PHP - Creation Of Image Album Website
I was not sure on which board to put this thread, so please move it if there is a more appropriate board.
I am wanting to create a website that displays images which I have put into a database. I would like the number of images on one page not to exceed 10, for example, and when the number of images in the database exceeds this number, a new page will be created. I would like it to function similar to Google Images where the number of pages is dictated by the number of images in the database. I cant figure out how to achieve this, so any pointers or thoughts would be great Thanks, Matlab Similar Tutorialsi do think is a small mistake i making can u please have a look if u can ok here's the problem i am trying to delete an album within the album should also delete the photos related to that album this what i tried gives this error Delete image failed. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 3 my tables are table albums fields album_id, album_name, album_owner, sub_album table photos fields, photo_id, photo_name, photo_extension, photo_proper photo_owner, photo_date, photo_comments, photo_size, album_id photo_proper is the name image stored in folder <?php define('ROOT_DIR', './'); define('PROPER', TRUE); /** * include common files */ include_once(ROOT_DIR. 'includes/common.inc.php'); // No album id has been selected if (isset($_GET['albums'])) // get the album name since we need to display // a message that album 'foo' is deleted $result = mysql_query("SELECT album_id, album_name, album_owner, sub_album FROM albums WHERE album_id = $album_id") or die('Delete image failed. ' . mysql_error()); if (mysql_num_rows($result) == 1) { $row = mysql_fetch_assoc($result); $album_id = $row['album_id']; $album_name = $row['album_name']; // get the image filenames first so we can delete them // from the server $result = mysql_query("SELECT photo_id, photo_id FROM photos WHERE album_id = $album_id") or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { define("GALLERY_IMG_DIR", "./photos/"); unlink(GALLERY_IMG_DIR . $row['photo_proper']); unlink(GALLERY_IMG_DIR . 'thumbs/' . $row['photo_proper']); } $result = mysql_query("DELETE FROM photos WHERE album_id = $album_id") or die('Delete image failed. ' . mysql_error()); $result = mysql_query("DELETE FROM album WHERE album_id = $album_id") or die('Delete album failed. ' . mysql_error()); // album deleted successfully, let the user know about it echo "<p align=center>Album '$album_name' deleted.</p>"; } else { echo "<p align=center>Cannot delete a non-existent album.</p>"; } ?> try 2 error line 5 <?php define('ROOT_DIR', './'); define('PROPER', TRUE); /** * include common files */ include_once(ROOT_DIR. 'includes/common.inc.php'); // No album id has been selected if (isset($_GET['albums'])) { // get the image file name so we // can delete it from the server $sql = "SELECT album_id, album_name, album_owner, sub_album FROM albums WHERE album_id = {$_GET['albums']}"; $result = mysql_query($sql) or die('Delete photo failed. ' . mysql_error()); if (mysql_num_rows($result) == 1) { $row = mysql_fetch_assoc($result); // get the image filenames first so we can delete them // from the server $sql = "SELECT photo_id, photo_proper FROM photos WHERE photo_id = {$_GET['photos']}"; $result = mysql_query($sql) or die('Delete photo failed. ' . mysql_error()); if (mysql_num_rows($result) == 1) { $row = mysql_fetch_assoc($result); define("GALLERY_IMG_DIR", "./photos/"); // remove the image and the thumbnail from the server unlink(GALLERY_IMG_DIR . $row['photo_proper']); unlink(GALLERY_IMG_DIR . 'thumbs/' . $row['photo_proper']); // and then remove the database entry $sql = "DELETE FROM photos WHERE photo_id = {$_GET['photos']}"; $result = mysql_query("DELETE FROM album WHERE album_id = $album_id") or die('Delete album failed. ' . mysql_error()); // album deleted successfully, let the user know about it echo "<p align=center>Album '$album_name' deleted.</p>"; } else { echo "<p align=center>Cannot delete a non-existent album.</p>"; } } } ?> I am attempting to create an image and I have run into a snag. I want to add both text and another image to the image I am making. Here is the code so far minus unnecessary parts: header ("Content-type: image/png"); $name = $fetch_user['username']; $rank = $fetch_rank['rank_title']; $img_url = 'path''. $fetch_rank['rank_image']; $im = @imagecreatefrompng("$img_url") or die("Cannot Initialize new GD image stream"); // try changing this as well $font = 4; $width = imagefontwidth($font) * strlen($string) ; $height = imagefontheight($font) ; $im = imagecreatefrompng("image.png"); $x = imagesx($im) - $width ; $y = imagesy($im) - $height; $backgroundColor = imagecolorallocate ($im, 255, 255, 255); $textColor = imagecolorallocate ($im, 255, 255, 255); //imagestring ($im, $font, $x, $y, $string, $textColor); imagestring ($im, $font, 100, 10, $name, $textColor); imagestring ($im, $font, 100, 22, $rank, $textColor); imagepng($im); Everything works UNTIL I added in this part: $im = @imagecreatefrompng("$img_url") or die("Cannot Initialize new GD image stream"); Any idea what I am doing wrong? All that comes up is "Cannot Initialize new GD image stream"; and I know the URL's work. I am trying to create a basic image with GD library, but my browser says that the image cannot be displayed because it contains errors... I have no idea why. Here's my code: Code: [Select] switch($_GET['case']){ case "progressbar": header('Content-Type: image/jpeg'); $barWidth = 6; $barHeight = 14; $barPadding = 2; $imageWidth = ($barWidth * 10) + (10 * $barPadding); $imageHeight = ($barHeight + ($barPadding * 2)); $img = @imagecreate($imageWidth, $imageHeight); $imgBorder = imagecolorallocate($img, 0, 0, 0); //imagefilledrectangle($img, 1, $imageHeight -1, $imageWidth - 1, $imageHeight - 1, $imgBorder); $output = "Hello, world!"; imagestring($img, 1, 4, 4, $output, $imgBorder); imagejpeg($img); imagedestroy($img); break; } I have a file called image.php that i use to create images and I use the URL image.php?case=progressbar. Hey guys I have a small issue, i have an upload that resizes the image into thumbnail by max width and ratios the width based on that. Here is my code
What I am wanting to do is instead upload the image with a max height and ratio the width proportionally. What variables do I have to reverse? This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=326412.0 Hello, There is something wrong with my statement it is not returning sql queries results based on matching the url based based userid to match with the table user_id and get the Ablum id and Album name: Please hlep: Code: [Select] if(isset($_GET['userid'])) { $db =& JFactory::getDBO(); $user =& JFactory::getUser(); $id = $user->id; $album_id = $_GET['userid']; $query = 'SELECT user_id, id, format_id, year, name FROM #__muscol_albums WHERE user_id = ' . $album_id; //$query = 'SELECT user_id FROM #__muscol_albums WHERE id = ' . $album_id ; $result = mysql_query($query) or die('Error, No Album Search failed'); list($name, $user_id, $id, $year) = mysql_fetch_array($result); echo $id; //exit; } I'm wanting to extract an image from an external website and save to a location on my web server automatically. The scenario: 1. A user has an order form on our website with a field to paste the URL of a website in; 2. The user visits an external website page, which includes an image; 3. The user copies the URL out of the address bar, goes back to our website and pastes the link into the order form; 4. The user clicks next, and this extracts the image from the external website and uploads it to our website so the image can be seen alongside the order. My questions a a. Can this be done? b. If it can't physically download the image and copy to my web server, can I read the html source of the external website and grab the full URL to use to show the image on my website? Thanks for your help. Whats is the best way to create a forum using php? Should I use any frameworks? If not is there any guide/tutorial/book that descibe this procedure? Thank you. i have mysql community server 5.5 workbench 5.2 php 5.3.8 apache http server 2.2 summary: can i write the same code using less variables for the file name and content? okay so i managed to write a code that creates a filename, and each new filename is 1 higher. colony1 colony2 colony3. it also uses the same mysql query created variable to write idcol as 1 higher in the new file than in the previous written file. however to write the code i had to use a buttload of variables. here is the code, the resultant filename and the resultant content of the created file. is there a way to write such a code with less than 15 variables? the problem seems to be that when writing text to a file, i cannot insert certain characters outside of a variable into the final variable which is put in the $string variable input into the fwrite() function. File Name: Colony$.php where $=the last entry in the idcol column in table coltest. File Code: Code: [Select] <?php $idcol = $; include('Colony0.php'); ?> </body> </html> where $=the last entry in the idcol column in table coltest. Code: Code: [Select] <?php $dbhost = 'localhost:3306'; $dbuser = 'root'; $dbpass = 'root'; $dbname = 'aosdb'; $conn = mysql_connect($dbhost,$dbuser,$dbpass) or die ('Error connecting to mysql'); mysql_select_db($dbname); $query = "SELECT idcol FROM coltest ORDER BY idcol DESC LIMIT 1"; $result = mysql_query($query); $id = mysql_result($result, 0); $vcol = "Colony"; $vcolp = ".php"; $File = "$vcol$id$vcolp"; $FileHandle = fopen($File, 'w') or die("can't open file"); fclose($FileHandle); $File = "$vcol$id$vcolp"; $myFile = $File; $fh = fopen($myFile, 'w') or die("can't open file"); $idw = "$"; $icol= "idcol"; $ique = "?"; $iphp = "Colony0.php"; $ief = "<"; $ieb = ">"; $irese = "= "; $iresn = ";"; $iii= "include('$iphp');"; $ihtml ="</body> </html>"; $hph = "php"; $string = "$ief$ique$hph $idw$icol $irese$id$iresn $iii $ique$ieb $ihtml"; fwrite($fh, $string); fclose($fh); ?> With the following code i can create an xml file. Code: [Select] <?php $myXML = new SimpleXMLElement("<myroot></myroot>"); $title= $myXML->addChild('title'); $title->addAttribute('number','12'); $titleName= $title->addChild('titleName', 'title1'); $titleLink= $title->addChild('titleLink', 'link1'); Header('Content-type: text/xml'); echo $myXML->asXML(); ?> But when i check the validity of xml file http://www.validome.org/xml/validate/ This error occurs: Can not find declaration of element 'myroot'. I suppose that the problem is missing of !DOCTYPE and !ELEMENT lines. How can i create valid XML documents with PHP automatically?? Is it possible to make it without writing doctype and element types for the whole element types of xml by hand $title, $titleName and $titleLink Thank you Hello guys, I have created a mini image hosting website. Well, I have successfully coded the file upload, including security to allow certain image extensions and size as a beginner in PHP. However, only one thing remains is the image link. You can view the website on this address http://mini-image-hosting.99k.org/ where it is currently hosting on a free web hosting account with a free sub-domain. Right now, only the image can be uploaded and is being stored in a directory. Nevertheless, I want that when the person uploads an image, he gets also the link, for example: http://mini-image-hosting.99k.org/xxx.jpg something like that. Can you help me for this? I am not sure what to look for to even get started. If some one could help point me in the right direction that would be appreciated.
I created a customer system that tracks contracts and invoices ect. Where do I start if I would want a customer to be able to go to the website, sign up, and their own version of the system would be created.
Do I just write a script that creates a new database with the correct credentials? Do all of the users access the same files but just have different database. Do I have to copy the directory of files to a specific folder for each customer?
I know there must be tons of websites that once you sign up for example a calendar app that keeps your data seperate from everyone else.
Any help is appreciated.
Thanks
Jack
Hi, I am wanting to check to see if a folder exists, if not create it. PHP on a windows machine so I would like to know how to handle the folder separators. I am wanting to use a path which includes the drive letter, such as... $foldername ="C:\folder1\images\$checkfolder" I was thinking about using something like this: if(!is_dir($foldername)) mkdir ("$foldername",777); What would be the best way of tackling this using windows paths? Hello, I was thinking of using this image upload script on my website: http://www.white-hat-web-design.co.uk/blog/resizing-images-with-php/ The question I have are the following: - 1. Do I need to take in consideration the $type="image/pjpeg" on this script or would it work anyway? - 2. When is the pjpeg actually used, is it only on certain PC browsers. Where can I find info about this? Thanks, df Day
I'm using Codeigniter and inside the framework I create a file with PHP code If I try to point to it, the web-server doesn't deliver due to permission issues.
All this is done in a 'register function' where I create some fields in a SQL
If I put the user credentials into the database 'by hand' and likewise when I
I'm pretty sure this issue has been existed before and I'd appreciate any help.
Thanks for your help in forward. Gee Edited August 17, 2020 by bogusI've been stuck on this for about a week so I was wondering if I can get some help from you guys! Basically I'm exporting an array of values to a CSV file, within that array are two other arrays containing data I need. What I need to work out is how I can create a CSV with headers for these values, then with the nested arrays also print their values into the same rows as the first array. Example: $List = array ( 'Product ID' => '10', 'Customer Address' => '123 Fake Street', array( array('Product Name'=>'Product1', 'Product Price'=>'10.00', 'Product Reference'=>'HGJEN'), array('Product Name'=>'Product2', 'Product Price'=>'5.00', 'Product Reference'=>'HGJTN'), array('Product Name'=>'Product3', 'Product Price'=>'10.00', 'Product Reference'=>'HGJNN'), ), array( array('Product Customisation Name'=>'Additional Info', 'Customisation Value'=>'Things are great.'), array('Product Customisation Name'=>'Image Upload', 'Customisation Value'=>'Logo.jpg'), ), 'Telephone Number'=>'999', ); To be exported to something looking like this: Product ID Customer Address Product Name Product Price Product Reference Product Cus Name Cus Value Telephone Number 10 123 Fake Street Product2 5.00 HGJTN Additional Info Things are great 999 10 123 Fake Street Product1 10.00 HGJEN Image Upload Logo.jpg 999 10 123 Fake Street Product3 10.00 HGJNN 999 If anyone could help me out in anyway I would be so appreciative, I'm sure this ones going to end up killing me! Hi All I am trying to insert a feed on my site that automatically updates the football league etc like this one http://news.bbc.co.uk/sport1/hi/football/eng_div_1/table/default.stm is there a way to do this please? thanks Hi So I have successfully set up a website and database where a user can create a listing and view listings on a listing details page. The viewing can only be done, however, only from either from either running a search query from the search form or from clicking one of the listings from the listing table. The listings are not being indexed from the search engines from the direct page, instead from the listing table which is about 3,000 listings, not very user friendly to land on this page and have to go through the table or use the table filter. Here's what I am trying to accomplish: A web page that has a dynamic presence / URL like a Wordpress page: http://www.mysite.com/listing_title/ The title of the page has the listing information in it: This page is about a $listing_title $listing_item listing number $listing_number that is online! Each individual Listing / page is indexed on the search engines and has a direct link to it I am familiar with PHP and have created this site but I am far from an expert! Please give me a code example or link to one because if you just say "do this to this and that" you will totally loose me. Thank you so much for any help. |