|
PHP - Moved: Help Please - Photos Are Not Showing
This topic has been moved to Third Party PHP Scripts.
http://www.phpfreaks.com/forums/index.php?topic=354562.0 Similar TutorialsIf you goto http://www.actionfx.net/bfd/photos.php you will see that im trying to show an rss feed from http://picasaweb.google.com/data/feed/base/user/114565750484201639035?alt=rss&kind=album&hl=en_US&access=public My code works well if its just text on the rss but it does not work on this particular rss. I want the photos to show up on my page also. Im hoping someone here can be so kind to help me out with this coding. Below is my code. Code: [Select] <?php $feed_url = "http://picasaweb.google.com/data/feed/base/user/114565750484201639035?alt=rss&kind=album&hl=en_US&access=public"; // INITIATE CURL. $curl = curl_init(); // CURL SETTINGS. curl_setopt($curl, CURLOPT_URL,"$feed_url"); curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); curl_setopt($curl, CURLOPT_CONNECTTIMEOUT, 0); // GRAB THE XML FILE. $xmlTwitter = curl_exec($curl); curl_close($curl); // SET UP XML OBJECT. // Use either one of these, depending on revision of PHP. // Comment-out the line you are not using. //$xml = new SimpleXMLElement($xmlTwitter); $xml = simplexml_load_string($xmlTwitter); // How many items to display $count = 10; // How many characters from each item // 0 (zero) will show them all. $char = 100; foreach ($xml->channel->item as $item) { if($char == 0){ $newstring = $item->description; } else{ $newstring = substr($item->description, 0, $char); } if($count > 0){ //in case they have non-closed italics or bold, etc ... echo"</i></b></u></a>\n"; echo" <div style='font-family:verdana; font-size:.12;'> <b>{$item->title}</b><br /> $newstring ... <span class='redlink' id='redlink' style='redlink'> <a href='{$item->guid}'>read more</a> </span> <br /><br /> </div> "; } $count--; } ?> This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=341970.0 This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=314084.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=327211.0 This topic has been moved to CSS Help. http://www.phpfreaks.com/forums/index.php?topic=317486.0 This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=330251.0 This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=357605.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=325313.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=326995.0 I have some thousands of photos about nature I ll let visitors/members to see them one by one, but I dont want to show them the same photo again after they visit 1 week later How can I do this ? What I think as a solution is; For members; I can store the ids (like "everest01") of the photos that member has visited , and show user the most visited photos that he/she has not see for next visit. But what I m wondering is, how will I take the photos from DB ? select * from photos WHERE id not in ( $thousandsofvisitedphotoids ) ?? I m stuck here ? For visitors ( not members ) ; I can set a cookie that keeps the ids of visited photos.. when visitor visits the website again, I take the cookie and sent to $thousandsofvisitedphotoids and make a query again ? I m stuck here, How you guys do this ? what's the logic of this ? How should I go about keeping track of photos that a user has rated, since I only want the user to rate the photo once? Can I store arrays in a mysql database? thanks, George Hi! I am creating a small project somewhat like a photo gallery as my first practice page. I just would like to know how can I alter photos or file in php. Or should I just do a delete and then upload a new one? Thanks in advance... I am pretty new to PHP and am trying to create a simple (so I assumed) page to takes data from one html page(works fine) and updates a MYSQL Database. I am getting no error message, but the connect string down to the end of the body section is showing up as plain text in my browser window. I do not know how to correct this. I have tried using two different types of connect strings and have verified my names from the HTML page are the same as listed within the php page. Suggestions on what I need to look for to correct would be great. I have looked online, but so far all I am getting is how to connect, or how to create a comment, so I thought I would try here. Thank you for any assistance I may get!! - Amy - Code: [Select] <body><font color="006600"> <div style="background-color:#f9f9dd;"> <fieldset> <h1>Asset Entry Results</h1> <?php // create short variable names $tag=$_POST['tag']; $serial=$_POST['serial']; $category=$_POST['category']; $status=$_POST['status']; $branch=$_POST['branch']; $comments=$_POST['comments']; if (!$tag || !$serial || !$category || !$status || !$branch) { echo "You have not entered all the required details.<br />" ."Please go back and try again."; exit; } if (!get_magic_quotes_gpc()) { $tag = addslashes($tag); $serial = addslashes($serial); $category = addslashes($category); $status = addslashes($status); $branch = addslashes($branch); $comments = addslashes($comments); } //@ $db = new mysqli('localhost', 'id', 'pw', 'inventory'); $db = DBI->connect("dbi:mysql:inventory:localhost","id","pw") or die("couldnt connect to database"); $query = "insert into assets values ('".$serial."', '".$tag."', '".$branch."', '".$status."', '".$category."', '".$comments."')"; $result = $db->query($query); if ($result) { echo $db->affected_rows." asset inserted into Inventory."; } else { echo "An error has occurred. The item was not added."; } $db->close(); ?> </fieldset> </div> </body> My photo files are not being displayed in my table? They get sent to the mySQL database, then the server and it does grab all the other variables in the table and displays them, but the .jpg's are not shown, instead theres just the file name?? Code: [Select] <?php error_reporting(E_ALL); ini_set("display_errors", 1); echo '<pre>' . print_r($_FILES, true) . '</pre>'; //This is the directory where images will be saved $target = "/home/users/web/b109/ipg.removalspacecom/images/COMPANIES"; $target = $target . basename( $_FILES['upload']['name']); //This gets all the other information from the form $company_name=$_POST['company_name']; $basicpackage_description=$_POST['basicpackage_description']; $location=$_POST['location']; $postcode=$_POST['postcode']; $upload=($_FILES['upload']['name']); // Connects to your Database mysql_connect("server****", "username***", "password****") or die(mysql_error()) ; mysql_select_db("DB") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO `Companies` (company_name, basicpackage_description, location, postcode, upload) VALUES ('$company_name', '$basicpackage_description', '$location', '$postcode', '$upload')") ; echo mysql_error(); //Writes the photo to the server if(move_uploaded_file($_FILES['upload']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['upload']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } ?> "upload" is the variable that isnt displaying in my table how i want it to? Have you guys any ideas how to get it displayed correctly? I have a page for image uploads and I just realized it will only work if a user already has on picture uploaded. If they don't it won't work. The ones that do fail some photos anyway which is probably that they don't pass the image check but when I put echoes in there to trace what happens any user with an empty gallery can't upload a photo because the page says there is no file in the $_FILES['image']['name'] variable. Here are the initial conditions and the form (leaving out the image processing etc since that works): Code: [Select] if (!isset($_SESSION['user'])) die("<br /><br /> You need to log in to view this page"); $user = sanitizeString($_SESSION['user']); $view = sanitizeString($_GET['view']); $dir = './grafik/users/'.$user.'/big/'; $files = scandir($dir); $len = count($files); $nr= $len-1; $maxPhotos = 8; if ($view == $user) { echo "view is user"; if(!file_exists("grafik/users/$user")) { mkdir("grafik/users/$user"); mkdir("grafik/users/$user/big/");} } if (!isset($_FILES['image']['name'])) echo "There is no file <br />$dir - $user - $nr"; if (isset($_FILES['image']['name'])) { echo "<br />...is a file <br />$dir - $user - $nr"; $photoName="$dir$user$nr.jpg"; move_uploaded_file($_FILES['image']['tmp_name'], $photoName); $typeok = TRUE; .... <form method='post' action='gallery.php?view=$user' enctype='multipart/form-data'> Upload another photo: <br /> Max $maxPhotos allowed, max filesize 2Mb <br /> <input type='file' name='image' size='10' /><br /> <input type='submit' value='Upload' /> </form> I can't see why it wouldn't let me but I have a feeling someone here knows why. I have users becoming members and allowed them to upload their own photos. But when they try to upload 5MB photos, it takes time to upload the photo, and sometimes server gives a timeout error. I have searched and found javascripts that uploads to the server but I have noticed that it has security problems. So how do you let users to upload photos ? I was thinking of a board where you can bookmark the photos from Facebook.
It may require an App and a "Facebook Login" for the website, so one can bookmark photos from Facebook. The website would have additional features for the photos.
I thought of an "Add to ..." function.
Is this possible with the Facebook API?
Edited by glassfish, 07 October 2014 - 02:31 PM. Does anyone know of a good tutorial on uploading a picture file to a folder using php and copying the name to the database in mysql? Resizing photos on upload is helpful also...If you know it works...some I have tried do not work. Not asking for someone to write code for me but info or tutorial would be nice. Or maybe a code that has worked for you that is similar that I could learn from and edit... I can upload the actual photo into the database but it slows it way down. I have heard of loading the name only and resizing the photo and sending the actual photo to a file folder on the server. The few codes I have tried were not successful. Thanks for any guidance. I appreciate it. Hello. My website has a photo gallery of thumbnails that is created by reading all photo files in a specified directory. Here is my function that builds the array which is ultimately displayed in the gallery...
<?php
function getPhotoFilesArray($photoPath){
/**
* Takes path to photo-directory, and returns an array containing photo-filenames.
*/
// Initialize Array.
$photoFiles = array();
// Check for Photo-Directory.
if (is_dir($photoPath)){
// Photo-Directory Found.
// Open Directory-Handle.
$handle = opendir($photoPath);
// Open Photo-Directory.
if($handle){
// Initialize Key.
$i = 1001;
// Iterate through Photo-Directory items.
while(($file = readdir($handle)) !== FALSE){
// Return next Filename in Directory.
// Define fullpath to file/folder.
$fullPath = $photoPath . $file;
// Populate Array.
if(!is_dir($fullPath)
&& preg_match("#^[^\.].*$#", $file)){
// Not Directory.
// Not Hidden File.
// Add to array.
$photoFiles[$i] = $file;
$i++;
}//End of POPULATE ARRAY.
}//End of ITERATE THROUGH PHOTO-DIRECTORY ITEMS
closedir($handle);
}//End of OPEN PHOTO-DIRECTORY
}else{
// Photo-Directory Not Found.
// Redirect to Page-Not-Found.
header("Location: " . BASE_URL . "/utilities/page-not-found");
// End script.
exit();
}//End of CHECK FOR PHOTO-DIRECTORY
return $photoFiles;
}//End of getPhotoFilesArray
?>
Everything works fine locally in DEV, but when I uploaded my website (and photos) onto a webserver, the photos are appearing in a backwards order in PROD. This is annoying, because I want the photos displayed chronologically from oldest (first) to newest (last). I'm not sure where the problem is happening, because each photo was taken with my camera and by nature of the camera, photo names are incremented by one, so IMG_001.jpg would have been taken FIRST, followed by IMG_002.jpg, IMG_003.jpg, and so on. How can I fix things so the photos are displayed in the order they were physically taken AND match how things display locally in DEV? Thanks!
Hi friends, I have two mysql db tables, photos and album, I would like to list photos by album how do i do that ? CREATE TABLE `album` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `album_name` varchar(95) NOT NULL, `album_desc` text NOT NULL, PRIMARY KEY (`id`) ); CREATE TABLE `photos` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `album_id` int(11) NOT NULL, `thumb_name` varchar(255) NOT NULL, `photo_name` varchar(250) NOT NULL, PRIMARY KEY (`id`) ) |
|||||||