Use opendir(), readdir(), and closedir() -
http://www.php.net/manual/en/function.opendir.php
HTH,
Jason Soza
----- Original Message -----
From: "Ashley M. Kirchner" <[EMAIL PROTECTED]>
Date: Monday, May 13, 2002 9:50 am
Subject: [PHP] Reading dir contents...
>
> Take this function:
>
> function randomImg() {
> $numargs = func_num_args();
> $ImgArray = array($numargs-1);
> $arg_list = func_get_args();
> $ImgDir = $arg_list[0];
> srand((double)microtime()*1234567);
> $img = $arg_list[rand(1,count($arg_list)-1)];
> while(!$img) {
> $img = $ImgArray[rand(1,count($ImgArray))];
> }
> return "$ImgDir/$img";
> }
>
>
> ...it assumes two things: $arg_list[0] is a variable (which
> is always
> passed) that ends up being $ImgDir, and then there's $arg_list[1-
> n] where 'n'
> is >1 . Right now, I'm calling this function like so:
>
> <? echo randomImg("home", "bg01.gif", "bg02.gif", "bg03.gif"); ?>
>
> However, I'd like to change it so that I don't have to specify
> the image
> names ($arg_list[1..n]), but just the first variable. I'd like
> for it to grab
> whatever it finds in that folder (passed as the variable), and
> use. This way I
> can have any number of images in that folder without having to
> edit my pages
> every time to add or remove more. And what if the image name
> changes, once
> again I have to manually fix the function call in my pages.
>
> Can PHP do this; open that directory, read its contents and
> use it (if so,
> how?) Or do I have to go to Perl for this?
>
> --
> W | I haven't lost my mind; it's backed up on tape somewhere.
> +----------------------------------------------------------------
> ----
> Ashley M. Kirchner <mailto:[EMAIL PROTECTED]> . 303.442.6410
> x130 IT Director / SysAdmin / WebSmith .
> 800.441.3873 x130
> Photo Craft Laboratories, Inc. . 3550 Arapahoe
> Ave. #6
> http://www.pcraft.com ..... . . . Boulder, CO 80303,
> U.S.A.
>
>
>
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