Bash – Execute Random Command

bashrandom

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

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

Best Answer

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

Related Solutions

Bash – random number needed

See man bash, PARAMETERS section, Shell Variables subsection:

RANDOM Each time this parameter is referenced, a random integer between
       0 and 32767 is generated.  The sequence of random numbers may be
       initialized by assigning a value to RANDOM.  If RANDOM is unset,
       it loses its special properties, even if it is subsequently
       reset.

To restrict the value to the 1..6 interval use the $((RANDOM%6+1)) expression. 1

Or you can use /dev/urandom as follows:

tr -dc '1-6' < /dev/urandom | head -c 1

Bash – My script produces the same output when using $RANDOM

It doesn't. But $RANDOM returns big numbers (between 0 and 32767) which, especially for words of limited lengths, shows the same result, as the head portion probably returns all the results of the grep (for 3, there are only 819 matches in my /usr/share/dict/words).

Better solution might be to shuffle the results:

egrep "^.{$num}$" /usr/share/dict/words | sort -R | tail -n 1

where -R means --random-sort (a GNU sort extension).

Best Answer

Related Solutions

Bash – random number needed

Bash – My script produces the same output when using $RANDOM

Related Question