Execute random command

Question

I want my bash script to execute random command of given below. For example

[mysterious command] ("command1", "command2", "command3")

This reminds me of https://github.com/timofurrer/russian-roulette/blob/master/russian-roulette — Ismael Miguel, Aug 04 '16 at 18:10
@IsmaelMiguel BSD rm saves the day: rm: illegal option -- - ;-) — Kusalananda, Aug 04 '16 at 18:57

Kusalananda · Answer 1 · 2016-08-04T21:23:37.613

Put your commands in an array.

cmds=( "cmd1" "cmd2" "cmd3" )

$RANDOM is a random number and ${#cmds[@]} evaluates to the length of your array (3 in this example). $(( RANDOM % ${#cmds[@]} )) will be a random number between 0 and one less than the length of the array cmds, i.e. 0, 1, or 2.

i=$(( RANDOM % ${#cmds[@]} ))

Doing the following would pick the string out of $cmds corresponding to the index $i and execute it as a command.

${cmds[i]}

or all in one go (which looks a bit horrible):

${cmds[RANDOM % ${#cmds[@]}]}

Zachary Brady · Accepted Answer · 2016-08-04T15:44:12.493

3

This should accomplish what you're looking for.

COMMANDS=("command1" "command2" "command3")
$(shuf -n1 -e "${COMMANDS[@]}")

Takes the array and uses shuf to generate a random command.

UPDATE: Changed shuf command per @steeldriver

edited Aug 04 '16 at 15:44

answered Aug 04 '16 at 15:36

Zachary Brady

4,240

can i add more commands than 3? – Aug 04 '16 at 15:37
Yeah, I had a typo in the commands, You should be able to put as many as you want, the index is based off the length of the array. – Zachary Brady Aug 04 '16 at 15:38
2

If your shuf has the -e option, you can skip the indexing and just do $(shuf -n1 -e "${commands[@]}") – steeldriver Aug 04 '16 at 15:41
@steeldriver brilliant. Learn something new every day, updating my answer. – Zachary Brady Aug 04 '16 at 15:42

Execute random command

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