Execute random file from a folder in a bash script

Question

I'm newbies with bash scripts, so please, be indulgent ;-) !!!

I would like to read php files randomly in a pre-defined folder (there are 30 php files in this folder).

My current script:

#!/bin/sh
curl "/myfolder/myfile.php" &
exit 0

With my research i've already done, i've found some examples but i'm not sure of anything with my little skills.

I know i must use an "for ls" and then do something like "echo $ ((1 + RANDOM% 30))", but i'm not sure !

Could you help me please?

@GilbertOOI Did you notice [SO: Random number from a range in a Bash Script](http://stackoverflow.com/questions/2556190/random-number-from-a-range-in-a-bash-script)? This combined with an [array](http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_10_02.html) should work. — Scheff's Cat, Mar 13 '17 at 06:38
Thanks a lot Scheff ! i'll try to make a good things with your links! — GilbertOOl, Mar 13 '17 at 06:45

score 3 · Accepted Answer · answered Mar 13 '17 at 12:23

You can just use the "sort" tool with the option "-R" to sort randomly. In a given directory you can run

ls -1 /etc/ | sort -R | head -1

to pick one file randomly out of all. The "etc" directory is just an example.

Your code should then look like this:

curl `ls -1 /my_folder/ | sort -R | head -1` &

But I don't understand how you call "curl" with a file? This will not execute the PHP code in the file.

1 Answers1