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POHow to display image from server (newest-oldest) to a new php page?
 

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  1. POHow to display image from server (newest-oldest) to a new php page?
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    <p>i have a form that create images (using imagejpeg, imagettftext and imagecreatefromjpeg) and save it on a folder in the server. What i need to do is to display ALL created images to another page, the newest on the top and so that the oldest is at the bottom. i have a form called formdisplay.php but it's just displaying a broken image and not newest to oldest. hope you can help me with this. really need to get this working. thanks in advance for your help.</p> <p>i have read the posts but none of those worked for me.</p> <p><a href="https://stackoverflow.com/questions/15056053/pull-dedicated-images-from-folder-show-image-filename-strip-part-of-filenam">Pull dedicated images from folder - show image + filename (strip part of filename + file-extension)</a></p> <p><a href="https://stackoverflow.com/questions/6557980/how-can-i-display-latest-uploaded-image-first-phpcss">How can I display latest uploaded image first? (PHP+CSS)</a></p> <p><a href="https://stackoverflow.com/questions/8359916/getting-images-from-the-server-and-display-them">getting images from the server and display them</a></p> <p><a href="https://stackoverflow.com/questions/12795385/displaying-images-from-folder-in-php">Displaying images from folder in php</a></p> <p><a href="https://stackoverflow.com/questions/14257682/display-image-from-server-embedding-php-in-html">display image from server embedding php in html</a></p> <p>formcreatesave.php</p> <pre><code> imagecopymerge($im, $img2, 10, 350, 0, 0, imagesx($img2), imagesy($img2), 100); $date_created = date("YmdHis");//get date created $img_name = "-img_entry.jpg"; //the file name of the generated image $img_newname = $date_created . $img_name; //datecreated+name $img_dir =dirname($_SERVER['SCRIPT_FILENAME']) ."/". $img_newname; //the location to save imagejpeg($im, $img_dir , 80); //function to save the image with the name and quality imagedestroy($im); </code></pre> <p>formdisplay.php</p> <pre><code>$dir = '/home3/site/public_html/Master/uploader/uploader'; $base_url = 'http://mysite.com/Master/uploader/uploader/20131027024705-img_entry.jpg'; $newest_mtime = 0; $show_file = 'BROKEN'; if ($handle = opendir($dir)) { while (false !== ($file = readdir($handle))) { if (($file != '.') &amp;&amp; ($file != '..')) { $mtime = filemtime("$dir/$file"); if ($mtime &gt; $newest_mtime) { $newest_mtime = $mtime; $show_file = "$base_url/$file"; } } } } print '&lt;img src="' .$show_file. '" alt="Image Title Here"&gt;'; </code></pre> <p>please feel free to edit my code. thanks :)</p>
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    1. COplace `print '<img src="' .$show_file. '" alt="Image Title Here">';` below `$show_file = "$base_url/$file";`
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      1. POHow to display image from server (newest-oldest) to a new php page?
      1. USel Dude
    2. COhi @E L i have tried your suggestion. but its still returning broken img. this time, 3 broken images. please take note that the image i was trying to display was created and saved using imagejpeg and imagecreatefromjpeg. hope you can give me further help. thanks
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      1. POHow to display image from server (newest-oldest) to a new php page?
      1. USJben Kaye
    3. COThat's because your `$base_url` has tailing filename. Make it `$base_url = 'http://birdieonawire.com/Amadeus/uploader/uploader/';` (is it really `uploader/uploader/` dirs there?)
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      1. POHow to display image from server (newest-oldest) to a new php page?
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