kestas Posted July 31, 2006 Share Posted July 31, 2006 Hi! Need help with imagejpeg() function. It should be outputting image to a file or a browser. Outputting to a file works fine, but when trying to output an image to the browser image is shown as a text(bunch of symbols with header in front). Is there anything that I don`t understand about it? Thanks in advance! Will include my code: <? session_start(); $maindir="C:\web\docs\\\thumbs"; $mydir = opendir($maindir); $exclude = array("index.php", "convert.php", "..", ".", "thumbs", "nuotr", "Thumbs.db") ; while($fn = readdir($mydir)) { if ($fn == $exclude[0] || $fn == $exclude[1]|| $fn == $exclude[2]|| $fn == $exclude[3]|| $fn == $exclude[4]|| $fn == $exclude[5]|| $fn == $exclude[6]) continue; $myimage="$fn"; echo"$fn<br>"; $photo = ImageCreateFromJPEG("C:\web\docs\\thumbs\\$fn"); ImageJPEG($photo); } closedir($mydir); ?> Quote Link to comment https://forums.phpfreaks.com/topic/16076-output-of-imagejpeg-function/ Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.