Re: Show Filename using Wildcards

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On 3/29/07, Rahul Sitaram Johari <sleepwalker@xxxxxxxxxxxxxxxx> wrote:
Ave,

This Works!!

   <?php
   exec("find /Users/rjohari/Documents/XFER/espi -type f -name
".$row['PHONE']."*.vox", $files);
   foreach ($files as $value) {
       echo substr($value,35)."<br>";
   }
   ?>

THANKS!

I recommend using basename($value) instead of substr($value,35)
If the directory changes, the basename would still return valid
values, while substr wouldn't

Tijnema

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Rahul Sitaram Johari
CEO, Twenty Four Seventy Nine Inc.

W: http://www.rahulsjohari.com
E: sleepwalker@xxxxxxxxxxxxxxxx

³I morti non sono piu soli ... The dead are no longer lonely²



On 3/29/07 4:31 PM, "Peter Lauri" <lists@xxxxxxxxxxx> wrote:


> [Peter Lauri - DWS Asia]
>
> Hi,
>
> Assuming you are on a linux you could try:
>
> exec("find /the/path/to/the/place/where/you/should/start/searching -type f
> -name 515515515*.ext", $files);
>
> Then the $files will be an array with the found files matching the search.
>
> Best regards,
> Peter Lauri
>
> www.dwsasia.com - company web site
> www.lauri.se - personal web site
> www.carbonfree.org.uk - become Carbon Free
>
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