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I want to run one of the two commands C1 and C2 at random. How do I do that on commandline (bash)? Will appreciate if a one-liner is possible.

3 Answers 3

39
if (( RANDOM % 2 )); then C1; else C2; fi
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  • 2
    Bear in mind that if you want a 1 in N odds of running C1, you must probably use (( RANDOM % N == 0 )) rather than (( RANDOM % N )). It's only the in special case of N=2 that these have identical probability.
    – Asclepius
    Jul 8, 2015 at 3:25
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    Plainly put, of what practical use would be RANDOM % $N without == 0? IMO, for example with N=100, it is much more pragmatic to want a 1 in 100 odds (delivered with == 0) than a 99 in 100 odds (delivered without == 0).
    – Asclepius
    Jul 8, 2015 at 20:41
21

In your special case:

C$((RANDOM%2+1))

will work :) And hey, it's the shortest answer!

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  • Why does it need two sets of parentheses around it? I am just not very familiar with whatever you used. Jul 3, 2013 at 18:16
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    @AJMansfield $((1+1)) -> 2, eg. $(( )) calculates what's inside.
    – Tyilo
    Jul 3, 2013 at 18:46
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    It would be even shorter if the OP chose C0 and C1. Maybe he's not a programmer?
    – ott--
    Jul 3, 2013 at 19:51
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    I'm pretty sure that C1 and C2 were placeholders rather than actual commands to run. :-)
    – Chris Down
    Jul 6, 2013 at 3:42
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    That's cute. +1
    – sudo
    Jan 18, 2017 at 22:47
6

You can do something like this in Bash:

$ (( RANDOM%2 == 0 )) && C1 || C2

This will generate a random number, either 0 or 1. If it's a 0, then C1 runs, otherwise C2 runs if it isn't.

example

$ (( RANDOM%2 == 0 )) && echo 0 || echo 1
1

$ (( RANDOM%2 == 0 )) && echo 0 || echo 1
0

NOTE: The first character, $, is the prompt.

another example

If you're concerned with C1 having to fail so that C2 can run you could restructure the above like so:

(( RANDOM%2 == 0 )) && CMD=C1 || CMD=C2
$CMD

why use this over a if/then statement?

This answer has been criticized a bit but there's a method to my madness. Though this pattern may seem more obscure than a if/then I find it more readable and compact when doing something like the following:

#!/bin/bash
CMD=""
DIRS="/etc /home /www /data1 /data2 /var/log /var/spool/mail"
FILE="/backup/$(hostname)-$(date +'%m-%d-%y').tar.gz"
[ "$1" == "nas" ]  && CMD="lftp -u user,password -e 'cd /dump/; mput /backup/*; quit' nas.mylan.com" || :
[ "$1" == "scp" ]  && CMD="scp /backup/* scponly@dumpserver:incoming' username" || :
[ "$1" == "tape" ] && CMD='tar -cf /dev/st0 /backup/*' || :
[ "$CMD" == "" ]   && exit 1 || :
# make a backup
tar -zcvf  $FILE $DIRS
# Now depend upon circumstances run a backup command
$CMD

References

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  • 2
    This will run C2 if C1 fails, which violates the specification laid out in the question (namely, that only one of the commands is run). x && y || z is not equivalent to if x; then y; else z; fi.
    – Chris Down
    Jul 3, 2013 at 4:34
  • @ChrisDown - see mods.
    – slm
    Jul 3, 2013 at 4:48
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    The edit nominally fixes this, but it's really obscure. "If you're concerned with C1 having to fail so that C2 can run" doesn't capture the scenario at all. The issue is that if C1 is a command which sometimes returns a non-zero exit code, you will end up executing both C1 and C2 when that happens.
    – tripleee
    Jul 3, 2013 at 6:23
  • +1 but use ((...)) instead of [...] -- double parentheses are specifically for arithmetic expressions. Jul 3, 2013 at 9:45
  • @glennjackman - thanks I've made the change.
    – slm
    Jul 3, 2013 at 11:57

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