PHP - Changing An Image When Clicking On A Link
Hi
I'm trying to change an image when a certain link is clicked. The image is a logo of a company and the links that will be clicked is english, arabic, and Chinese.
So if someone clicks on Chinese then the logo changes to the Chinese version of the normal logo.
Here is my html
<?php $langArabic=''; $langChinese=''; $langSpanish=''; $langEng=''; $linkArabic=getSEOLink($page_id); $linkChinese=getSEOLink($page_id); $linkSpanish=getSEOLink($page_id); $linkEng=getSEOLink($page_id); $queryLink = mysql_query("SELECT * FROM $_SEO_TABLE WHERE `id_page`='".$page_id."'"); if(mysql_num_rows($queryLink) > 0){ $resultLink = mysql_fetch_object($queryLink); if($resultLink->url != ''){ $linkEng = $_HTTP_ADDRESS."".$resultLink->url; if($page_id == 1 || $page_id == '1'){ $linkEng = $_HTTP_ADDRESS; } } if($resultLink->url_arabic != ''){ $linkArabic = $_HTTP_ADDRESS."".$resultLink->url_arabic; } if($resultLink->url_chinese != ''){ $linkChinese = $_HTTP_ADDRESS."".$resultLink->url_chinese; } if($resultLink->url_spanish != ''){ $linkSpanish = $_HTTP_ADDRESS."".$resultLink->url_spanish; } } switch (urlExtension()) { case 'html': $langClass = ''; $langEng='langActive'; $langTag='en'; break; case 'ar': if($page->title_arabic == ''){ $langClass = '_arabic'; }else{ $langClass = '_arabic'; } $langArabic='langActive'; $langTag='ar'; break; case 'es': $langClass = ''; $langSpanish='langActive'; $langTag='es'; break; case 'cn': $langClass = ''; $langChinese='langActive'; $langTag='cn'; break; default: $langClass = ''; $langEng='langActive'; $langTag='en'; } ?> <a class="main-logo<?php echo $langClass;?>" href="javascript:void(0);"><img src="<?php echo $_HTTP_ADDRESS;?>images/logo-letters.png" /></a> <div class="language-selection-wrapper<?php echo $langClass;?> overlay-bg<?php echo $langClass;?> border-bottom-white<?php echo $langClass;?> padding-level-one<?php echo $langClass;?>"> <div class="language-inner-wrapper<?php echo $langClass;?>"> <a class="<?php echo $langEng;?>" href="<?php echo $linkEng;?>">English</a><span>|</span> <a class="<?php echo $langArabic;?>" href="<?php echo $linkArabic;?>">العربيّة</a><span>|</span> <a class="<?php echo $langSpanish;?>" href="<?php echo $linkSpanish;?>">Español</a><span>|</span> <a class="<?php echo $langChinese;?>" href="<?php echo $linkChinese;?>">中文</a> <div class="clearboth"></div> </div> Similar TutorialsI have a PHP application where I am trying to execute a PHP routine when the user clicks on a link to go to another page. The routine needs to execute before I use a header location command to send the user to the next location. I can't do this all with JavaScript, but it could be part of the solution if necessary. I'd prefer to just use PHP if I could, but I'm not sure that this is possible. Can anyone share with me a brief example of how I could do this? Thanks! So I am trying to execute an sql query by clicking a button or a link. Ultimately I want users to be able to click the "Fav" button and it then fades into "Added to favorites!" 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I however want to be able to click on something to execute that query how do i do that? I read a bunch of stuff on javascript and ajax but it got me nowhere... Help is greatly appreciated! Hi.. I have an issue on my project.I want to open a particular page of a PDF file by clicking a hyper link.How should I do it?any idea ? I used the following code <?php header("Content-Type: application/pdf"); $pdfFile="readme.pdf#page=10"; ?> <html> <head> <title>Untitled Document</title> </head> <body> <a href="http://<?php echo $pdfFile; ?>">Click Here</a> </body> </html> But it showing an alert like following File does not begin with '%PDF-' After clicking alert showing a black screen.. can u please send me the entire code? Thanks... Goal: To have a gallery that downloads images from the folder I previously uploaded to in a previous script. Bug: When I load the page the thumbnail comes up as broken and when I click on the thumbnail to get the bigger picture it comes up with the following error message: "Firefox doesn't know how to open this address, because the protocol (c) isn't associated with any program." <?php include 'db.inc.php'; //connect to MySQL $db = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASSWORD) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db(MYSQL_DB, $db) or die(mysql_error($db)); //change this path to match your images directory $dir ='C:/x/xampp/htdocs/images'; //change this path to match your thumbnail directory $thumbdir = $dir . '/thumbs'; ?> <html> <head> <title>Welcome to our Photo Gallery</title> <style type="text/css"> th { background-color: #999;} .odd_row { background-color: #EEE; } .even_row { background-color: #FFF; } </style> </head> <body> <p>Click on any image to see it full sized.</p> <table style="width:100%;"> <tr> <th>Image</th> <th>Caption</th> <th>Uploaded By</th> <th>Date Uploaded</th> </tr> <?php //get the thumbs $result = mysql_query('SELECT * FROM images') or die(mysql_error()); $odd = true; while ($rows = mysql_fetch_array($result)) { echo ($odd == true) ? '<tr class="odd_row">' : '<tr class="even_row">'; $odd = !$odd; extract($rows); echo '<td><a href="' . $dir . '/' . $image_id . '.jpg">'; echo '<img src="' . $thumbdir . '/' . $image_id . '.jpg">'; echo '</a></td>'; echo '<td>' . $image_caption . '</td>'; echo '<td>' . $image_username . '</td>'; echo '<td>' . $image_date . '</td>'; echo '</tr>'; } ?> </table> </body> </html> Any help appreciated. This topic has been moved to CSS Help. http://www.phpfreaks.com/forums/index.php?topic=321778.0 Hi I've got this database I created with fields ProductId ProductName Image I've managed to get it to list the ID,productname, and Image urls in a list. My next step is to have the image field actually display an image and make it clickable: heres what I've done so far: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("productfeed", $con); $result = mysql_query("SELECT * FROM productfeeds"); echo "<table border='0'> <tr> <th>Firstname</th> <th>Lastname</th> <th>Image</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; echo "<td>" . $row['ImageURL'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Heres what I want to do: Code: [Select] while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; // my changes beneath echo "<td>" . <a href="<?php echo $row['ImageURL'];?>"> <img src="<?php echo $row['LinkURL']; ?>"> </a>. "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Can you guys point me in the right direction? Many thanks Any chance you could help me out with some probably simple code ? I need to change an image when the script change state. So instead of $Streamstatus "off"or "on" I would like to change an image. Cheers Code: [Select] <?php $request = 'http://api.own3d.tv/liveCheck.php?live_id='; $arg = 'Zerkilive'; $session = curl_init($request.$arg); curl_setopt($session, CURLOPT_HEADER, false); curl_setopt($session, CURLOPT_RETURNTRANSFER, true); $response = curl_exec($session); curl_close($session); if (preg_match("/true/",$response, $result)) { $streamStatus="on"; $color="green"; } else { $streamStatus="off"; $color="red"; } echo $streamStatus; ?> Please keep in mind, im a newbi at php as well as these forums, and I just started using it this week. I have some background in C programing and know a little VB. I am currently trying to make a php based webpage that has 5 different images on it. I want each image to change with respect to what is written in a text file within the server space. So to better explain this, I have a text file that is named "status.txt". status.txt contents: active inactive active active active I want the first image to change to a different image if I change the first line of the text file to be inactive instead of active and the 2nd image to change if I change the 2nd line (inactive) to active and so forth on all the other images. I do know that i need to open the file, then read the file with fopen and fread and set what the contents that i read from the file equal to some variable (still not 100% sure im using the functions right). but im not sure how to set up the code so the image will change with respect to what is read from the text aka what the variable is equal too. I thought it would be a simple if statement but it seems more complicated then that. and futhermore, im not sure where to place the open file, should i do this in the header? so every time website refreshes, is opens the file and reads the txt file and changes with respect to that? Any help would be great! Thank you in advance. Hi! I am currently making a really simple photo blog. The query fetches every reference to the images which are stored on the server. I have a problem when it comes to dynamically cycling through these though. After running the query, I store each image reference in a twodimensional array like this: [["id"], ["date"], ["title"], ["filename"]] (an Array of Arrays). The image displays just fine by doing this: $pointer; //starting from the last element (image) in the array <img src="images/gallery/<?php echo $images[3][$pointer] ?>" alt="<?php echo $images[2][$pointer] ?>"> The problem is that, when I update the pointer, say for instance pointer++, the current image on the screen doesnt update. I know for a fact that the pointer updates, because I have echoed it. The basic idea is to have a previous and next link just under the image, so that the viewer can cycle the available images. Any help is greatly appreciated I have a line like this it prints text link but I prefer image link how should i edit it I would appreciate some feedback Code: [Select] $templates['etiket'] = array('name' => t('ETİKET'), 'module' => 'uc_invoice_pdf', 'path' => $templates_uc_invoice_pdf_path, 'pdf_settings' => $pdf_settings); Hi, I've read a lot of places that it's not recommended to store binary files in my db. So instead I'm supposed to upload the image to a directory, and store the link to that directory in database. First, how would I make a form that uploads the picture to the directory (And what kinda directories are we talking?). Secondly, how would I retrieve that link? And I guess I should rename the picture.. I'd appreciate any help, or a good tutorial (Haven't found any myself). Hello. Well I haven't any code for this, but its only a question. Do you think it would be possible for PHP, to make and image or draw one from just a blank JPG. Then placing text on the image. That is possible and quite simple. But do you think it is possible to say. Say if you have 3 different set of texts on the image. Click here for 1 Click here for 2 Click here for 3 And also for the script to link those text to different web addresses. But all still workable with a dynamic signature EG [ img ] [ / img ] Basically its a dynamic signature with links on it is what i am trying to say. Do you think its possible? Or am I thinking to big. What I'm trying to do is have it match up the image file name wiht the same name as the link. Sometimes it does but sometimes obviously the image is random. Any ideas? Code: [Select] function spotlight($dbc){ $query = " SELECT * FROM characters WHERE characters.statusID = 1 AND characters.styleID = 1 ORDER BY RAND() LIMIT 1"; $result = mysqli_query($dbc, $query); $row = mysqli_fetch_array($result); $shortName = $row[ 'shortName' ]; $labels = array('shortName'); $img = array(); if($handle = opendir('images/spotlight/')) { $count = 0; while (false !== ($file = readdir($handle))) { if(strlen($file) > 2){ $img[$count] = $file; $count++; } } } echo "<a href='bio.php?shortName=" . $shortName . "'><img src='/images/spotlight/".$img[rand(0, (count($img)-1))]."' alt=Spotlight border=0 /></a>"; } It is easy to get image or link by DomDocument, but I did not find a way to get image with its target link. Imagine a html as Code: [Select] <div class=image> <a href='http://site.com'><img src='imagelink.jpg'></a> </div>How to get both the image link and href? $dom = new DOMDocument(); @$dom->loadHTML($html); $xpath = new DOMXPath($dom); $hrefs = $xpath->evaluate("/html/body//div[@class='image']"); for ($i = 0; $i < $hrefs->length; $i++) { $href = $hrefs->item($i); Now to get the image and its href, we need first getElementsByTagName('a') and getElementsByTagName('img') but they do not work inside foreach. What's your idea? Greetings! I have a website www.lanceronlinejobs.com/our_franchises.php I have a franchise images. I want to display each franchise record from database whenever a user click on franchise link. Here is my code. ourfranchises.php code: Code: [Select] <?php include('fconnection.php'); $sql = mysql_query("SELECT * FROM tbl_franchise ORDER BY id DESC") or die(mysql_error()); $row3 = mysql_fetch_array($sql); ?> <a href="franchiseDetails.php?city=<?php echo $row3['city'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>---------------------------------------------------------------------- franchiseDetails.php code: Code: [Select] <?php $id = $_GET['id']; $query = "SELECT * FROM tbl_franchise WHERE id = '$id' ORDER BY id DESC LIMIT 1"; $result = mysql_query($query); if (!$result) { echo "NO RECORD FOUND"; } else { while($row3 = mysql_fetch_array($result)): ?> Manager Name: <?php echo $row3['manager_name'];?> Please help me. Any help would be appreciated. Thanks. I have an image which is a complete image that has 4 titles(aquatic weeds,green algae,water quality,bottom muck) Can i write a code somehow that when i hover over those images or title i can link it to different hyperlinks? 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'm wondering if anyone can help me find/create a script which allows a preview of a link to be generated? Basically I need a user to be able to simply enter a url and click submit and then a preview of that link will be shown in the form of some text and an image. Very similar to what Facebook does when a user shares a link. I think they use the description tag of the site in general. Can anyone point me in the right direction in doing this? I need to be able to do it in php/javascript alone without the use of jquery or plugins. Hello i kinda new to programing mostly learning myself stuf there so i need some help to figure out. so i have lets say static link who always same and i write something in input field and press search button. so i get new link and immediately going to dat link, which means i leaving my site. in my case i working with image so my problem, is it somehow possible to display image in same page while writing someting in input field when presing button and image just change down below not going directly to link ? P.S. sory for my english
i have this code
<html> <body>
<center>
<form action="#" method="POST">
</form>
<?php require_once 'linkgenerator.php' ?>
PHP link generator code
<?php
Edited March 18, 2020 by eagle101 |