Random number from a range in a Bash Script

Question

Hi,

I need to generate a random port number between 2000-65000 from a shell script. The problem is $RANDOM is only a 16bit number, so im stuck!

PORT=$(($RANDOM%63000+2001)) would work nicely if it wasn't for the size limitation.

Does anyone have an example of how I can do this, maybe by extracting something from /dev/urandom and getting it within a range?

Thanks.

Answer 1

The simplest general way that comes to mind is a perl one-liner:

perl -e 'print int(rand(65000-2000)) + 2000'

You could always just use two numbers:

PORT=$(($RANDOM + ($RANDOM % 2) * 32768))

You still have to clip to your range. It's not a general n-bit random number method, but it'll work for your case, and it's all inside bash.

If you want to be really cute and read from /dev/urandom, you could do this:

od -A n -N 2 -t u2 /dev/urandom

That'll read two bytes and print them as an unsigned int; you still have to do your clipping.

Answer 2

shuf -i 2000-65000 -n 1

Enjoy!

Edit: The range is inclusive.

Answer 3

and here's one with Python

randport=$(python -S -c "import random; print random.randrange(2000,63000)")

and one with awk

awk 'BEGIN{srand();print int(rand()*(63000-2000))+2000 }'

Answer 4

On Mac OS X and FreeBSD you may also use jot:

jot -r 1  2000 65000

Answer 5

Here's another one. I thought it would work on just about anything, but sort's random option isn't available on my centos box at work.

 seq 2000 65000 | sort -R | head -n 1

Answer 6

You can do this

cat /dev/urandom|od -N2 -An -i|awk -v f=2000 -v r=65000 '{printf "%i\n", f + r * $1 / 65536}'

If you need more details see Shell Script Random Number Generator.