I am trying to access files on a different drive but the same server to where the script is running.
but I only get the error
I have tried the following combinations
The system is Windows 2003 server.
Has anybody encountered this?
but I only get the error
Code:
failed to open dir: Invalid argument in C:\apache\xampp\htdocs\path\to\script
Code:
if(opendir('S:/folder/'))
echo 'found';
if(opendir('S:/folder))
echo 'found';
if(opendir('S:\folder\\'))
echo 'found';
if(opendir('S:\folder))
echo 'found';
if(opendir('//10.1.1.11/S/folder/'))
echo 'found';
if(opendir('//10.1.1.11/S/folder))
echo 'found';
if(opendir('\\\10.1.1.11\S\folder\\'))
echo 'found';
if(opendir('\\\10.1.1.11\S\folder))
echo 'found';
Has anybody encountered this?
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