# Run commands at random

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.

``````if (( RANDOM % 2 )); then C1; else C2; fi
``````
• 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. Commented Jul 8, 2015 at 3:25
• 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`). Commented Jul 8, 2015 at 20:41

``````C\$((RANDOM%2+1))
``````

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

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

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

• 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`. Commented Jul 3, 2013 at 4:34
• @ChrisDown - see mods.
– slm
Commented Jul 3, 2013 at 4:48
• 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. Commented Jul 3, 2013 at 6:23
• +1 but use `((...))` instead of `[...]` -- double parentheses are specifically for arithmetic expressions. Commented Jul 3, 2013 at 9:45
• @glennjackman - thanks I've made the change.
– slm
Commented Jul 3, 2013 at 11:57