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PHP - Displaying Images From A Directory
ok I got the images to upload to the directory on the server. now I got some code to try to display the image on the same page after you submit it to the directory. so it's like, submit form-> images goes in server directory->echoes back out on same page but just above the upload form. please any help greatly appreciated.
thanks. here is the code so far. Code: [Select] <?php // An Example about listing Images. $dir = "uploads/"; $odir = opendir($dir); while($file = readdir($odir)){ if(filetype($file) == "image/JPEG") { echo $file; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <p>UPLOAD YOUR PORTFOLIO AND PRICES</p> <p> </p> <p> <form enctype="multipart/form-data" action="uploader.php" method="POST"> <input type="hidden" name="MAX_FILE_SIZE" value="100000" /> Choose a file to upload: <input name="uploadedfile" type="file" /><br /> <input type="submit" value="Upload File" /> </form></p> </body> </html> Similar TutorialsI am trying to display images on a html page alphabetically in descending order by their filename. <?php $dir = "./plate_cache"; $handle = opendir($dir); while (($file = readdir($handle))!==false) { if (strpos($file, '.png',1)) { echo "<img id=\"plates\" src=\"/tags/plate_cache/$file\"></a><br />;"; } } closedir($handle); } ?> Any help is greatly appreciated! Hi, I have a list of images from a directory that are in order by their "id". How could I display one at a time on a page--starting from the beginning, and have it so the user can click on the image and go to the next one in line. Hi, I am trying to echo out images on my server's directory "uploads". The images are blank and not showing up. Please help if anyone can. Thank you. here is the page as it is. http://ealike.com/simple_uploader/uploader.php and here is the code. Code: [Select] <?php // Where the file is going to be placed $target_path = "uploads/"; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } //echo out all images $dir = "uploads/"; if (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { echo "<img src='$file' alt='$file' height='200' width='200' /> " ; } closedir($dh); } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> </body> </html> How can i display all the images in a certain directory and echo the image names. So if a new image is uploaded, it will still display without edited the gallery.php file. Cheers. So, We are trying to create a animation using the gifs in a directory by using the first 10 images. We just cannot figure out how to do the animation.. needs to have a 3 second delay between each frame Heres the code at the moment. Which creates the starting images and also sorts the images in the directory by time.. <?php $time = date("YmdHis"); $handle_data = file_get_contents('http://www.bungie.net/Stats/Halo3/Nightmap.ashx'); $img = new Imagick(); $img->readImageBlob($handle_data); $img->writeImage('nightmap/'.$time.'.gif'); foreach (glob('nightmap/*.gif') as $f) { # store the image name $list[] = $f; } sort($list);# sort is oldest to newest, $gif_0 = array_pop($list); $gif_1 = array_pop($list); $gif_2 = array_pop($list); $gif_3 = array_pop($list); $gif_4 = array_pop($list); $gif_5 = array_pop($list); $gif_6 = array_pop($list); $gif_7 = array_pop($list); $gif_8 = array_pop($list); $gif_9 = array_pop($list); ?> Hi, i want to make an image gallery, this is the first step. uploading files. below is the code I am using , but don't i have to include info about the server? i dont know what to include since it is the server I am working with and not a database, any help greatly appreciated. thanks Code: [Select] <?php // Where the file is going to be placed $target_path = "uploads/"; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> </body> </html> I currently have a file upload form that uploads images to a folder on my server called "attachments". This is the directory path: www.mysite.com/docs/attachments/ However, I want the images to upload to www.mysite.com/data/attachments/ instead. I have included the relevant part of the PHP below. I have tried changing this: $cur_dir = $cur_dir."/".$dir; to this: $cur_dir = $cur_dir."www.mysite.com/data/attachments/".$dir; and changing this: $dirs = explode("/","attachments//"); to this: $dirs = explode("www.mysite.com/data/","attachments//"); and also changing this: $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; to this: $_values[$file[0]."_real-name"] = "www.mysite.com/data/attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; But nothing has worked so far. Any suggestions would be appreciated. foreach($files[$form] as $file){ $str = $file[1]; if (eval("if($str){return true;}")) { $_values[$file[0]] = $_FILES[$file[0]]["name"]; $dirs = explode("/","attachments//"); $cur_dir ="."; foreach($dirs as $dir){ $cur_dir = $cur_dir."/".$dir; if (!@opendir($cur_dir)) { mkdir($cur_dir, 0777);}} $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; copy($_FILES[$file[0]]["tmp_name"],$_values[$file[0]."_real-name"]); @unlink($_FILES[$file[0]]["tmp_name"]); }else{ $flag=true; if ($_isdisplay) { //$ExtFltr = $file[2]; //$FileSize = $file[4]; if (!eval("if($file[2]){return true;}")){echo $file[3];} if (!eval("if($file[4]){return true;}")){echo $file[5];} $_ErrorList[] = $file[0]; } } } I am trying to display images on a html page alphabetically in descending order by their filename. This script inserts all the images in a directory on the page. Just can't figure out how to maintain this with a sorting option <?php $dir = "./plate_cache"; $handle = opendir($dir); while (($file = readdir($handle))!==false) { if (strpos($file, '.png',1)) { echo "<img id=\"plates\" src=\"/tags/plate_cache/$file\"></a><br />;"; } } closedir($handle); } ?> I have a PHP script that uploads images to a folder on my server (attachments folder). Currently the folder sits within my webroot and is publicly accessible (I have to use chmod 777 due to permissions issue). So, I created the "attachments" folder outside of my webroot (so that it is not publicly accessible), but I do not know how to set the path in the PHP code to upload it to that "attachments" folder outside of the webroot. As you see in the snippet of PHP code below, the code currently uploads the the "attachments" folder within the www (webroot) directory. How do I make it upload to the "attachments" folder OUTSIDE of the www (webroot) directory? foreach($files[$form] as $file){ $str = $file[1]; if (eval("if($str){return true;}")) { $_values[$file[0]] = $_FILES[$file[0]]["name"]; $dirs = explode("/","attachments//"); $cur_dir ="."; foreach($dirs as $dir){ $cur_dir = $cur_dir."/".$dir; if (!@opendir($cur_dir)) { mkdir($cur_dir, 0777);}} $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; copy($_FILES[$file[0]]["tmp_name"],$_values[$file[0]."_real-name"]); @unlink($_FILES[$file[0]]["tmp_name"]); }else{ $flag=true; if ($_isdisplay) { //$ExtFltr = $file[2]; //$FileSize = $file[4]; if (!eval("if($file[2]){return true;}")){echo $file[3];} if (!eval("if($file[4]){return true;}")){echo $file[5];} $_ErrorList[] = $file[0]; } } } Thanks Hi I'm trying to setup Pagination at the bottom of the page after showing 10 images in the following script but have no clue where to start and how to implement it into the script. My images generator comprises
an array of image names extracted form an images table from a database using a select statement
a random number generator,
and a string that builds the correct pathname for the selected file.
To select one of the images for display, i generate a random number between 1 and the length of the array.
Though the generator is working, I noticed one of the random numbers is throwing up this error:
Warning: getimagesize(images/): failed to open stream: No such file or directory in
An inspection of the array reveals 2 array elements (representing my number of images) but one array element is NULL ( the first entry in the banner table
image_generator.php
require_once('connection.inc.php');
$sql = 'SELECT `filename` FROM banner';
$result = $mysqli->query($sql, MYSQLI_STORE_RESULT) or die(mysqli_error());
$row = $result->fetch_array(MYSQLI_ASSOC);//an array of image names
$count = $result->num_rows;
for ($i = 1; $i <= $count; ++$i)
{
$row[$i] = $result->fetch_array(MYSQLI_ASSOC);
}
$i = rand(1, $count); //a random number generator,
//The random number is used in the final line to build the correct pathname for the selected file.
$selectedImage = "images/{$row[$i]['filename']}";
if (file_exists($selectedImage) && is_readable($selectedImage))
{
$imageSize = getimagesize($selectedImage);
}
var_dump($row)
array (size=1)
'filename' => string 'ginsomin2.jpg' (length=13)
null
RANDON IMAGE DISPLAY
require_once 'image_generator.php';
<div id="banner" class="wrapper clearfix">
<img src="<?php echo $selectedImage; ?>" alt="banner">
</div>
Kindly advice how i may proceed from here?
Thanks.
I had this page working fine where it retrieves a set of images from the database depending on what thumbnail was click on the previous page. Now I am trying to add an edit and delete button to each of the pictures and it will not display any of it now! This is the code I am using: Code: [Select] <?php $id = $_GET[id]; ?> <?php $max_items = 16; /* max number of news items to show */ require_once('config.php'); $db = mysql_connect (DB_HOST,DB_USER,DB_PASSWORD); if(!$db) { die('Failed to connect to server: ' . mysql_error()); } mysql_select_db (DB_DATABASE,$db); function displayNews($all = 0) { global $db, $max_items; if ($all2 == 0) { /* query for news up to $max_items */ $query13 = "SELECT image,id,description,projectcode,thumbnail " . "FROM project WHERE projectcode='$id' ORDER BY id DESC LIMIT $max_items"; } else { /* query for all news */ $query13 = "SELECT image,id,description,projectcode,thumbnail " . "FROM project WHERE projectcode='$id' ORDER BY id DESC"; } $result13 = mysql_query($query13) or print("<p>Error fetching entries from the database, error: " . "Statement: " . $query13 . "</p>" . mysql_error()); while ($row13 = mysql_fetch_array($result13)) { $title = htmlentities ($row13['description']); $news = nl2br (strip_tags ($row13['projectcode'], '<a><b><i><u>')); $image = $row13['image']; $thumbnail = $row13['thumbnail']; $id = $row13['id']; /* display the data */ echo "<div class='portfolioproject'>";?> <div class='boxgrid captionfull'> <a id="<?php echo $id ?>" class="example_group" href="images/gallery/<?php echo $image ?>" title="<?php echo $title ?>" rel="gallery"><img alt="" src="images/gallery/<?php echo $thumbnail ?>" width="122px" height="122px"</a> </a>} <?php //Check whether the session variable SESS_MEMBER_ID is present or not if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) { echo "<form class='editbtn' action='editportfolio.php' method='POST'>"; echo "<input type='hidden' name='idf' value='$id' />"; echo "<input src='images/editbtn.png' type='image' value='Edit' />"; echo "</form>"; echo "<form class='editbtn' action='deleteportfolio.php' method='POST'>"; echo "<input type='hidden' name='ide' value='$id' />"; echo "<input src='images/delbtn.png' type='image' value='Delete' />"; echo "</form>"; } else { echo ""; } echo "</div>"; } /* if not displaying all news, give a link to do so */ if ($all == 0) { echo "<div class='show_all'><a class='show' href=\"{$_SERVER['PHP_SELF']}" . "?action=all\">Show all</a></div>\n"; } } /* this is where the script decides what do do */ echo "\n"; switch($_GET['action']) { case 'all': displayNews(1); break; default: displayNews(); } echo "\n"; ?> <?php //Check whether the session variable SESS_MEMBER_ID is present or not if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) { echo "<span class='show_all'>"; echo "<a class='show' href='admincp.php'> Admin</a>"; echo "</span>"; echo "<span class='show_all'>"; echo "<a class='show' href='logout.php'> Logout</a>"; echo "</span>"; } else { echo ""; } ?> No error messages are displayed, only the 'show all' button, and the 'admin' and 'logout' buttons if signed in! Any help on this would be much appreciated! Thank you Martin So I am trying to display an image from a database (I know it's not the best idea but it's what I was told to do). Anyway I'm having problems displaying it. Here is the code for displaying the image. $cat = $_GET['cat']; include 'includes/openDbConn.php'; $sql = 'SELECT * FROM CushionsCategories WHERE CushionCategory = "'.$cat.'"'; echo $sql; $result = mysql_query($sql); echo '<table>'; while($val = mysql_fetch_assoc($result)){ $cush = 'SELECT * FROM Cushion WHERE SKU = "'.$val['CushionSKU'].'"'; echo $cush; $cushres = mysql_query($cush); $cushval = mysql_fetch_assoc; $img = 'SELECT * FROM Images WHERE SKU = "'.$val['CushionSKU'].'"'; echo $img; $imgres = mysql_query($img); $imgval = mysql_fetch_assoc($imgres); if( $i % 3 == 0 ) { echo '</tr><tr>'; } echo '<td><img src="image.php?sku='.$val['CushionSKU'].'" name="'.$imgval['FileName'].'" description="'.$imgval['Description'].'" /></td>'; } echo '</tr></table>'; and the page that gets the image: $sql = 'SELECT Image FROM Images WHERE SKU = "'.$sku.'"'; //echo $sql; $result = mysql_query($sql) or die("Invalid query: " . mysql_error()); // set the header for the image header("Content-type: image/jpeg"); echo mysql_result($result, 0); Thanks for any help in advance. Hello I am using this script right here http://www.nearby.org.uk/sphinx/search-example5-withcomments.phps and am trying to show images from my db but I can only get it to show the first image. Does anyone know how to make it show all images if images even exist for that particular search result? I changed the query to $CONF['mysql_query'] = ' SELECT l.link_id AS id, l.title AS title, l.fulltxt AS body, l.url AS url, m.media_id AS im_id, m.title AS im_title, m.thumb_link AS im_t_link FROM search1_links AS l LEFT JOIN search1_media AS m ON (m.link_id = l.link_id) WHERE l.link_id IN ($ids) '; and added if ($row['im_id']) { echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> '; } right here //Actully display the Results print "<ol class=\"results\" start=\"".($currentOffset+1)."\">"; foreach ($ids as $c => $id) { $row = $rows[$id]; $link = htmlentities(str_replace('$id',$row['id'],$CONF['link_format'])); print "<li><a href=\"$link\">".htmlentities($row['title'])."</a><br/>"; if ($CONF['body'] == 'excerpt' && !empty($reply[$c])) print ($reply[$c])."</li>"; else if ($row['im_id']) { echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> '; } print htmlentities($row['body'])."</li>"; } print "</ol>"; if ($numberOfPages > 1) { print "<p class='pages'>Page $currentPage of $numberOfPages. "; printf("Result %d..%d of %d. ",($currentOffset)+1,min(($currentOffset)+$CONF['page_size'],$resultCount),$resultCount); print pagesString($currentPage,$numberOfPages)."</p>"; } Can anyone help me to be able to display all images if they exist? Thanks. Hey guys, I wrote a code for displaying images in columns, but for some reason it's not working. Can you take a look and tell me what I'm doing wrong? $i = 0; if($i % 4 == 0){ echo ' <td width="25%" align="center" valign="top"> <a href= "'.$slika_velika.''.$slika_v.'" rel="lytebox" title="'.$naziv.' - '.$mjesto.' - '.$medalja.'" target="_blank"><img src="'.$slika_mala.''.$slika_m.'"width="200" height="150" border="0"></a><br> </td> <td width="25%" align="center" valign="top"> '.$naziv.' - '.$mjesto.' </td> <td width="25%" align="center" valign="top"> '.$medalja.' </td><br /> '; echo '</tr><tr>'; } $i++; } The output I'm getting now is one image below another. I would like to display them one next to another. Please help!!! I am trying to display two images per row. Will add in table tags later so it will look nicer. Anyway, it is leaving off the second image. So, I have 6 images and image number two is blank.
$dirname = "Files/$listingid/images/";
$images = glob($dirname."*");
/*
$html .= '<table width="100%">';
foreach($images as $image){
$html .= "<tr><td><center>";
$html .="<img src=\"$image\" width=\"300\">";
$html .= "</a></center></td>";
$html .= "</td></tr>";
}
$html .= "</table>";
*/
$countImages = count($images) ;
$imagesPerRow = 2;
for ($i = 0 ; $i < $countImages; $i++) {
//display image here
$image = $images[$i] ;
$html .= "<img width = \"200\" src='$image'>" ;
if ($i % $imagesPerRow == 0) {
//have displayed an entire row
$html .= '<br>' ;
}
}
I'm having real problems with a rss box that I'm trying to incorporate onto the homepage of our website... The code I'm using as a basis I got from here - http://www.dynamicdrive.com/dynamicindex18/rssdisplaybox/index.htm It worked okay displaying title, date and content, however, we wanted it to display the first image of the blog entry as well. In the outputbody.php (the file that dictates what is shown in the rss box) I incorporated this code (shown in bold EDIT: with the bold tags around it! ) - Code: [Select] <?php //Function for ouputting the body of each RSS item displayed (inside loop)- DynamicDrive.com //For syntax pf bpdu, see: http://simplepie.org/docs/installation/from-scratch/ and http://simplepie.org/docs/reference/ //Function by default defines 3 different body outputs (templates). Modify or add additional templates as desired [b] function get_image($text) { $imageurl=""; preg_match('/<\s*img [^\>]*src\s*=\s*[\""\']?([^\""\'>]*)/i' , $text, $matches); $imageurl=$matches[1]; if (!$imageurl) { preg_match("/([a-zA-Z0-9\-\_]+\.|)youtube\.com\/watch(\?v\=|\/v\/)([a-zA-Z0-9\-\_]{11})([^<\s]*)/", $text, $matches2); $youtubeurl = $matches2[0]; if ($youtubeurl) $imageurl = "http://i.ytimg.com/vi/{$matches2[3]}/1.jpg"; } return $imageurl; }[/b] function shortencontent($content, $len=40){ if(strlen($content) > $len){ return substr($content, 0, $len - 6) . ' . . .'; } else { return $content; } } function outputbody($item, $template=""){ if ($template=="" || $template=="default"){ //DEFAULT TEMPLATE ?> <div class="rsscontainer"> <div class="rsstitle"><a href="<?php echo $item->get_permalink(); ?>"><?php echo $item->get_title(); ?></a></div> <div class="rssdate"><?php echo $item->get_date('d M Y g:i a'); ?></div> <div class="rssdescription"><?php echo shortencontent($item->get_description(), 400); ?></div> [b]<div class="rssimage"><?php echo get_image(); ?></div>[/b] </div> <?php } //end default template else if ($template=="titles"){ //"TITLES" TEMPLATE ?> <div class="rsscontainer"> <div class="rsstitle"><a href="<?php echo $item->get_permalink(); ?>" target="_new"><?php echo $item->get_title(); ?></a></div> <div>Category: <?php echo $item->get_category(); ?></div> </div> <?php } //end titles template else if ($template=="titlesdates"){ //"TITLESDATES" TEMPLATE ?> <div class="rsscontainer"> <span class="rsstitle"><a href="<?php echo $item->get_permalink(); ?>"><?php echo $item->get_title(); ?></a></span> <span class="rssdate"><?php echo $item->get_date('m/d/y g:i a'); ?></span> </div> <?php } //end titlesdates template else if ($template=="mytemplatename"){ //"mytemplatename" TEMPLATE ?> //DEFINE ADDITIONAL CUSTOM TEMPLATE(s) USING SAME LOGIC STRUCTURE AS ABOVE //For syntax of template body, see SimplePie docs: http://simplepie.org/docs/installation/from-scratch/ and http://simplepie.org/docs/reference/ <?php } else die ("No template exists with such name!"); } //Closing function bracket ?> While it creates an imagebox nothing is shown and when you open the empty image box in the browser it displays the following URL - http://www.oururl.%20.%20.%20.</div>%20%20%20%20%20%20%20%20<div%20class= I have little PHP knowledge but don't really know where to start with this... please can anyone help or point me in the right direction. I appreciate that it is frustrating dealing with someone with limited knowledge but please have a little patience with me... Cheers people! Hi All take a look at http://www.buypromo.co.uk/index.php in IE 7 or 8 the images at the bottom of the page are not showing up, whereas therest are find, any ideas? thanks Guys, i need your help,i have one table employee with empno=number,image=blob, images uploaded sucessfully and insert into database but when i am trying to retrieve records image are not displaying instead of it some encrypted form shows please help me Hi! I want the image of my site to be clickable so that it shows the next image contained in the result set of the sql query. Code: [Select] <?php require_once('includes/connection.inc.php'); $conn = dbConnect(); $sql = 'SELECT filename FROM images ORDER BY image_id DESC LIMIT 1'; $msg = ''; if(!$sql) { echo "Something went wrong with the query. Please try again."; } $result = $conn->query($sql) or die(mysqli_error()); if(mysqli_num_rows($result) == 0) { header('Location: empty_blog.php'); $conn->close(); } else { $row = $result->fetch_row(); $image = $row[0]; } ?> <?php include('header.php'); ?> <div class="content main"> <img id="image" src="images/<?php echo $image; ?>"/> </div> <?php include('includes/footer.inc.php'); ?> All the filenames of the images are contained correctly in the $result variable, all I need to do now is to fetch the next image in the set. How would I go about this? Any help is greatly appreciated! |
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