Play a random song with PHP array_rand

Question

I'm trying to have a page select a song at random from a dir of .mp3 files and play it. This is what I have so far http://pastebin.com/Jypx80RD I get this error:

Second argument has to be between 1 and the number of elements in the array

For this line: $random = array_rand($files, 2);

As the error message explains, your second argument of `array_rand` isn't valid, because it isn't within the legal range. Have you tried fixing this? If so, what is the problem you are encountering? — FThompson, Mar 23 '14 at 21:58
Possible duplicate of [Select random file using OPENDIR()](http://stackoverflow.com/questions/12119304/select-random-file-using-opendir) — SpYk3HH, Jul 14 '16 at 15:29
http://stackoverflow.com/questions/12119304/select-random-file-using-opendir/38361953#38361953 — SpYk3HH, Jul 14 '16 at 15:29

Lee Salminen · Answer 1 · 2014-03-23T22:13:02.460

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Try this:

$random = $files[rand(0, count($files) - 1)];

edited Mar 23 '14 at 22:13

answered Mar 23 '14 at 22:02

Lee Salminen

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Your method has a chance to produce errors. `rand` has an inlusive upper bound. It should be, `$random = $files[rand(0, (count($files)-1))];` in order to prevent the index from being out of bounds. – Spencer D Mar 23 '14 at 22:12
Not a problem. Glad to help. – Spencer D Mar 23 '14 at 22:13
@LeeSalminen When I changed it to this, it load the src as `http://domain.com/dir/Array` – Sharkie Mar 23 '14 at 22:24
Try var_dump($random) and append the corresponding array key to the end of the variable – Lee Salminen Mar 23 '14 at 22:35

score 0 · Answer 2 · answered Mar 23 '14 at 22:05

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This should do it:

$random = array_rand($files);

$files looks like it may be empty, from what I had to guess though.

answered Mar 23 '14 at 22:05

Mike Weir

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Play a random song with PHP array_rand

2 Answers2