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PHP - Trouble Doing Update Set For Editing Photos ...
Hello:
I am trying to figure out how to do an UPDATE SET for a photo gallery based on the "photo_id" ... not working to well ... Currently, it just uploads a photo but updates everything. This is the code: Photo_Edit.php Code: [Select] <?php $photo_id = $_REQUEST['photo_id']; ?> ... <?php $query=mysql_query("SELECT * FROM gallery_photos WHERE photo_id = $photo_id") or die("Could not get data from db: ".mysql_error()); while($result=mysql_fetch_array($query)) { $photo_id=$result['photo_id']; $photo_filename=$result['photo_filename']; $photo_caption=$result['photo_caption']; } ?> <form enctype="multipart/form-data" action="a_Photo_Upload2.php" method="post" name="upload_form"> <input type='hidden' name='photo_id' value='<?php echo $photo_id; ?>' /> <table width='600' border='0' id='tablepadding' align='center'> <tr> <td width='50'> Photo: <td width='550'> <input name='photo_filename[]' id='my_photo' type='file' /> <div style='clear: both;'></div> Current photo: <div style='clear: both;'></div> <img src="../gallery/tb_<?php echo $photo_filename; ?>" width="100px" class='imgCenter' /> <div style='clear: both;'></div> <a href="javascript:void(0);" onmouseover="tooltip.show('<img src=../gallery/tb_<?php echo $photo_filename; ?> width=500 />');" onmouseout="tooltip.hide();">Larger view</a> </td> </tr> <tr> <td> Description: </td> <td> <textarea name='photo_caption[]' cols='100' rows='10'><?php echo $photo_caption; ?></textarea> </td> </tr> <tr> <td> <input type='submit' name='submit' value='Edit Product' /> </td> </tr> </table> </form> Goes to this page, where I am having the problems: a_Photo_Upload2.php Code: [Select] <?php echo $photo_id; ?> <?php // initialization $result_final = ""; $counter = 0; // List of our known photo types $known_photo_types = array( 'image/pjpeg' => 'jpg', 'image/jpeg' => 'jpg', 'image/gif' => 'gif', 'image/bmp' => 'bmp', 'image/x-png' => 'png' ); // GD Function List $gd_function_suffix = array( 'image/pjpeg' => 'JPEG', 'image/jpeg' => 'JPEG', 'image/gif' => 'GIF', 'image/bmp' => 'WBMP', 'image/x-png' => 'PNG' ); // Fetch the photo array sent by preupload.php $photos_uploaded = $_FILES['photo_filename']; // Fetch the photo caption array $photo_caption = $_POST['photo_caption']; while( $counter <= count($photos_uploaded) ) { if($photos_uploaded['size'][$counter] > 0) { if(!array_key_exists($photos_uploaded['type'][$counter], $known_photo_types)) { $result_final .= "File ".($counter+1)." is not a photo<br />"; } else { $new_id = mysql_insert_id(); $filetype = $photos_uploaded['type'][$counter]; $extention = $known_photo_types[$filetype]; $filename = $new_id.".".$extention; mysql_query( "UPDATE gallery_photos SET photo_filename='".addslashes($filename)."' WHERE photo_id='".addslashes($new_id)."'" ); // Store the orignal file copy($photos_uploaded['tmp_name'][$counter], $images_dir."/".$filename); // Let's get the Thumbnail size $size = GetImageSize( $images_dir."/".$filename ); if($size[0] > $size[1]) //{ //$thumbnail_width = 100; //$thumbnail_height = (int)(100 * $size[1] / $size[0]); //} //else //{ //$thumbnail_width = (int)(100 * $size[0] / $size[1]); //$thumbnail_height = 100; //} { //$thumbnail_width = 690; //$thumbnail_height = (int)(500 * $size[1] / $size[0]); $old_width = $size[0]; $old_height = $size[1]; $thumbnail_width = 690; $thumbnail_height = ($old_height * $thumbnail_width / $old_width); } else { $thumbnail_width = (int)(690 * $size[0] / $size[1]); $thumbnail_height = 500; } // Build Thumbnail with GD 1.x.x, you can use the other described methods too $function_suffix = $gd_function_suffix[$filetype]; $function_to_read = "ImageCreateFrom".$function_suffix; $function_to_write = "Image".$function_suffix; // Read the source file $source_handle = $function_to_read ( $images_dir."/".$filename ); if($source_handle) { // Let's create an blank image for the thumbnail //$destination_handle = ImageCreate ( $thumbnail_width, $thumbnail_height ); $destination_handle = imagecreatetruecolor( $thumbnail_width, $thumbnail_height ); // Now we resize it ImageCopyResized( $destination_handle, $source_handle, 0, 0, 0, 0, $thumbnail_width, $thumbnail_height, $size[0], $size[1] ); } // Let's save the thumbnail $function_to_write( $destination_handle, $images_dir."/tb_".$filename ); ImageDestroy($destination_handle ); // $result_final .= "<img src='".$images_dir. "/tb_".$filename."' style='margin-right: 20px; width: 100px;' />"; } } $counter++; } // Print Result echo <<<__HTML_END $result_final __HTML_END; ?> Can anyone show me how to make this work, OR show me a way to do this ... Thank you. Similar TutorialsI need help trying to figure out why my form won't write the database it is supposed to - i checked the connection to the database and it works and the user seems to have permission to edit database - the error I get is "Error: User not added to database." from "register.php". Can someone please look over my code and see if the problem is coming from somewhere within?
I created a connection file (connect.php)
<?
session_start();
// Replace the variable values below
// with your specific database information.
$host = "localhost";
$user = "master";
$pass = "hidden";
$db = "user";
// This part sets up the connection to the
// database (so you don't need to reopen the connection
// again on the same page).
$ms = mysql_pconnect($host, $user, $pass);
if ( !$ms )
{
echo "Error connecting to database.\n";
}
// Then you need to make sure the database you want
// is selected.
mysql_select_db($db);
?>
Then there is the php script (register.php):
<?php
session_start();
// connect.php is a file that contains your
// database connection information. This
// tutorial assumes a connection is made from
// this existing file.
require('connect.php');
// If the values are posted, insert them into the database.
if (isset($_POST['email']) && isset($_POST['password'])){
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
$password = $_POST['password'];
$query = "INSERT INTO `member` (firstname, lastname, email, password) VALUES ('$firstname', '$lastname', '$email' '$password')";
$result = mysql_query($query);
if ( !mysql_insert_id() )
{
die("Error: User not added to database.");
}
else
{
// Redirect to thank you page.
Header("Location: surveylanding_no-sidebar.html");
}
}
?>
Here is the HTML form:
<form name="htmlform" method="post" class="form" action="register.php">
<p class="firstname">
<input type="text" name="firstname" id="firstname" />
<label for="firstname">First Name</label>
</p>
<p class="lastname">
<input type="text" name="lastname" id="lastname" />
<label for="lastname">Last Name</label>
</p>
<p class="email">
<input type="email" name="email" id="email" />
<label for="email">Email</label>
</p>
<p class="Password">
<input type="password" name="password" id="password" />
<label for="password">Password</label>
</p>
<p class="submit">
<input type="submit" value="Register"/>
</p>
</form>
Hi guys, Sorry - this might more belong in the SQL section but I think it is PHP related. I am trying to update a table with: $sql="UPDATE $tbl_name SET (gender, description) VALUES ('$gender', '$description') WHERE image = ('$photoname')"; $result=mysql_query($sql); Needless to say, its not working. Only odd thing is that the $photoname variable is sent with a hidden field on the previous form. I can echo all 3 varaibles though and get the right results. I just dont know why it isnt updating - seems pretty standard to me. Can anyone help? Thanks! Having update statement trouble... The error I'm getting is... "Could not query the database - : " Not sure what I'm missing, I had this working before also, but now it's not. Here is my code for the update page... Code: [Select] <html> <head> <title>Update User</title> </head> <body> <form method="post" action="update_user2.php"> <?php $dbc = mysqli_connect('localhost', 'se266_user', 'pwd', 'se266') or die(mysql_error()); //delete users echo '<b>Delete or Update User</b>.<br />'; if (isset($_POST['remove'])) { foreach($_POST['delete'] as $delete_id) { $query = "DELETE FROM users WHERE course_id = $delete_id"; mysqli_query($dbc, $query) or die ('can\'t delete user'); } echo 'user has been deleted.<br />'; } if (isset($_POST['update'])) { $course_id = $_POST['course_id']; $course_name = $_POST['course_name']; $student_id = $_POST['student_id']; $query = "UPDATE users SET course_name ='". $course_name ."' WHERE course_id = $course_id"; $updres = mysqli_query($query); if(!$updres) { die(" Could not query the database - : <br/>". mysqli_error() ); } else { echo 'course has been updated.<br />'; } } //display users info with checkbox to delete $query = "SELECT * FROM users"; $result = mysqli_query($dbc, $query); while($row = mysqli_fetch_array($result)) { echo '<input type="checkbox" value="' .$row['course_id'] . '" name="delete[]" />'; echo ' ' .$row['course_name'] .' '. $row['student_id']; echo '<br />'; } mysqli_close($dbc); ?> <form method="POST" action="update_user2.php"> <label for="course_id">Course ID:</label> <input type="text" id="course_id" name="course_id" /> <br /> <label for="course_name">Course Name:</label> <input type="text" id="course_name" name="course_name" /> <br /> <label for="course_name">Student ID:</label> <input type="text" id="student_id" name="student_id" /> <br /> <input type="submit" name="remove" value="Remove" /> <input type="submit" name="update" value="Update" /> <br><br> </form> </body> </html> Hi, I am having trouble coding for mysqli update. Please, somebody tell me the correct way. I'm trying to update the "lastused" (current date) field in "emailtbl". Somebody please tell the best way to code this. Below is the message and following, the current code: Fatal error: Call to undefined function curdate() in C:\xampp\htdocs\home\lastused.php on line 14
$db = new mysqli('localhost', 'root', 'pass', 'mydb');
if($db->connect_errno > 0)
{die('Unable to connect to database [' . $db->connect_error . ']');}
$sql = <<<SQL
SELECT *
FROM `emailtbl`
WHERE `id` = '$id'
SQL;
if(!$result = $db->query($sql))
{die('There was an error running the query [' . $db->error . ']');}
$lastused = $_POST['lastused'];
$lastused = curdate();
echo "last date accessed is ".$data['lastused'];
$result->free();
$db->escape_string('This is an unescape "string"');
$db->close();
?>
$update
= mysqli_query($dbconnect, "UPDATE emailtbl SETlastused = curdate() WHERE id ='$id'"); if($update == false) { die("UPDATE FAILED: ".mysqli_error($dbconnect)); } echo "$lastused is the last date this account was accessed"; First time poster, and very amateur php coder. I am trying to delete items that in a list using ajax. I can't figure out what I am doing wrong. Any help would be greatly appreciated!
Here is a snippet of my javascript...
$('.delete-item').click(function() {
var itemID = $(this).data('itemID');
var clear = 1;
$.ajax({
url: 'includes/delete-item.php',
method: 'POST',
data: {
itemid:itemID,
clear:1
},
success: function(data) {
$('.content').load('includes/lists.php')
}
})
})
and here is the relevant php... $item = $_POST['itemid']; $clear = $_POST['clear']; $clearSQL = "DELETE FROM `List_Items` WHERE `item_ID` = $item"; $cleared = mysqli_query($connect, $clearSQL); So i have this CMS that is used by lots of users and generally they do have rights to edit each others article. So now i'm thinking how can i at least warn them that somebody else is already editing it. Just that. maybe with sessions? but what happens when a user just closes the windows, will it destroy it. i taught about storing the value to mySQL but again what if the user closes the windows and so on. i never did this and i could use a boost. thanks guys 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 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 This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=354562.0 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. 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. 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 $maxPhotos allowed, max filesize 2Mb <br /> <input type='file' name='image' size='10' /><br /> <input type='submit' value='Upload' /> </form> I can't see why it wouldn't let me but I have a feeling someone here knows why. Hi friends, I have two mysql db tables, photos and album, I would like to list photos by album how do i do that ? CREATE TABLE `album` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `album_name` varchar(95) NOT NULL, `album_desc` text NOT NULL, PRIMARY KEY (`id`) ); CREATE TABLE `photos` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `album_id` int(11) NOT NULL, `thumb_name` varchar(255) NOT NULL, `photo_name` varchar(250) NOT NULL, PRIMARY KEY (`id`) ) Hello. My website has a photo gallery of thumbnails that is created by reading all photo files in a specified directory. Here is my function that builds the array which is ultimately displayed in the gallery...
<?php
function getPhotoFilesArray($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 $photoFiles;
}//End of getPhotoFilesArray
?>
Everything works fine locally in DEV, but when I uploaded my website (and photos) onto a webserver, the photos are appearing in a backwards order in PROD. This is annoying, because I want the photos displayed chronologically from oldest (first) to newest (last). I'm not sure where the problem is happening, because each photo was taken with my camera and by nature of the camera, photo names are incremented by one, so IMG_001.jpg would have been taken FIRST, followed by IMG_002.jpg, IMG_003.jpg, and so on. How can I fix things so the photos are displayed in the order they were physically taken AND match how things display locally in DEV? Thanks!
My photo files are not being displayed in my table? They get sent to the mySQL database, then the server and it does grab all the other variables in the table and displays them, but the .jpg's are not shown, instead theres just the file name?? Code: [Select] <?php error_reporting(E_ALL); ini_set("display_errors", 1); echo '<pre>' . print_r($_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( $_FILES['upload']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } ?> "upload" is the variable that isnt displaying in my table how i want it to? Have you guys any ideas how to get it displayed correctly? I need help with my webpage here, how do you get so if you are at http://www.blabla.com/account.php and then click on a photo, it will go to www.blabla.com/photo.php?id=1 but still be at account.php? Just like facebook shows their photos. I dont know how to think nor to get it work. Would appreciate some help! ThNXX 1n 4dv4nc3 // Machram! I am looking at making a photogallery system for an online community based site. I know that we need to have unique id's , photo names and albums. What i am looking for is: Would it be recommended to create a directory for each user once they are registerd? How would you go about naming and sorting photogralleries? How would you go about naming you photos? How would you do it? I was thinking about a uuid kind of code: ( from php.net) Code: [Select] class UUID { public static function v3($namespace, $name) { if(!self::is_valid($namespace)) return false; // Get hexadecimal components of namespace $nhex = str_replace(array('-','{','}'), '', $namespace); // Binary Value $nstr = ''; // Convert Namespace UUID to bits for($i = 0; $i < strlen($nhex); $i+=2) { $nstr .= chr(hexdec($nhex[$i].$nhex[$i+1])); } // Calculate hash value $hash = md5($nstr . $name); return sprintf('%08s-%04s-%04x-%04x-%12s', // 32 bits for "time_low" substr($hash, 0, 8), // 16 bits for "time_mid" substr($hash, 8, 4), // 16 bits for "time_hi_and_version", // four most significant bits holds version number 3 (hexdec(substr($hash, 12, 4)) & 0x0fff) | 0x3000, // 16 bits, 8 bits for "clk_seq_hi_res", // 8 bits for "clk_seq_low", // two most significant bits holds zero and one for variant DCE1.1 (hexdec(substr($hash, 16, 4)) & 0x3fff) | 0x8000, // 48 bits for "node" substr($hash, 20, 12) ); } public static function v4() { return sprintf('%04x%04x-%04x-%04x-%04x-%04x%04x%04x', // 32 bits for "time_low" mt_rand(0, 0xffff), mt_rand(0, 0xffff), // 16 bits for "time_mid" mt_rand(0, 0xffff), // 16 bits for "time_hi_and_version", // four most significant bits holds version number 4 mt_rand(0, 0x0fff) | 0x4000, // 16 bits, 8 bits for "clk_seq_hi_res", // 8 bits for "clk_seq_low", // two most significant bits holds zero and one for variant DCE1.1 mt_rand(0, 0x3fff) | 0x8000, // 48 bits for "node" mt_rand(0, 0xffff), mt_rand(0, 0xffff), mt_rand(0, 0xffff) ); } public static function v5($namespace, $name) { if(!self::is_valid($namespace)) return false; // Get hexadecimal components of namespace $nhex = str_replace(array('-','{','}'), '', $namespace); // Binary Value $nstr = ''; // Convert Namespace UUID to bits for($i = 0; $i < strlen($nhex); $i+=2) { $nstr .= chr(hexdec($nhex[$i].$nhex[$i+1])); } // Calculate hash value $hash = sha1($nstr . $name); return sprintf('%08s-%04s-%04x-%04x-%12s', // 32 bits for "time_low" substr($hash, 0, 8), // 16 bits for "time_mid" substr($hash, 8, 4), // 16 bits for "time_hi_and_version", // four most significant bits holds version number 5 (hexdec(substr($hash, 12, 4)) & 0x0fff) | 0x5000, // 16 bits, 8 bits for "clk_seq_hi_res", // 8 bits for "clk_seq_low", // two most significant bits holds zero and one for variant DCE1.1 (hexdec(substr($hash, 16, 4)) & 0x3fff) | 0x8000, // 48 bits for "node" substr($hash, 20, 12) ); } public static function is_valid($uuid) { return preg_match('/^\{?[0-9a-f]{8}\-?[0-9a-f]{4}\-?[0-9a-f]{4}\-?'. '[0-9a-f]{4}\-?[0-9a-f]{12}\}?$/i', $uuid) === 1; } } // Usage // Named-based UUID. $v3uuid = UUID::v3('1546058f-5a25-4334-85ae-e68f2a44bbaf', 'SomeRandomString'); $v5uuid = UUID::v5('1546058f-5a25-4334-85ae-e68f2a44bbaf', 'SomeRandomString'); // Pseudo-random UUID $v4uuid = UUID::v4(); 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! |
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