hammerslane
11-17-2003, 10:46 AM
I know this is possible, but was just wondering what the right function was.
I'm basically running a variable called $CLIENT from 1.html to 2.html.
1.html is just a drop down box and whatever client you select (for example BARCLAYS BANK), the variable would be BAR.
2.php takes the variable, and looks in a directory called /FOLDERS for a folder named BAR (in this case) and then displays whatever files are in the folder.
The problem is, that if there is a client name and a variable added to the drop down box, there will be a Warning: opendir(\localhost\clients) [function.opendir]:
failed to open dir: Invalid argument error.
Is it possible to run an IF statement to check whether the folder exists, and if it doesn't, create a folder of the variable name in /FOLDER ?
I hope I've explained this properly... any pointers in the general right direction are accepted gratefully.
I'm basically running a variable called $CLIENT from 1.html to 2.html.
1.html is just a drop down box and whatever client you select (for example BARCLAYS BANK), the variable would be BAR.
2.php takes the variable, and looks in a directory called /FOLDERS for a folder named BAR (in this case) and then displays whatever files are in the folder.
The problem is, that if there is a client name and a variable added to the drop down box, there will be a Warning: opendir(\localhost\clients) [function.opendir]:
failed to open dir: Invalid argument error.
Is it possible to run an IF statement to check whether the folder exists, and if it doesn't, create a folder of the variable name in /FOLDER ?
I hope I've explained this properly... any pointers in the general right direction are accepted gratefully.