Thanks for your reaction,
You are right about your conclusion how I think the code should work. I am
indeed trying to run the script off one server (my_domain1) and then
attempting to "feed" the function a directory structure off another server
(my_domain2).
I suppose the function opendir() only
The function is correct, BUT, it looks like you are trying to run this
script off one server and then attempting to "feed" the function a
directory structure off another server.
Your code will work "as-is" if you run the script off my_domain2, but it
won't work if you run it off my_domain asking f
Hello,
For my website I use some PHP code for navigation. Therefore I use a
directory structure which contains some navigation files the visitor can
open. The directory structure looks like this:
my_domain
|
- /navigation
| - file_1.html
| - file_2.ht
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