Running `docker run` from bash script fails. Command does not fail on Command Line

Question

If I run the following from the command line.

docker run -t repo:tag ls -l

the command succeeds just fine. However, if I invoke the same from within a bash script I get the following ERROR:

docker: Error response from daemon: OCI runtime create failed: container_linux.go:348: starting container process caused "exec: \"ls -l\": executable file not found in $PATH": unknown.

What about the bash script causes this error?

Without seeing the exact invocation of the script, I'd say that you're incorrectly quoting the arguments so that "ls" "-l" has become "ls -l". — erik258
– erik258, Commented Dec 31, 2018 at 19:48
Have you tried removing the -t flag when running in the bash script? — Tim Klein
– Tim Klein, Commented Dec 31, 2018 at 19:52
@DanFarrell yes "ls" "-l" was being treated as "ls -l". Please put this as an answer? — Matthew Hoggan
– Matthew Hoggan, Commented Dec 31, 2018 at 20:48

erik258 · Accepted Answer · 2019-01-01 01:27:50Z

1

"exec: \"ls -l\": executable file not found in $PATH"

From the error I can tell that when you invoke docker, you somehow invoke with ls -l including space as one argument. Something like,

docker run -t repo:tag "ls -l" # wrong

or perhaps

cmd="ls -l" 
docker run -t repo:tag "$cmd" # wrong

The shell to parse the docker command must see ls and -l as separate parameters so that the argument -l is distinguished from the ls executable name.

cmd="ls -l" 
docker run -t repo:tag $cmd #works

answered Jan 1, 2019 at 1:27

erik258

16.4k2 gold badges32 silver badges47 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Running `docker run` from bash script fails. Command does not fail on Command Line

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related