PHP - Go Through Images In A Folder.
Hi, I want to be able to click on the photo and go to the next one in a folder. I have this code already, I just am not quite sure how to finish it.
-George Code: [Select] <?php $count = $_GET['count']; $dir = "images"; $names = array(); $handle = opendir($dir); while ($name = readdir($handle)){ if(is_dir("$dir/$name")) { if($name != '.' && $name != '..') { echo "directory: $name\n"; } } elseif ($name != '.DS_Store' ) { $names[] = $name; } } closedir($handle); $numberofitems = count($names); $numberofitems--; if ($count <= $numberofitems){ echo "<p>"; echo "<img src='images/".$names[$count]."'>"; } else {echo "end";} ?> Similar TutorialsI have this script from http://lampload.com/...,view.download/ (I am not using a database) I can upload images fine, I can view files, but I want to delete them. When I press the delete button, nothing happens
http://www.jayg.co.u...oad_gallery.php
<form>
<?php $dir = dirname(__FILENAME__)."/images/gallery" ; $files1 = scandir($dir); foreach($files1 as $file){ if(strlen($file) >=3){ $foil = strstr($file, 'jpg'); // As of PHP 5.3.0 $foil = $file; $pos = strpos($file, 'css'); if ($foil==true){ echo '<input type="checkbox" name="filenames[]" value="'.$foil.'" />'; echo "<img width='130' height='38' src='images/gallery/$file' /><br/>"; // for live host //echo "<img width='130' height='38' src='/ABOOK/SORTING/gallery-dynamic/images/gallery/ $file' /><br/>"; } } }?> <input type="submit" name="mysubmit2" value="Delete"> </form>
any ideas please?
thanks
I made an upload image system, the images are stored in a folder, while the image name is stored in database. When I try to execute the image name, it works successfully, but when I try to disply the image from the folder by using the image name in the database, I fail each time. The folder where the images are being stored is named: saveimage Here is the query: Code: [Select] <?php mysql_connect('localhost', 'root', '') or die ('Connection Failed'); mysql_select_db('imagedatabase'); $images = mysql_query("SELECT * FROM img WHERE email='$lemail'"); while($row = mysql_fetch_array($images)) { echo "<img src='saveimage/'".$row['img_description']; } ?> Can someone tell me how I can remove or delete an image from a folder on a server using PHP? I tried this: Code: [Select] unlink("http://midwestcreativeconsulting.com/jhrevell/wp-content/themes/twentyten_3/upload/" . $location); before my delete MySQL statement, but I keep getting this error: Warning: unlink() [function.unlink]: http does not allow unlinking in /home/midwestc/public_html/jhrevell/wp-content/themes/twentyten_3/removejewelry.php on line 22 Can anyone help and tell me how I can make it work? I'm trying to write code that will let me pull 10 out of 15 images out of a folder and display them on my site. The images are all different, and I don't want dupes to show. So far, I have the following code figured out: --------------- $s = array ("image.jpg", "image2.jpg"); // as many images as you want $n = rand(1,len($s)); // randomly pick a number between 1 and the length of the array echo "<img src='". $s[$n] .'">"; // create an image tag for the randomly selected imagine (value of the randomly defined key) array_pop($s, $n); // This piece isn't right, it needs to EXTRACT and delete the $n array element. // Next random image $n = rand(1,len($s)); echo "<img src='". $s[$n] .'">"; --------------- Any ideas what array_pop should be to work properly? Thank you for the help! The problem: I'm trying to create a page which outputs images from a folder which I have been able to do, but the problem I'm having is not being able to get the page to display the most recent image according to file modification date at the top. The first set of code below outputs the image timestamps in descending order, from newest to oldest which is what I want, but as soon as I change/add a couple lines of code (Shown in the second lot of code) to get the image file name along with the timestamp, the echoed list (timestamp and file names) gets muddled up in a random order. In short; As soon as the file names are retrieved with the timestamp, the list goes from being organised descendingly, to not. Show timestamp only code (1st lot of code): <?php //Open images directory $ignore = array("..","."); $dir = opendir("images1"); $images = array(); $sortedimages = array(); //List files in images directory while (($file = readdir($dir)) !== false) if (!in_array($file, $ignore)) $images[] = $file; foreach ($images as $image) { $filetime = filemtime("images1/$image"); $sortedimages[] = $filetime; } rsort($sortedimages); foreach ($sortedimages as $sorted) { //foreach ($sorted as $key => $value) //{ echo "$sorted<br/>"; } //} closedir($dir); ?> The show timestamp and file name code (2nd lot of code): <?php //Open images directory $ignore = array("..","."); $dir = opendir("images1"); $images = array(); $sortedimages = array(); //List files in images directory while (($file = readdir($dir)) !== false) if (!in_array($file, $ignore)) $images[] = $file; foreach ($images as $image) { $filetime = filemtime("images1/$image"); $sortedimages[] = array($filetime => $image); } rsort($sortedimages); foreach ($sortedimages as $sorted) { foreach ($sorted as $key => $value) { echo "$key and $value<br/>"; } } closedir($dir); ?> The changes made in the 2nd script from the 1st: //Changed: $sortedimages[] = $filetime; ---> $sortedimages[] = array($filetime => $image); //Included the previously commented out: foreach ($sorted as $key => $value) { } //Changed: echo "$sorted<br/>"; ---> echo "$key and $value<br/>"; Thanks for any help! I have searched around and found this code that suggests that I should be able to read an image file and echo it directly in to the page without hyperlinking to the file that is outside the public folder, but i get the error message that it is not there even though the file is. Warning: getimagesize() [function.getimagesize]: Unable to access "/home/****/upload/AAABBBCCC.JPG" in /home/****/public_html/client.php on line 40 Code: [Select] $image = "AAABBBCCC.JPG"; $path= "/home/****/upload/"; $details = getimagesize($path . $image); header ('Content-Type: ' . $details['mime']); echo file_get_contents($path . $image); Im using this code to call all the images in a folder: Code: [Select] $handle = opendir(dirname(realpath(__FILE__)).'/images/'); while($file = readdir($handle)){ if($file !== '.' && $file !== '..'){ echo '<img src="admin/img/uploads/'.$file.'" border="0" />'; } } My html says the images are present but they aren't visable on screen: Code: [Select] <div id="contentbody"> <img src="admin/img/uploads/send-button-sprite copy.png" border="0" /> <img src="admin/img/uploads/test" border="0" /> <img src="admin/img/uploads/counter.jpg" border="0" /> <img src="admin/img/uploads/send-button-sprite.png" border="0" /> </div> Any help is much appreciated! Hi
I echo out images from a folder. The problem is that I have a html file in the same folder but I don't want to echo that out. How can I write my code to get rid of the html in the output?
Here my code:
<?php $filetype = '*.html'; $dirname = substr('$filetype'); $i=0; if (isset($_POST['submit2'])) { //get the dir from POST $selected_dir = $_POST['myDirs']; //now get the files from within the selected dir and echo them to the screen foreach(glob($selected_dir . DIRECTORY_SEPARATOR . '*') as $dirname) { echo substr($dirname, 0, -4); echo '<img src="'.$dirname.'" />'; echo "<label><div class=\"radiobt\"><input type='radio' name='radio1' value='$i'/></div></label>"; } } ?>P.S. when I echo out : substr($dirname, 0, -4); I want to get ride of the html filename there too. How can I do so something like this? Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this? thanks Hey guys, i'm new to this site and would need some help with coding. So i'm making a car part website which should has brand/model search. It's a dropdown search which will get the brand / model from database and should display all the parts for that exact model, from folder. Database structure which i have is id, master, name. ( Here's some pictures to clear out what i'm doing. http://imgur.com/a/7XwVd ) So the problem is that it does not get the images from folder named ex: *_audi_a3.jpg. And link to codes what have been written already. Parts.php: http://pastebin.com/q6vdypge Update.php: http://pastebin.com/DymhGQ17 Search_images.php: http://pastebin.com/LF5Q0i8f Core.js: http://pastebin.com/bgc0y4TS I don't know what's wrong with it sadly. One thing i noticed when i used firebug it gives error on category when searching error:true. Hope you guys understand what i mean here, and also all help is appreciated. And move this post if it's in wrong place. Thanks! Edited by aeonius, 17 October 2014 - 12:15 PM. Hello once again, i got this code that takes all images from a folder and displays them close to each other. Heres the code: <?php $dir = 'uploads/thumbs/watermarkedthumbs/'; $file_display = array ('jpg', 'jpeg', 'png', 'gif'); if (file_exists($dir) == false) { echo 'tt'; } else { $dir_contents = scandir($dir); foreach ($dir_contents as $file) { $file_type = strtolower(end(explode('.', $file))); if ($file !== '.' && $file !== '..' && in_array($file_type, $file_display) == true) { echo '<img src="', $dir, '/', $file, '" alt="', $file, '" />'; } } } ?> How do i add margins in between them? i wanna make it smth like 'margin-left:10px, margin-top-10px'. Also, how do i add an 'overflow'? i mean, my images currently are displayed in a div, and i wanna make it so that if the folder has lets say 9 images, it would only display 6 of them, and the rest would be displayed in the same div after i click a button or smth (like '1' for the first div and '2' would appear if theres an overflow, and when i click '2' the rest of the images would appear). To make these adjustments do i need to make a different php file or do i change something in this code? thanks in advance. I got this script: But it give me error, file_get_contents cannot open stream. I need to add the FTP connection with user/pass paramaters. then look in set http url, to get the file contents(images) and transfer to ftp server location. Can Anyone take alook and tell me if I am going down the right path and how to get there. Please Code: [Select] function postToHost($host, $port, $path, $postdata = array(), $filedata = array()) { $data = ""; $boundary = "---------------------".substr(md5(rand(0,32000)),0,10); $fp = fsockopen($host, $port); fputs($fp, "POST $path HTTP/1.0\n"); fputs($fp, "Host: $host\n"); fputs($fp, "Content-type: multipart/form-data; boundary=".$boundary."\n"); // Ab dieser Stelle sammeln wir erstmal alle Daten in einem String // Sammeln der POST Daten foreach($postdata as $key => $val){ $data .= "--$boundary\n"; $data .= "Content-Disposition: form-data; name=\"".$key."\"\n\n".$val."\n"; } // Sammeln der FILE Daten if($filedata) { $data .= "--$boundary\n"; $data .= "Content-Disposition: form-data; name=\"".$filedata['name']."\"; filename=\"".$filedata['name']."\"\n"; $data .= "Content-Type: ".$filedata['type']."\n"; $data .= "Content-Transfer-Encoding: binary\n\n"; $data .= $filedata['data']."\n"; $data .= "--$boundary--\n"; } // Senden aller Informationen fputs($fp, "Content-length: ".strlen($data)."\n\n"); fputs($fp, $data); // Auslesen der Antwort while(!feof($fp)) { $res .= fread($fp, 1); } fclose($fp); return $res; } $postdata = array('var1'=>'today', 'var2'=>'yesterday'); $filedata = array( 'type' => 'image/png', 'data' => file_get_contents('http://xxx/tdr-images/images/mapping/dynamic/deals/spot_map') ); echo '<pre>'.postToHost ("localhost", 80, "/test3.php", $postdata, $filedata).'</pre>'; I want to copy everything in templates/blue to the folder code/ However: shell_exec("cp -r 'templates/blue' 'code'"); Creates a folder called blue inside code. I tried cp -r 'templates/blue/*' 'code', but that didn't do anything. Any ideas? Hey, I'm trying to build up a simple framework for my projects, and need a little help! How can I find folders using PHP? I have one file called global.php which will include a folder called "plugins". Inside "plugins" will be all the actual site files. I do not want a index.php?page=pluginfilehere... I want strictly say.. example.com/site/index.php. So inside plugins folder, I will have another folder called site, which inside that will be the .php files for it. But, keep in mind, I will not just have a folder called site in the plugins.. It could be called account, or user whatever I am building! Thanks in advance. - Justin Hi, I just connect to my server through ftp. i want to edit php.ini file. which is placed in /etc/php.ini "etc" folder. but i'm unable to see etc folder. there are only two folders one is cgi-bin and other is www. can anyone help me to find out etc folder. i've check to show hidden files in filezilla. Thanks Hey, I hope you will understand what I am saying, because english isn't my main language -.-, Anyway, I need to create a folder navigation, what I mean is like you got folder on your comp : C:\navThing\ (and many sub folders and files) what I want to do is list all files from the current directory where the user is. if he is in C:\navThing\ show dir and files in C:\navThing\ only... if he is in C:\navThing\lol\ show dir and files in C:\navThing\lol only... but I don't know how to get the user position, for the moment I'm using $_GET, but well, I don't think it's really secure... ( xxx.php?path=lol/ ) if anybody can think of any other way, where the user can't modify it tell me, I thougth session would be good, but what if he use previous... anyway any suggestion? Hi all, I'm trying to echo all the files in the folder 'pics' <?php $dir = "pics/"; // Open a directory, and read its contents if (is_dir($dir)){ if ($dh = opendir($dir)){ while (($file = readdir($dh)) !== false){ echo '<img src="pics/' .$file . '" width="120px" height="120px" />'; } closedir($dh); } } ?>
The above code outputs the following. (there is only one image in the folder) So the code gets the image correctly and works with multiple images in the folder, but also produces two broken pathways at the beginning. <img src="pics/." width="120px" height="120px" /><img src="pics/.." width="120px" height="120px" /><img src="pics/j1.jpg" width="120px" height="120px" />
My question is why, I can't figure out what is happening. Any pointers or explanations would be appreciated. Cheers for your time.
Paul
<?php $zip = new ZipArchive; $res = $zip->open('test.zip'); if ($res === TRUE) { $zip->renameName('currentname.txt','newname.txt'); $zip->close(); } else { echo 'failed, code:' . $res; } ?> I can rename a text file name but I can't rename a folder name.Can you help me? |