how to show No of images in directory?
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how to show No of images in directory?
I need to use php codes to show how many images r dr in particular directory.. so please help me on dat.. ASAP 
Re: how to show No of images in directory?
You can count the files in a specific directory like that:
Code: Select all
function countFiles($directory){
$glob = glob($directory . "*.*");
if ($glob != false) {
return count($glob);
}
return 0;
}
Re: how to show No of images in directory?
In other news, ~genix2011's code will work, but should be tweaked:[url=http://forums.devnetwork.net/viewtopic.php?t=30037]Forum Rules[/url] Section 1.1 wrote:11. Please use proper, complete spelling when posting in the forums. AOL Speak, leet speak and other abbreviated wording can confuse those that are trying to help you (or those that you are trying to help). Please keep in mind that there are many people from many countries that use our forums to read, post and learn. They do not always speak English as well as some of us, nor do they know these aberrant abbreviations. Therefore, use as few abbreviations as possible, especially when using such simple words.
Some examples of what not to do are ne1, any1 (anyone); u (you); ur (your or you're); 2 (to too); prolly (probably); afaik (as far as I know); etc.
1) glob() can return an empty array, a populated array, or boolean FALSE on error. Generally I'd have the empty array and the error condition return different values, so you can tell if the directory is actually empty, or if a problem happened.
2) The function as it stands will return all files/sub-directories in a directory. If you only want images, you need to pass the parameters to glob a bit differently
Code: Select all
function countFiles($directory){
$glob = glob($directory . "*.{jpg,gif,png,etc}",GLOB_BRACE);
if ($glob !== false) {
return count($glob);
}
return false;
}Real programmers don't comment their code. If it was hard to write, it should be hard to understand.