put curl in different places, or would /usr/bin/curl be a reliable cross-platform location? Commented Nov 26, 2020 at 13:23
  • Not only can other distros put curl elsewhere, they can also use a different installer than apt. Commented Nov 26, 2020 at 15:21

1 Answer 1

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which returns a failure if it doesn't find the program, so you can

which curl &> /dev/null || sudo apt install curl

Note that if someone has a different program in $PATH named curl, the script can still fail later.

-eq compares numbers, not strings.

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haha that's a creative way to remain shell-portable, thanks!

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