output of imagejpeg() function
Posted: Mon Jul 31, 2006 12:22 am
Everah | Please use
Everah | Please use
Code: Select all
,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]
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:Code: Select all
<? 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);
?>Everah | Please use
Code: Select all
,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]