Boerke Posted March 2, 2006 Share Posted March 2, 2006 i have a form that let's the user give in an url with pictures, so what i want to do is read that url and display the pictures that were found, but is there a way to do this? because with opendir/readdir i get errors, and also when i change my htaccess file with the url_open stuff, it also doesn't work Quote Link to comment Share on other sites More sharing options...
Boerke Posted March 2, 2006 Author Share Posted March 2, 2006 anyone that can send me in the good direction? Quote Link to comment Share on other sites More sharing options...
kenrbnsn Posted March 2, 2006 Share Posted March 2, 2006 Please post your code. Quote Link to comment Share on other sites More sharing options...
Boerke Posted March 2, 2006 Author Share Posted March 2, 2006 oki read the url folder( for example: [a href=\"http://www.site.com/fotodirectory/\" target=\"_blank\"]http://www.site.com/fotodirectory/[/a] ):$directory = $_POST['fotodir'];then i'm using a script i found on the internet to test it[code]$imgdir = "$directory"; $allowed_types = array('png','jpg','jpeg','gif'); $dimg = opendir($imgdir); while($imgfile = readdir($dimg)) { if(in_array(strtolower(substr($imgfile,-3)),$allowed_types)) { $a_img[] = $imgfile; sort($a_img); reset ($a_img); } } $totimg = count($a_img); // total image number for($x=0; $x < $totimg; $x++) { $size = getimagesize($imgdir.'/'.$a_img[$x]); // do whatever $halfwidth = ceil($size[0]/2); $halfheight = ceil($size[1]/2); echo 'name: '.$a_img[$x].' width: '.$size[0].' height: '.$size[1].'<br />'; }[/code]but the error i getfailed to open dir: not implemented in ....php on line ...andreaddir() not valid argumentso i know that i am supposed to give a local folder, and that's why i asked if there is a method that opens the url instead of just a dir Quote Link to comment Share on other sites More sharing options...
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