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How can I check if a program exists from a Bash script?

How would I validate that a program exists, in a way that will either return an error and exit, or continue with the script?

It seems like it should be easy, but it's been stumping me.

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Doesn't work. test -x $(which ls) returns 0, as does test -x $(which sudo), even though ls is installed and runnable and sudo is not even installed within the docker container I'm running in.

algal
– algal

2019年02月20日 02:26:23 +00:00
Commented Feb 20, 2019 at 2:26
2

@algal You need to use quotes I think, so test -x "$(which <command>)"

JoniVR
– JoniVR

2019年04月02日 12:19:10 +00:00
Commented Apr 2, 2019 at 12:19
@algal Perhaps ls is aliased? I dont think it would work if the command has parameter.

AnthonyWC
– AnthonyWC

2019年04月02日 15:38:36 +00:00
Commented Apr 2, 2019 at 15:38
I can't vouch for this answer, but I'd also recommend quotes. $ test -x $(which absent_cmd) returns test: too many arguments, while test -x "$(which absent_cmd)" is properly parsed and results in exit code 1.

acorello
– acorello

2020年12月17日 18:13:38 +00:00
Commented Dec 17, 2020 at 18:13
2

@acorello, the test: too many arguments suggests that which absent_cmd returns a multi-word error message. An executable file having the same name as the message could exist. 😁 Using which this way is just a bad idea.

Palec
– Palec

2021年12月23日 14:57:15 +00:00
Commented Dec 23, 2021 at 14:57

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How can I check if a program exists from a Bash script?

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