PHP - Can't Dynamically Create Images
We recently migrated over to a new AMP platform that I built from source and now we can't dynamically generate images. It seems like GD is properly included into PHP (although I didn't know then I could do it using dynamic modules):
My PHP configure is as follows: ================================================== './configure' '--prefix=/export/appl/pkgs/php/v5.3.9' '--with-apxs2=/export/appl/pkgs/httpd/latest/bin/apxs' '--with-mysql=/export/appl/pkgs/mysql/latest' '--with-mysqli=/export/appl/pkgs/mysql/latest/bin/mysql_config' '--with-gd=/export/appl/pkgs/libgd/latest' '--with-pear' '--with-png-dir=/export/appl/pkgs/libpng/latest' '--with-jpeg-dir=/export/appl/pkgs/jpeg/latest' '--with-curl=/export/appl/pkgs/curl/latest' '--with-freetype-dir=/export/appl/pkgs/freetype/latest' '--with-mhash=/export/appl/pkgs/mhash/latest' '--with-mcrypt=/export/appl/pkgs/libmcrypt/latest' '--enable-pcntl' '--enable-soap' '--enable-mbstring' '--with-zlib-dir=/export/appl/zlib/v1.2.5' '--with-ldap' ================================================== The GD section in phpinfo is as follows: ================================================== GD Support enabled GD Version 2.0 FreeType Support enabled FreeType Linkage with freetype FreeType Version 2.4.6 GIF Read Support enabled GIF Create Support enabled JPEG Support enabled libJPEG Version unknown PNG Support enabled libPNG Version 1.5.5 WBMP Support enabled Directive Local Value Master Value gd.jpeg_ignore_warning 0 0 ================================================== I get broken image icons in IE and Chrome, and I get an "image cannot be displayed" msg in FF. I'd appreciate someone clubbing me with a cluestick. - Joe Similar TutorialsI have a slideshow that I am having to modify for my needs. The data file uses an array and i need this to be dynamic. I need to loop through all images in a directory and create the array. I have the directory being stored in a variable already. Now i just need to loop though all images in that directory to create the array like you see in the below code. You can also see the directory variable at the top. Can someone give me a hand with this please? Code: [Select] <?php echo $slideimagefolder; //http://localhost:82/images/groupphotos/1/1/ $slides = array( // --- Start slide list --- // slide elements array( "slidelink" => "http://www.frontpageslideshow.net", "title" => "Frontpage Slideshow 1.7", "category" => "About Frontpage Slideshow", "tagline" => "Image taken from the movie \"Shoot 'em up\"", "text" => "\"Frontpage SlideShow\" is the most eye-catching way to display your featured articles, stories or even products in your php based website or CMS, like Time.com, Joost.com or Yahoo! Movies do. \"Frontpage SlideShow\" creates an uber-cool slideshow with text snippets laying on top of images. These \"slides\" are being rotated one after the other with a nice fade effect. The slideshow features navigation and play/pause buttons, so the visitor has complete control over the slideshow's \"playback\"! And best of all, Frontpage Slideshow can be skinned!", "slideimage" => "ao4.jpg" ), // slide elements array( "slidelink" => "http://www.frontpageslideshow.net", "title" => "Use Frontpage Slideshow on any PHP based site!", "category" => "About Frontpage Slideshow", "tagline" => "Image taken from the movie \"The Kingdom\"", "text" => "JoomlaWorks has developed this modification of Frontpage Slideshow to work on every website that supports PHP as a minimum requirement. We call this modification the \"Static PHP\" version of Frontpage Slideshow. It's ideal for use on non-Joomla!/Mambo websites, like for example your corporate PHP based website or your Wordpress blog or Drupal website! You can obviously use this version on any CMS that is based on PHP!", "slideimage" => "the_kingdom_20070820114258369.jpg" ), // slide elements array( "slidelink" => "http://www.frontpageslideshow.net/content/view/14/37/", "title" => "FPSS is Search Engine Friendly!", "category" => "About Frontpage Slideshow", "tagline" => "Image taken from the movie \"Invaders\"", "text" => "Unlike Flash based slideshows, Frontpage Slideshow uses unobtrusive javascript and some CSS wizardry only. The content of the slides is laid out as html code, which means it can be \"read\" by search engines. The proper usage (and order) of h1, h2, p (and more) tags will make sure Google (or any other search engine) regularly \"scans\" your latest/featured items.", "slideimage" => "TVD_TR_02183_2.jpg" ), // slide elements (***TIP***: copy this data block and paste it below itself to add more slides) array( "slidelink" => "http://www.joomlaworks.gr", "title" => "About JoomlaWorks", "category" => "", "tagline" => "Image taken from the movie \"Transformers\"", "text" => "JoomlaWorks is a team of professional web designers and developers dedicated to delivering high-quality extensions and templates for Joomla! (the most popular open source Content Management System (CMS) worldwide) and Mambo (award winning CMS). JoomlaWorks has established a solid reputation in the Joomla! & Mambo communities, having developed some of the most popular and innovative free & commercial extensions & templates, since 2006.", "slideimage" => "2007_transformers_014_1.jpg" ), // --- End slide list --- ); ?> Friends, I have a PHP script that populates a calendar based on an array as follows: $days = array(2=>array('javascript:showWhatsOnText(2);','linked-day'), 3=>array('javascript:showWhatsOnText(3);','linked-day'), 4=>array('javascript:showWhatsOnText(4);','linked-day')); echo generate_calendar(2011, 4, $days, 1, '#'); where, for instance, the 2, 3 4 are the days of the month, By extracting a series values from a MySQL database how would I generate this array dynamically. E.g I have startday, startmonth, name and text fields in the database and would need to loop through them all to create the array. I tried playing around with something like this without any luck: $queryCat1 = "SELECT * FROM $dbCalendarEvents ORDER BY id ASC"; // WHERE visible = '1' $result1 = mysql_query($queryCat1) or die (mysql_error()); $num1 = mysql_num_rows($result1); $i=0; $daysArray = array(); while ($i < $num1) { $startday=mysql_result($result1,$i,"startday"); $startmonth=mysql_result($result1,$i,"startmonth"); $name=mysql_result($result1,$i,"name"); $name=mysql_result($result1,$i,"name"); $copy=mysql_result($result1,$i,"copy"); $daysArray[$i] = $startday=>array('/weblog/archive/2004/Jan/02','linked-day'); $i++; } Your help is appreciated. Hi, I want to create a .html file dynamically on the server. Below is the code for that but i was not able to create the file. I have given all the read, write and execute permission on the folder. $filerand = rand(); $fileName = "test.html"; $filepath = $_SERVER['DOCUMENT_ROOT'] . "\\testfolder\\" . $fileName . ".html"; echo $filepath; $bodytxt = "Welcome to my webpage"; $ourFileHandle = fopen($filepath, 'w') or die("can't open file"); fwrite($ourFileHandle, $bodytxt); fclose($ourFileHandle); Thanks in advance Hi guys i have to create text field & enter data and store in the data base. here im able to create text field but couldn't insert the data. so could anyone please check this code for me. <body> <form method="POST" action="cell.php"> Enter the number of question <input type="text" name="question"> <input type="submit"> <?php $value=$_POST['question']; for($i=0;$i<$value;$i++) { echo "Question NO:"; echo '<input type="text" name="qno">'."\t"; echo "Enter Marks:"; echo '<input type="text" name="marks">'."\t"; echo "<br>"; } ?> </form> <form name="form1" method="post" action="cellresult.php"> <label> <input type="submit" name="Submit" value="Submit"> </label> </form> </body> cellresult.php <body> <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("test", $con); $sql="INSERT INTO cell (QNO,MARKS) VALUES ('$_POST[qno]','$_POST[marks]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> </body> Hi, i coded a very simple forum "website" with registration, login, and a place where everyone posts comments. I wanted to make some sort of profile page for each user so they would each have their own URL too (e.i. test/barney or test/simpsons). each of these pages would have different content, depending on what barney or simpsons puts on that page. i've been looking everywhere trying to find documentation but i don't think i know how to search it correctly. I checked a lot of mod_rewrite documentation but i don't understand if i'm suppose to call a php function to create a profile page or something. Any guidance would be greatly appreciated Thanks! Hi, This problem has been driving me crazy all day. I am relatively new to PHP. I basically am trying to populate a database with data from site users. I am using session variables to store their data temporarily as they navigate through the sign up process. A user will input how many 'categories' they wish to populate on page 1. Page 2 will then ask them to specify the details of each category. Eg Category 1: Title, Description, Amount. Category 2: Title, etc. So far I have been able to do this, I now want to store what they have input in session variables. My thoughts were to take the number of categories they have sepcified and create that number of arrays using a loop. Each array will store the details on each category. My code is as follows: Code: [Select] $count=$_POST['count']; //Get how many categories were added //Create an array for each category for ( $counter = 0; $counter <= $count; $counter++){ ${'Categoryarr'.$counter} = array(); }; for ( $counter = 0; $counter <= $count; $counter++){ $Categoryarr[$counter][1]=$_POST['amount_'.$count]; $Categoryarr[$counter][2]=$_POST['desc_'.$count]; $Categoryarr[$counter][3]=$_POST['title_'.$count]; }; When I output the code, I seems to have created the specified number of arrays, but has populated all of them with the same data from the last category. Does anyone know where I am going wrong? Thanks, Bernard How to create a pdf dynamically and that pdf should sent as a email attachment.
Hi, I have a table row that has a dropdown and two textboxes in it. I would like to use a button that allows me to add another row beneath the existing row. It cannot be added to the bottom of the table as there is further content beneath. The content of the dropdown comes from a database query. The dropdown on the new row should have the selection in the first row greyed out. I expect this will need to be a js function but i dont really know where to start with it being dynamic content. This is the table as it stands if this provides clarity of my intentions - picture below echo "<form method='post' id='staffOrderForm'>"; $stmt->close(); echo "<table id='staffOrderTable' class='table table-striped table-bordered mt-3 text-center'>"; foreach($daterange as $date){ echo" <tr> <th class='table-dark' colspan='3'>".$date->format("l - jS F Y")."</th> </tr> <tr> <th class='' colspan='3'>Management </th> </tr> <tr> <th class='col-4'>Name</th> <th class='col-4'>Start Time</th> <th class='col-4'>End Time</th> </tr> <tr> <td> <select class='custom-select managerSelect'>"; $managerRoleId = 1; $stmt = $conn->prepare(" SELECT user_firstname, user_lastname, user_id FROM ssm_user WHERE user_role_id = ? "); $stmt->bind_param("i", $managerRoleId); $stmt->execute(); $stmt->store_result(); $stmt->bind_result($ufn, $uln, $uid); while($stmt->fetch()){echo "<option>".$ufn." ".$uln."</option>";}; echo"</select> </td> <td><input class='form-control' type='' name=''></td> <td><input class='form-control' type='' name=''></td> </tr> <tr> <th colspan='3'>Chefs</th> <tr/> <tr> <th class='col-4'>Name</th> <th class='col-4'>Start Time</th> <th class='col-4'>End Time</th> </tr> <tr> <td> <select class='custom-select chefSelect'>"; $chefRoleId = 2; $stmt = $conn->prepare(" SELECT user_firstname, user_lastname, user_id FROM ssm_user WHERE user_role_id = ? "); $stmt->bind_param("i", $chefRoleId); $stmt->execute(); $stmt->store_result(); $stmt->bind_result($ufn, $uln, $uid); while($stmt->fetch()){echo "<option>".$ufn." ".$uln."</option>";}; echo"</select> </td> <td><input class='form-control' type='' name=''></td> <td><input class='form-control' type='' name=''></td> </tr>"; } echo "</table></form>"; I appreciate that i will likely have to add a button with an onclick but from there i am pretty lost. As always i appreciate the help provided. hey i am using a MySql database and i need to create a dynamic HTML table with one of its columns as checkboxes.so i have to create multiple checkboxes.but these checkbox values are to be stored in a mysql table and then later retrieved when form reloads.and depending on previous state when form was submitted, the newly created checkboxes have to be checked in the same manner.so how do i store multiple checkbox values in my table and also how do i retrieve them? please help. Hi there, I hope somebody can help me creating a loop for the following code: What it currently does is, showing only 1 image, while in fact there are 3 of them. I've combined a loginsystem with slideviewer. When i use the original php code (or a part of it) it only shows one listing. I need it to list all of them. <div id="mygalone" class="svw"> <ul> <li> <?php $result = mysql_query("SELECT reference FROM user_photos WHERE`profile_id`='".$row['id']."'"); while ($row2 = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<a href=\"".$_GET['username']."/pics/".$row2['reference']."\"> <img src=\"".$_GET['username']."/pics/thumbs/".$row2['reference']."\"></a><br/><br/>"; } } ?> </li> </div> What can be stripped down to this: <?php { $result = mysql_query("SELECT reference FROM user_photos WHERE`profile_id`='".$row['id']."'"); while ($row2 = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<img src=\"".$_GET['username']."/pics/".$row2['reference']."\">"; } } ?> You would asume that this is allready a loop, but it does not act like it. Best regards, Martijn (the netherlands) Hello,
I'm embarking on a pretty ambitious task and I need some bits of information here and there.
One of the functions I need to achieve is to build a query or array of photos based on a background image and user input.
So imagine that I have a box and within that box is a column of three rows.
I need to have three different bits of data be placed into each of the subsequent rows and then an image is taken of these three pieces of data overlaying the background photo.
Then stored somewhere with an incremented identifier to be pulled later.
I think I can already begin to imagine how it would work but what eludes me is a "screenshot" function to generate the images. I'm looking for .png or .jpg end result files with fixed width/height and item placement.
Thank you for any help
Hello everyone, I am working on a form that is similar to a shopping cart system and I am thinking of creating a button that submits the checked value and saves them to a $_SESSION variable. And also a link that links to a cart.html that takes the values of a $_SESSION variable. I am have trouble figuring what tag/attribute should I use in order to achieve that.
Right now my code attached below submits the checked values to cart.html directly. However I want my submit button to save the checked box to a $_SESSION variable and STAY on the same page. And then I will implement a <a> to link to the cart.php.
I researched a little bit about this subject and I know it's somewhat related to ajax/jquery. I just wanted to know more about it from you guys. I appreciate your attention for reading the post and Thanks!
Below is the form that I currently have:
<form name= "finalForm" method="POST" action="cart.php"> <input type="Submit" name="finalSelected"/> <?php foreach($FinalName as $key => $item) {?> <tr> <td><input type="checkbox" name="fSelected[]" value="<?php echo htmlspecialchars($FinalID[$key])?>" /> <?php echo "$FinalID[$key] & $item";?> </td> </tr> <?php } ;?>Below is the code for cart.php <?php require ('connect_db.php'); if(isset($_POST['finalSelected'])) { if(!empty($_POST['fSelected'])) { $chosen = $_POST['fSelected']; foreach ($chosen as $item) echo "aID selected: $item </br>"; $delimit = implode(", ", $chosen); print_r($delimit); } } if(isset($delimit)) { $cartSQL = "SELECT * from article where aID in ($delimit)"; $cartQuery = mysqli_query($dbc, $cartSQL) or die (mysqli_error($dbc)); while($row = mysqli_fetch_array($cartQuery, MYSQLI_BOTH)) { $aTitle[] = $row[ 'name' ]; } } ?> <table> <?php if(isset($delimit)) { $c=0; foreach($aTitle as $item) {?> <tr> <td> <?php echo $aTitle[$c]; $c++;?> </td> </tr> <?php }}?> </table> I 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
Hi, This line of code works just fine for me: $sql_quest = 'SELECT id, username FROM user ORDER BY id ASC LIMIT 0, 30'; I want to be able to set the integer values dynamically and I thought something like this would work: $sql_quest = 'SELECT id, username FROM user ORDER BY id ASC LIMIT' . $my_int_value . ',' . $my_int_value + 30; But it doesn't... Any ideas? I've searched around the internet and can't find a good method to do this. To start, I have a database with a field called 'fpath' that stores the filepath (root-relative) to uploaded PDF's. In this case, they're scanned personnel files. What I'd like to do is be able to download relevant PDF's based on a generated report... such as all PDF's of a certain category or all PDF's for a certain user. If there's a specific PEAR package or other script out there that you know works well, I can research it on my own.. but each search result is bringing up a different method to do it, which makes me nervous. Thanks! Is there a way to dynamically print the url of a web page once it loads? If so, how? This is for metadata purposes. Thanks! Hello - I'm (very) new to php and working on a project and wondering if anyone can point the direction? (Not necessarily give the answer, but any helpful direction). I need to take YouTube video id's, (built into an array), and post the various video clips. Also the <object> code is in an php array. (Code Below.. hard to explain). Do I need to convert the array text into a string (as shown), and how do I take that string text and paste it into multiple youtube object posts? Note that the original $embedCode is commented out, '//), so that it can be referenced. Any advice is appreciated! <?php $number = 0; $youtubeIds = array("MX0D4oZwCsA","2v-v3jh-Cco","WuYDSa4BRaw","zwo48StJSr4","exOxUAntx8I"); // $embedCode = '<object width="480" height="385"><param name="movie" value="http://www.youtube.com/v/__YOUTUBE_ID__?fs=1&hl=en_US"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/__YOUTUBE_ID__?fs=1&hl=en_US" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="385"></embed></object>'; $string = implode($youtubeIds); $embedCode = '<object width="480" height="385"><param name="movie" value="http://www.youtube.com/v/'<?php $string?>'?fs=1&hl=en_US"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/'<?php $string?>'?fs=1&hl=en_US" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="385"></embed></object>'; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en" xmlns:fb="http://www.facebook.com/2008/fbml" xmlns:og="http://opengraphprotocol.org/schema/"> <head> <title>Youtube Videos</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <meta name="language" content="en" /> </head> <body> <h1>Youtube Videos</h1> <p><?php for($i = 0; $i < count($youtubeIds); $i++) { echo $embedCode; } ?> </p> </body> </html> Hello
I am using fpdf to generate pdf files based on certain csv files. Now the client wants certain fields of the generated pdf file to be 'interactive' .
Is this possible ? Googling about editable (interactive pdf's) only confused me . any help ?
btw by interactive I mean fields that can be typed into directly by the client and then saved into the pdf .
Hi there, so recently I have been trying to dynamically set the title on a page with PHP I have tried Code: [Select] <title><?ech"$username";?>'s Profile</title> the result is "'s Profile" note that the username variable is being passed thru a file that is required. Any help is appreciated I am trying to set the HTML Page Title dynamically, but this code from my book doesn't work... <title> <?php // Dynamically set Page Title. if (isset($page_title)){ echo $page_title; } else { // Default Page Title. echo 'Knowledge is Power: And It Pays To Know'; } ?> </title> What is wrong with it? (All I see is the code in the Webpage/Window Title...) TomTees |