CODEWITHSUNDEEP
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Hellzzar
Hellzzar

Reputation: 195

bash is not executed 'at -f foo.sh' command, even with #!/bin/bash shebang

I'm using the 'at' command in order to create 3 directories, just a dumb bash script:

#!/bin/bash
for i in {1..3}
do
mkdir dir$i
done

Everything is ok if I execute that script directly on terminal, but when I use 'at' command as follows:

at -f g.sh 18:06

It only creates one directory named dir{1..3}, taking interval not as an interval but as a list with one element {1..3}. According to this I think my mistake is using bash script due to at executes commands using /bin/sh but I'm not sure. Please tell me if I'm right and I would appreciate some alternative to my code since even it is useless I'm curious to know what's wrong with at and bash.

Upvotes: 2

Views: 228

Answers (1)

Barmar
Barmar

Reputation: 780818

The #! line only affects what happens when you run a script as a program (e.g. using it as a command in the shell). When you use at, it's not being run as a program, it's simply used as the standard input to /bin/sh, so the shebang has no effect.

You could do:

echo './g.sh' | at 18:06

Upvotes: 2

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