PHP - Dynamically Create,store And Retrieve Values From Multiple Checkboxes
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. Similar TutorialsHi everyone, I am having trouble passing/displaying the values inside of a selected list. I created a add/remove list using Jquery and I tried to display the values passed using foreach and for loops but it is still not working. The values I am trying to get are $existing_mID[$j], which is inside of the option value attribute. Please kindly let me know what should I do in order to get the values and I really appreciate your help. <?php $selected = $_POST['selectto']; if(isset($selected)) { echo "something in selected<br />"; for ($i=0;$i<count($selected);$i++) echo "selected #1 : $selected[$i]"; foreach ($selected as $item) echo "selected: item: $item"; } ?>This is the form Hello people, I have created an application in Code Charge where my site users can login and play games online.The game code is run via flash and html. The Game has 5 stages and from a page in my application called transfertogame.php , i transfer them to the game page. Now i want to be able to track each players progress via database so that when they log back in and they get to the transfertogame.php page, it can check the page they were on before and immediately take them to the page without starting from page 1. Any help will be appreciated as i am very poor with php scripting This is the session ID i guess is being used from my application page //Initialize Method @4-537EA73F function Initialize() { if(!$this->Visible) return; $this->ds->Parameters["sesUserID"] = CCGetSession("UserID"); } //End Initialize Method //Validate Method @4-7E1FC38C function Validate() { $Validation = true; $Where = ""; $this->CCSEventResult = CCGetEvent($this->CCSEvents, "OnValidate"); return (($this->Errors->Count() == 0) && $Validation); } //End Validate Method Hello guys, I am a new programmer and i am building a new website. I would like to give me your advice for the following problem: I want to build a webpage, in which there will be a <div id="book-content"> ...............</div> part. Inside this div i would like to dynamically display pages from a book. Each page will have text, scripting code blocks, blocks with the output of each scripting code, and images. There will be a bar on the left of the webpage in which the user can select which page of the book he wants to load. For example...My web page will look something like this... Code: [Select] <HTML> <HEAD> </HEAD> <BODY> <DIV ID="PAGE-HEADER"> //LOGO OF THE WEB PAGE </DIV> <DIV ID="PAGE-MENU"> //PAGE MENU </DIV> <DIV ID="BOOK-INDEX" WITH FLOAT:LEFT > //HERE A WILL SHOW THE CHAPTERS OF THE BOOK AND THE PAGES OF EACH CHAPTER </DIV> <DIV ID="BOOK-PAGE"> //THE PAGE SELECTED FROM THE PREVIOUS 'BOOK-INDEX' MENU WILL BE DISPLAYED HERE WITH A MYSQL QUERY ****** </DIV> </BODY> </HTML> Then in a mysql database, i would like to have records with a text field, that will contain for example the followng: <h1> Chapter 1: bla bla </h1> <p> in this chapter we will speak about bla bla bla.... </p> <div id="code"> int main() { int x,y; x=2; y=3; x=x+y; } </div> <p> this will outpout the following:</p> <div id="code output"> x=5! </div> I would like to get this html code from the database, and then show it in the <div id=BOOK-PAGE> div. But i dont want to use php eval(). Also, if i store the code to a file and then include it, i will have too may files(equal to the book's number of pages etc 100). Any ideas? Hi, I want to be able to let user type in xml text and it will be parsed and uploaded to db, but it doesn't work, it just keeps redirecting me back to this form below! here is html form: Code: [Select] <html> <body> <form method='post' method='form.php'> <p> <textarea name="pastedXML" rows="10" cols="30"> Please paste your xml file here. </textarea> </p> <p> <input type="submit" value="Convert to SQL" name="textXML" /> </p> </form> </body> </html> Here is script (form.php;I just want to retrieve the contents typed in text area to store to variable...how??) Code: [Select] <?php //get the text in textarea and shred it! if(isset($_POST['textXML'])) print $_POST['pastedXML']; ?> Any help much appreciated! Hi, my PHP code currently outputs the results from a users search by querying a backend postgresql database. As result/row is retuned to the user i would like to be able to detect whether or not they have checked a checkbox at the end of each rown in which case multiple rows can then be deleted from the database upon the user clicking a 'delete multiple records' button. I have no problem in being able to display the checkboxes on the webpage but i am a little unsure as to how to refernce them and detect which ones the user has checked. Given my database stores hostnames/IP addresses would it be best to name each checkbox to reflect the hostname or name the value it returns when checked to reflect the hostname. Given the above and that I also store the results of a users search in a $_SESSION['hosts'][][] array how can I then tie the two together? Thanks for reading. Hello all! I have a page that dynamically generates checkboxes in a for() loop if $i<$product_qty, while it is generating the checkboxes the name for them is set like this $product_id_$i. In part of the next page that processes the checkboxes, I have a part that regenerates the names for those checkboxes using a for() loop again and the product_qty and then checks to see whether the checkbox is empty or not. Everything looks right in the for() loop, yet the if() that checks whether it isset or not just ignores the ones that are set. If I make a static if with the actual name of the checkbox the if() comes back as true. I've hit a brickwall, can anyone see any errors in my code? Code is below. // Get Old Order $get_order = @mysql_query("SELECT * FROM orders WHERE order_id = {$_POST['order_id']}"); $order = @mysql_fetch_assoc($get_order); // Get Old Order Items $products = $order['products']; //breaking products text down for display $prod = array(); $_products = explode('|', $products); foreach ($_products AS $p) $prod[] = explode(',', $p); if(empty($prod)) { header("Location: tracking.php"); die(); } /* // Create New Order @mysql_query("INSERT INTO orders SET customer_id = {$order['customer_id']}, order_status = {$order['order_status']}, order_date = '{$order['order_date']}', order_date_paid = '{$order['order_date_paid']}', order_shipping = '{$order['order_shipping']}', order_shipping_fee = '{$order['order_shipping_fee']}', order_insurance = '{$order['order_insurance']}', order_insurance_fee = '{$order['order_insurance_fee']}', order_insurance_total = '{$order['order_insurance_total']}', order_grand_total = '{$order['order_grand_total']}', order_date = '{$order['order_date']}', order_filled = '{$order['order_filled']}', order_ship_date = '{$order['ship_date']}'"); $get_new_order = @mysql_query("SELECT MAX(order_id) AS order_id FROM orders"); $new_order_id = @mysql_result($get_new_order, 'order_id', 0); */ // Add Items to New Order & Remove Items from Old Order $new_items = array(); foreach($prod as $p2) { for($i = 0; $i < $p2[0]; $i++) { if(!empty($_POST[$p2[3].'_'.$i])) { $new_items[$p2[3]]++; } } } if(isset($_POST['50_4'])) { echo "hi"; } TIA! Jonathan Ok so i want to grab an id from the checkbox then grab the option drop down associated with that check box and update a mysql row here is my code so far any help is awesome help taaaanks guys <?php mysql_connect("localhost", "root", "root") or die(mysql_error()); mysql_select_db("db1") or die(mysql_error()); $query = "SELECT * FROM tickets ORDER BY id DESC"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)) { print "<input name='delete[]' value='{$row['id']}' type='checkbox'>"; mysql_connect("localhost", "root", "root") or die(mysql_error()); mysql_select_db("db1") or die(mysql_error()); $query2 = "SELECT * FROM admin"; $result2 = mysql_query($query2) or die(mysql_error()); print "<form name='namestoupdate' method='post' action='update.php'>\n"; print '<select>'; while($row2 = mysql_fetch_array($result2)) { if ($row2['priv']==prov){print '<option value="'.$row2['user'].'" name="prov['.$i++.']">'.$row2['user'].'</option>';} } print '</select>'; print "<input type='submit' value='submit' />"; print "</form>"; } ?> Visual aid Hi, I have a database and I need to generate xls file selecting only few columns. The following should happen on click of a trigger like button/link: 1) 6 columns from mysql database table are selected. The first row of xls is the field name which is NOT the same as mysql table column name (xls columns are explicitly defined) 2) This file is generated and saved inside filesystem on remote server. 3) Link to this file is displayed which can be downloaded from server. Please help how to do this. I have the code for creating csv files but that doesnt save the file on server, just shoots it to browser to download. function createCSV() { include("dbconnect.php"); $query = "SELECT `EmailAddress` , `Name` , `FirstName` , `LastName` FROM directtable;"; $rsSearchResults = mysql_query($query) or die(mysql_error()); $out = ''; $fields = mysql_list_fields($dbDatabase,'directtable'); $columns = mysql_num_fields($fields); // Put the name of all fields for ($i = 0; $i < $columns; $i++) { $l=mysql_field_name($fields, $i); $out .= '"'.$l.'",'; } $out .="\n"; // Add all values in the table while ($l = mysql_fetch_array($rsSearchResults)) { for ($i = 0; $i < $columns; $i++) { $out .='"'.$l["$i"].'",'; } $out .="\n"; } // Output to browser with appropriate mime type, you choose ;) header("Content-type: text/x-csv"); //header("Content-type: text/csv"); //header("Content-type: application/csv"); $now=date("d-m-Y,H:i:s", time()); header("Content-Disposition: attachment; filename=emails_".$now.".csv"); echo $out; } I AM WORKING ON A PROJECT , I NEED YOUR HELP. I WANT TO USE EXISTING SRC OF IFRAME + RETRIVED VALUE FROM DATABASE INTO IFRAME This is my data.php to retrive values from database into iframe without refresh web page.
<?php $conn = new mysqli('localhost', 'root', '', 'x'); if ($conn->connect_error) { die("Connection error: " . $conn->connect_error); } $result = $conn->query("SELECT number1 FROM users"); if ($result->num_rows > 0) { while ($row = $result->fetch_assoc()) { echo $row['number1'] . '<br>'; } } ?>
<?php include('data.php') ?> <html> <head> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> </head> <body> <div id="show"></div> <script type="text/javascript" src="jquery-3.1.1.min.js"></script> <script type="text/javascript"> $(document).ready(function() { setInterval(function () { $('#show').load('data.php') }, 1000); }); </script> <iframe id="w3" src="DATABASE RETRIVED VALUE" height="80px" width="300px" frameborder="0" style="border: 0; width:300px; height:80px; background-color: #FFF;"></iframe> </body> </html>
INDEX FILE CODES CAN SUCCESSFULLY RETRIVE VALUES FROM DATABASE BUT, I WANT TO STORE IT IN IFRAME SRC . VAR= URLS = 'HTTP://WWW.GOOGLE.COM' IFRAME SRC WILL BE VAR URLS + RETRIEVED VALUE FROM DATABASE
I want to store... - questionID - questionNo - questionText ...in one array? How do I do that?! Debbie i'm trying to get all the results from a query search into an array and them have another query search the results from that array againts a different table to get the right results. Here is my code: Code: [Select] $name = $_GET['name']; //lets say $name = 'california' $get_state = mysql_query("SELECT * FROM locations WHERE state LIKE '$name'") or die(mysql_error()); $num_states = mysql_num_rows($get_state); $location_array = array(); $i = 0; while($state = mysql_fetch_assoc($get_state)){ $location_array[$i][] = $state['id']; $i++; } $get_buildings = mysql_query("SELECT * FROM points WHERE location_id='".implode(",", $location_array)."' ORDER BY name ASC") or die(mysql_error()); However i can't get any results. I tried outputing the array by imploding it and all i got was "array, array, array, array," I hope someone here can help me out. This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=327616.0 Hello to everyone,
I have a strange problem and I'm not able to solve it by myself and I need your help!
I have three simple checkboxes that process via POST their value. But with PHP I cannot have an array with the three indexes (if everyone checked) but only an array with the first checkbox checked. The code is very simple:
<?php $scelta = isset($_POST['user_rights']) ? $_POST['user_rights'] : array(); if (isset($_POST["user_rights"])){ print_r($scelta);}else{ ?> <form method="post"><input type="checkbox" name="user_rights[]" value="0"> 0<br><input type="checkbox" name="user_rights[]" value="1"> 1<br><input type="checkbox" name="user_rights[]" value="2"> 2 <input type="submit"></form> <?php} ?>The output if I check all the checkboxes is: Array ( [0] => 0 ) If I try this code on another server it works (it returns the complete array), but if i run it on my web server, it doesn't (it returns only the first item). Do you have any experience with this error? Do you know if there is some configuration to change? Thanks in advance. Rob To generate the stockroom drop down list function einv_generateStockrmSelectDropdown($Stockrm,$field,$dropdown) { //connect to database base_connectDatabase(); echo "<select id=\"stockrm\" name=\"".$field."[]\" multiple=\"multiple\" style=\"align:left\" class=\"form-control\">"; if (isset($Stockrm) && ($Stockrm != "")) { $stockrmname = einv_getStockrmDetail($Stockrm); echo "<option value=\"". $Stockrm ."\">". $stockrmname['einv_stockrm_name'] ."</option>"; } else { $Stockrm = 0; } $getStockrmSQL = base_executeSQL("SELECT * FROM einv_stockroom WHERE einv_stockrm_id<>" . $Stockrm . " ORDER BY einv_stockrm_name"); while ($Stockrmdata_row = base_fetch_array($getStockrmSQL)) { if (base_num_rows($getStockrmSQL)!= 0) { echo "<option value=\"".$Stockrmdata_row['einv_stockrm_id']."\">".$Stockrmdata_row['einv_stockrm_name']."</option>"; } } echo "</select>"; echo "<script src=\"../terms_base/js/jquery.multiple.select.js\"></script>"; //Input some codes to split the dropdown and make it into the setSelects. By default, the first time = AutoSelectAll if(isset($dropdown) && ($dropdown != "")) { //dropdown = 1;2;3 $SDDArrays = explode(";", $dropdown); $countTrap0 = count($SDDArrays); $drop = "$('#stockrm').multipleSelect(\"setSelects\", ["; $counting = 1; foreach ($SDDArrays as &$value) { if ($countTrap0 != $counting) { $drop .= "'". $value . "',"; } else { $drop .= "'". $value . "'"; } $counting++; } $drop .= "]);\n"; } elseif (isset($dropdown)) { //dropdown= $drop = "$('#stockrm').multipleSelect(\"uncheckAll\")"; } else { // $drop = "$('#stockrm').multipleSelect(\"checkAll\")"; } echo "<script>\n"; echo "$(function() {\n"; echo "".$drop.""; echo "});\n"; echo "$(\"#stockrm\").multipleSelect({ onClose: function() { document.getElementById('search').click(); }});\n"; echo "$(\"#stockrm\").multipleSelect();\n"; echo "</script>"; //close the database base_closeDatabase(); }add stockroom function //This function is use to add new stock room //code = stock room code //name = stock room name //desc = stock room description //remark = stock room remark //cat = stock room category function einv_addStockrm($code,$name,$desc,$remark,$cat) { //connect to database base_connectDatabase(); $User = base_getUserDetail($_SESSION['uID']); base_executeSQL("INSERT INTO einv_stockroom (einv_stockrm_code, einv_stockrm_name, einv_stockrm_desc, einv_stockrm_remark, einv_stockrm_cat) VALUES ('" . $code . "', '" . $name . "', '" . $desc . "', '" . $remark . "', '" . $cat . "')"); base_addTransactionLog('Manage Stock Room', 'Add', " Stock Room Code = " . $code . " || Stock Room Name = " . $name . " || Stock Room Description = " . $desc . " || Stock Room Remark = " . $remark . " || Stock Room Category = " . $cat . " "); //go to stock room page echo '<script type="text/javascript">' . "\n"; echo 'window.location="../einventory/stockrm_list.php";'; echo '</script>'; //close the database base_closeDatabase(); } Edit stockroom function einv_editStockrm($srid,$code,$name,$desc,$remark,$cat) { //connect to database base_connectDatabase(); $User = base_getUserDetail($_SESSION['uID']); $Stockroom = einv_getStockrmDetail($srid); base_executeSQL("UPDATE einv_stockroom SET einv_stockrm_code='" . $code . "', einv_stockrm_name='" . $name . "', einv_stockrm_desc='" . $desc . "', einv_stockrm_remark='" . $remark . "', einv_stockrm_cat = '" . $cat . "' WHERE einv_stockrm_id=" . $srid . ""); base_addTransactionLog('Manage Stock Room', 'Edit', " Stock Room Code = " . $code . " || Stock Room Name = " . $name . " || Stock Room Description = " . $desc . " || Stock Room Remark = " . $remark . " || Stock Room Category = " . $cat . " "); //go to stock room page echo '<script type="text/javascript">' . "\n"; echo 'window.location="../einventory/view_stockrm.php?id='. $srid .'";'; echo '</script>'; //close the database base_closeDatabase(); } I have a dropdown list named **Stockroom** where it displays an array of values. Example: **Stockroom** 1. [A] 2. [b] 3. [C] When user clicks on [A], it will display relevant data fields that [A] possess (appears below stockroom ddl). And as user clicks on [b], there is an onchange function that will then show [b] data fields (fields in A is deleted and replaced with B). I am able to add the initial values however, as i want to edit and change the stockroom from [A] to [b] which will result in a whole new data to be stored, i am unable to do so. Any ideas? I believe i have to amend my edit stockroom function where i require a set of coding such as If the array is selected, i have to delete existing data and add new data in accordance to the selected ID. Edited by mac_gyver, 14 January 2015 - 07:05 AM. code tags around posted code please I am trying to iterate through a 3 dimensional array and echo out all the details. I am having a trouble with the following: Array ( [DT_Intrinsic_mob7] => Array ( [0] => Array ( [cli] => 11111111111 [min] => 533 ) [1] => Array ( [cli] => 2222222222 [min] => 536 ) [2] => Array ( [cli] => 33333333333 [min] => 541 ) ------- SNIP ----- [46] => Array ( [cli] => 5555555555 [min] => 5471 ) ) [DT_belltel_mob7] => Array ( [0] => Array ( [cli] => 999999999 [min] => 508 ) [1] => Array ( [cli] => 8888888888 [min] => 519 ) [2] => Array ( [cli] => 7777777777 [min] => 602 ) [3] => Array ( [cli] => 7975757375 [min] => 701 ) ) ) My code looks like this: foreach ($cli_data as $key => $value){ foreach ($cli_data[$key] as $key2 => $value2){ foreach($cli_data[$key][$key2] as $key3 => $value3){ echo "key3 is $key3 \n"; echo "value is $value3['cli']"; } } echo "\n\n"; } I keep getting the following error PHP Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /var/www/html/index.php on line Thanks Hi, Im battling here to get "only" certain values from a "akey" and "value" query.. I have a list of items around 200 id's strong, i only want to pull cirtain values from the list, the table structure is like so: id, akey, value // i don't know the values of the specific akey's or their id's but i do know the akey value. what i am trying to do is only pull the values that are in the "getConfigTitle's" list below (this list is only a sample the actual list is around 100 strong) here is my code sample: $query="SELECT akey,value FROM settings WHERE LOWER(akey) LIKE getConfigTitles();"; $settingsList=doSelectSql($query); foreach ($settingsList as $set) { $r=array(); $r['CONFIGTITLE']=getConfigTitles(); $r['KEY'] = $set->akey; $r['SETTING'] = $set->value; $rows[]=$r; } function getConfigTitles() { $titles=array(); $titles['search']= "search" ; $titles['viewprofile']= "viewprofile" ; $titles['editprofile']= "editprofile" ; $titles['dashboard']= "dashboard" ; $titles['preview']= "preview" ; return $titles; } Hi, I have this form which will create a checkbox list using data from my database and also determin if a checkbox had been checked before and check if it had. <form style="text-align:center" name="PrefRestaurant" id="PrefRestaurant" action="preferances_check.php" method="post"><table align="center"> <?php checkbox(id, name, restaurants, id); ?></table> <input type="submit" name="Prefer" id="Prefer" value="Επιλογή"/></form> function checkbox($intIdField, $strNameField, $strTableName, $strOrderField, $strMethod="asc") { $strQuery = "select $intIdField, $strNameField from $strTableName order by $strOrderField $strMethod"; $rsrcResult = mysql_query($strQuery); while ($arrayRow = mysql_fetch_assoc($rsrcResult)) { $testqry = "SELECT * FROM user_restaurant WHERE user_id = $_SESSION[UserId] AND restaurant_id = $arrayRow[id]"; $rsltestqry = mysql_query($testqry); $numrows = mysql_num_rows($rsltestqry); if ($numrows == 1) { echo "<tr align=\"left\"><td><input type=\"checkbox\" name=\"restaurant[]\" value=\"$arrayRow[id]\" checked/>$arrayRow[name]</td></tr>"; } else{ echo "<tr align=\"left\"><td><input type=\"checkbox\" name=\"restaurant[]\" value=\"$arrayRow[id]\" />$arrayRow[name]</td></tr>"; } } } Now the part which I can't get to work is when I'm trying to store the new values in my database. When I click the submit button I clear my database of any row that is related to the currently loggedin user and I want to store his new preferences (checked cheboxes). I've read that only the cheked checkboxes' values are POSTed so I did this (preferances_check.php) foreach($_POST['restaurant'] as $value) { $query="INSERT INTO user_restaurant VALUES ('$_SESSION[UserId]','$value')"; } But it is not working, nothing gets written in my table Could someone please enlighten me on this? Thnks! 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. I 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 --- ); ?> 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 |