PHP - Php Altering Photos
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... Similar TutorialsI have a table with lots of records. My table has the following field, with records containing either NULL or '1' that I need to modify. Hi, I have a file which contains settings for a user. Similar to wordpress except simpler like setting a default email for error messages or something. The idea is that it's used as an installation script for web-ware where someone fills in a form with their database details, my script will hard-code them into the settings file and then delete itself. I'll admit I did get the idea from wordpress. Only the way it's coded currently requires a certain combination of characters at the end and seems really messy. The script looks like this (and does work): $endpos = strpos ( $contents, '"; //replace with name//', $startpos); The settings file looks like this: $hard_name = "Smith"; //replace with name// As you can see I use the comment in the settings file, which follows what I'm actually changing, to find the end of the variable. I am sure there must be a better way of doing things but don't know what. It just seems really inefficient because if the comment is removed the script breaks. Any info and tips would be much appreciated. IceKat so I'm trying to change some specific columns in a row in a table of a MySQL database by calling the Doctrine_Query when an option is selected inside an admin panel of my website. Here are the details: Table name: chatUsers I need to find all rows with the person who has a username of: $chatUsers->username (The column inside chatUsers is called username) Once all those rows are found, change the value of all the row's column "type" to "user" Is this even possible? So far I have: [/list] Code: [Select] <?php function userChatReset(){ $query = Doctrine_Query::create()->update('db_chatUsers')->set('type', 'user')->where('username = '.$chatUsers->username); //execute query $rows = $query->execute(); echo $rows.' rows updated'; } ?> ...And I'm not sure where to go from there, or if that's even correct. Sorry in advance, I'm not very good with PHP yet. I found this online at http://php.net/manual/en/function.time.php: <?php function rel_time($from, $to = null) { $to = (($to === null) ? (time()) : ($to)); $to = ((is_int($to)) ? ($to) : (strtotime($to))); $from = ((is_int($from)) ? ($from) : (strtotime($from))); $units = array ( "year" => 31536000, // seconds in a year (12 months) "month" => 2628000, // seconds in a month (4 weeks) "week" => 604800, // seconds in a week (7 days) "day" => 86400, // seconds in a day (24 hours) "hour" => 3600, // seconds in an hour (60 minutes) "minute" => 60, // seconds in a minute (60 seconds) "second" => 1 // 1 second ); $diff = abs($from - $to); $suffix = (($from > $to) ? ("from now") : ("ago")); foreach($units as $unit => $mult) if($diff >= $mult) { $and = (($mult != 1) ? ("") : ("and ")); $output .= ", ".$and.intval($diff / $mult)." ".$unit.((intval($diff / $mult) == 1) ? ("") : ("s")); $diff -= intval($diff / $mult) * $mult; } $output .= " ".$suffix; $output = substr($output, strlen(", ")); return $output; } ?> ======================= Basically, if you type <?php echo rel_time("March 12, 2011 7:00 PM"); ?>, it will show you the time ago relative to the current time. But there's two problems. 1) The time it sends out is about 5hrs early (the actual time is 7:00pm, I set the rel time to 7:00pm, but the time it displays reads '5 hours ago, 1 minute, and 3 seconds ago'. I'm not much of a coder, but judging by the looks of it, it would appear correct... but apparently not. 2) My second gripe is that the code shows too many increments of date. What I mean is, if if I set the date to over a year ago, it would display the years,months,weeks,days,minutes,seconds. That is way too much increments. I just want it to display seconds, if I submitted it seconds ago, just display minutes if it was displayed minutes ago, and display days if it was submitted days ago. You get the idea. Friends, family, fellow coders, and bored people- can you help a nooblet? Thanks! Hi, I am extracting data from a mysql db to then edit and update back into the db and use below to list items first - Code: [Select] <?php // Connect to server and select database. mysql_connect('localhost', 'xxxxxxxxx', 'xxxxxxxxxxxxx') or die("Error: ".mysql_error()); mysql_select_db("xxxxxxxxx"); $sql="SELECT * FROM properties"; $result=mysql_query($sql); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="400" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="4"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Name</strong></td> <td align="center"><strong>Lastname</strong></td> <td align="center"><strong>Email</strong></td> <td align="center"><strong>Update</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td><? echo $rows['Property_Rating']; ?></td> <td align="center"><a href="update.php?id=<? echo $rows['ID']; ?>">update</a></td> </tr> <?php } ?> </table> </td> </tr> </table> <?php mysql_close(); ?> I then go next page where I show data and allow editing - <?php // Connect to server and select database. mysql_connect('localhost', 'xxxxxxxxx', 'xxxxxxxxxxxxx') or die("Error: ".mysql_error()); mysql_select_db("xxxxxxxxx"); // get value of id that sent from address bar $id=$_GET['id']; // Retrieve data from database $sql="SELECT * FROM properties WHERE id='$id'"; $result=mysql_query($sql); $rows=mysql_fetch_array($result); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <form name="form1" method="post" action="update_ac.php"> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td> </td> <td colspan="3"><strong>Update data in mysql</strong> </td> </tr> <tr> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> </tr> <tr> <td align="center"> </td> <td align="center"><strong>Property_Name</strong></td> <td align="center"><strong>roperty_Rating</strong></td> <td align="center"><strong>Property_Address</strong></td> </tr> <tr> <td> </td> <td align="center"><input name="Property_Rating" type="text" id="Property_Rating" value="<? echo $rows['Property_Rating']; ?>"></td> </tr> <tr> <td> </td> <td><input name="id" type="hidden" id="id" value="<? echo $rows['id']; ?>"></td> <td align="center"><input type="submit" name="Submit" value="Submit"></td> <td> </td> </tr> </table> </td> </form> </tr> </table> <? // close connection mysql_close(); ?> And then the update page - <?php // Connect to server and select database. mysql_connect('localhost', 'xxxxxxxxx', 'xxxxxxxxxxxxx') or die("Error: ".mysql_error()); mysql_select_db("xxxxxxxxx"); // update data in mysql database $sql="UPDATE properties SET Property_Name='$Property_Name' WHERE id='$id'"; $result=mysql_query($sql); // if successfully updated. if($result){ echo "Successful"; } else { echo "ERROR"; } ?> Everything seems ok but when I edit the data and submit I get the success message but data not changed in db. Any ideas? Thanks. MOD EDIT: code tags added. I found the code below which checks to make sure something the user entered into a text field is an actual URL. I am looking to confirm that something a user enters is an actual link to an IMAGE, so does anyone know how I could adjust this code below to do that? Basically, this code below would accept http://monkeys.com AND http://monkeys.com/images/monkey.jpg. I want to alter the code so that http://monkeys.com would NOT be accepted since it's not an image. Essentially, I want to make sure whatever the user enters ends with .jpg, .gif or .png. Anyone know how to do this? Code: [Select] url": { "regex": /^(https?|ftp):\/\/(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*@)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)|\/|\?)*)?$/, "alertText": "* Invalid URL" } 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 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 $thousandsofvisitedphotoids and make a query again ? I m stuck here, How you guys do this ? what's the logic of this ? Hello. My website has a photo gallery of thumbnails that is created by reading all photo files in a specified directory. Here is my function that builds the array which is ultimately displayed in the gallery...
<?php function getPhotoFilesArray($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 $photoFiles; }//End of getPhotoFilesArray ?>
Everything works fine locally in DEV, but when I uploaded my website (and photos) onto a webserver, the photos are appearing in a backwards order in PROD. This is annoying, because I want the photos displayed chronologically from oldest (first) to newest (last). I'm not sure where the problem is happening, because each photo was taken with my camera and by nature of the camera, photo names are incremented by one, so IMG_001.jpg would have been taken FIRST, followed by IMG_002.jpg, IMG_003.jpg, and so on. How can I fix things so the photos are displayed in the order they were physically taken AND match how things display locally in DEV? Thanks!
This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=354562.0 I was thinking of a board where you can bookmark the photos from Facebook.
It may require an App and a "Facebook Login" for the website, so one can bookmark photos from Facebook. The website would have additional features for the photos.
I thought of an "Add to ..." function.
Is this possible with the Facebook API?
Edited by glassfish, 07 October 2014 - 02:31 PM. My photo files are not being displayed in my table? They get sent to the mySQL database, then the server and it does grab all the other variables in the table and displays them, but the .jpg's are not shown, instead theres just the file name?? Code: [Select] <?php error_reporting(E_ALL); ini_set("display_errors", 1); echo '<pre>' . print_r($_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 ? 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. 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`) ) I'm talking like /uploads at the main folder where index.php or index.html goes
These photos theoretically are meant to be publicly viewed anyway so what is bad about that?
Let's say profile pictures.
I mean couldn't anyone scrape facebook and "steal" profile pictures? I don't know why but I'm just wondering.
I have a php page that creates a photo gallery with thumbnails. It is populated by code that reads all photo files from a specified photo directory. This was working fine in DEV, but now that I have uploaded to my test web server, the pictures are in reverse order. Not the end of the world, yet annoying, because they should be in chronological order. Files names are straight off my iPhone (e.g. IMG_2203.jpg, IMG_2204.jpg, IMG_2207.jpg) What is happening, and how can I fix this? Thanks! I need help with my webpage here, how do you get so if you are at http://www.blabla.com/account.php and then click on a photo, it will go to www.blabla.com/photo.php?id=1 but still be at account.php? Just like facebook shows their photos. I dont know how to think nor to get it work. Would appreciate some help! ThNXX 1n 4dv4nc3 // Machram! If 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--; } ?> |