Unexpected variable scope in PHP

Question

I would like to show 3 random images from database in my website. Below is its code:

$query = mysql_query ("SELECT id,url FROM tbl_gallery2"); if (mysql_num_rows($query) >= 3) { $my_array = array(); $last_array = array(); while ($r = mysql_fetch_row($query)) { $my_array[] = $r[1]; } function makernd () { $number = array_rand($my_array,1); if (in_array($number,$last_array)) { makernd(); } else { $last_array[] = $number; return $number; } } for($i = 1 ; $i < 3 ; $i++) { $item = makernd(); echo '<img src="./images/slider/'.$item.'.jpg" alt="" class="slider" />'; } }

But whenever I run this code, I get the error below:

Undefined variable: my_array in line ... // The first line of makernd() function.

But I expected $my_array to be an accessible array for this function.
What's the problem?

You either have lo pass the array to your function or define it as a global var. — jtheman
– jtheman, Commented Dec 1, 2012 at 14:28
Why are you conditionally defining the function inside the if? That doesn't add much other than confusion. — Michael Berkowski
– Michael Berkowski, Commented Dec 1, 2012 at 14:28

Steven Moseley · Accepted Answer · 2012-12-01 14:39:37Z

To simply fix your problem, you should pass $my_array to makernd() as a parameter:

$query = mysql_query ("SELECT id,url FROM tbl_gallery2"); if (mysql_num_rows($query) >= 3) { $my_array = array(); $last_array = array(); while ($r = mysql_fetch_row($query)) { $my_array[] = $r[1]; } function makernd ($my_array) { $number = array_rand($my_array,1); if (in_array($number,$last_array)) { makernd($my_array); } else { $last_array[] = $number; return $number; } } for($i = 1 ; $i < 3 ; $i++) { $item = makernd($my_array); echo '<img src="./images/slider/'.$item.'.jpg" alt="" class="slider" />'; } }

HOWEVER, I strongly suggest putting the randomization in MySQL, to

Simplify your code
Significantly improve the performance, and
Eliminate excessive loops & recursion in PHP

Example:

$sql = "SELECT id,url FROM tbl_gallery2 ORDER BY RAND() LIMIT 3"; $query = mysql_query ($sql); if (mysql_num_rows($query) >= 3) { while ($r = mysql_fetch_row($query)) { echo '<img src="./images/slider/' . $r[1] . '.jpg" alt="" class="slider" />'; } }

PS - I also suggest you update your code to use mysqli, as mysql is deprecated

PPS - I also suggest you look into mysqli_fetch_assoc so you can reference query results by name instead of index (e.g. $r['url'] instead of $r[1] - as if you ever change the order of your query, you will break references by index.

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