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Found 5 results

  1. I want to display 10 products from each category from database, I started to display the categories but how to make displaying the products from each one. Here is the code I make so far. I'm not sure am I doing it right with doble sql select , or it can be done only with one. <div class="inner shadow"> <?php $query = 'SELECT id, title_bg AS `title_cat` FROM categories'; $result = $this->db->query($query); ?> <?php foreach ($result->result() as $row): ?> <?php $title_cat = stripslashes($row->title_cat); ?> <div class="prod-sec"> <div class="prod-head"> <h2><?= $title_cat?></h2> <div class="clear"></div> </div> <?php $query1 = 'SELECT t1.id, t1.title_bg AS `Title`, t1.text_bg AS `Text`, t1.price, t1.discount, t1.category_id, t1.promo_page, t2.id AS FileID, t2.ext, FROM products t1 LEFT JOIN products_pictures t2 ON t1.id = t2.object_id LEFT JOIN categories t3 ON t3.id = t1.category_id WHERE t1.promo_page = 0 AND t1.is_active = 1 AND t3.title_bg = '$title_cat' ORDER BY RAND() LIMIT 10'; $result1 = $this->db->query($query1); ?> <div id="classeslist2"> <ul class="home_middle_products"> <?PHP foreach($result1->result() as $row1) { $f = 'files/products_first_page/' . $row1->id . '.jpg'; if(is_file(dirname(__FILE__) . '/../../' . $f)) { $img = site_url() . "files/products_first_page/".$row1->id.".jpg"; } else { $img = site_url() . "files/products/".$row1->id."/".$row1->FileID."_2.".$row1->ext; } $title = stripslashes($row1->Title); $text = character_limiter(strip_tags(stripslashes($row1->Text)),250); $title_url = getLinkTitle($title); $link = site_url()."products/product/".$row1->id."/{$title_url}"; ?> <li style="width: 185px; height: 270px; margin-left: 3px; margin-top: 10px;"> <div class="thumb"> <a href="<?=$link?>"><img src="<?=$img?>" alt="<?=$title?>" width="182" /></a> <div class="price"><?PHP echo product_price($row1, array('show_discount' => false,"show_old_price"=>false, 'show_label' => false, 'view' => 'no')); ?> </div> </div> <h2></h2> <h2><a href="<?=$link?>"><?=$title?></a></h2> <? $text = substr($text, 0, 100); ?> <? if (strlen($text) == 100) $text .= '...' ; ?> <p><?=$text?></p> </li> <?php } ?> <div class="clear"></div> </ul> </div> </div> <?php endforeach; ?> <div class="clear"></div> </div>
  2. Every month I have pdf file with 40 pages, and I upload file to folder on webserver. Every user on internal company website have his own page in that pdf file. How to display specific page to specific user, without manual embedding to user profile page? Or display it on designated page for that pdf file, but with selection for user to pick his own page from pdf? Better option would be that user can navigate whole archive of pdf and pick year, month and his page from pdf. User's page in pdf has always same page number and his own phone number becouse pdf is phone bill for company mobile phones.
  3. All I am trying to do is add a record on a page without the page refreshing. For that ajax is used. Here is the code. It does not add the record to mysql table. Can anyone tell me what I am doing wrong? record.php <!DOCTYPE HTML> <html lang="en"> <head> <script type="text/javascript" src="js/jquery-1.11.0.min.js"></script> <script type="text/javascript" > $(function() { $(".submit_button").click(function() { var textcontent = $("#content").val(); var name = $("#name").val(); var dataString = 'content='+ textcontent + '&name='+name; if(textcontent=='') { alert("Enter some text.."); $("#content").focus(); } else { $("#flash").show(); $("#flash").fadeIn(400).html('<span class="load">Loading..</span>'); $.ajax({ type: "POST", url: "action.php", data: dataString, cache: true, success: function(html){ $("#show").after(html); document.getElementById('content').value=''; $("#flash").hide(); $("#content").focus(); } }); } return false; }); }); </script> </head> <body> <?php $record_id = $_GET['id']; // getting ID of current page record ?> <form action="" method="post" enctype="multipart/form-data"> <div class="field"> <label for="title">Name *</label> <input type="text" name="name" id="name" value="" maxlength="20" placeholder="Your name"> </div> <div class="field"> <label for="content">content *</label> <textarea id="content" name="content" maxlength="500" placeholder="Details..."></textarea> </div> <input type="submit" name="submit" value="submit" class="submit_button"> </form> <div id="flash"></div> <div id="show"></div> </body> </html> action.php if(isset($_POST['submit'])) { if(empty($_POST['name']) || empty($_POST['content'])) { $error = 'Please fill in the required fields!'; } else { try { $name = trim($_POST['name']); $content = trim($_POST['content']); $stmt = $db->prepare("INSERT INTO records(record_id, name, content) VALUES(:recordid, :name, :content"); $stmt->execute(array( 'recordid' => $record_id, 'name' => $name, 'content' => $content )); if(!$stmt){ $error = 'Please fill in the required fields.'; } else { $success = 'Your post has been submitted.'; } } catch(Exception $e) { die($e->getMessage()); } } }
  4. So I've followed this code, corrected about 12 error, talked to my hosting and I am so done. I have tried everything but the error won't go away. The code is pasted below. As it's really late any help would be appreciated that would make my day the next day... <?PHP //connect to server $connect = mysql_connect("localhost","cello10_import","brigite27"); //connection to database mysql_select_db("cello10_import"); //query the database $query = mysql_query("SELECT * FROM users WHERE cover_image = 'http://d1w7fb2mkkr3kw.cloudfront.net/assets/images/book/small/9781/2500/9781250038821.jpg' "); //fetch the results / convert into array WHILE($rows = mysql_fetch_array(query)): $cover_image = $rows['cover_image']; $title = $rows['title']; $author = $rows['author']; echo "$cover_image<br>$title<br>$author<br><br><br>"; endwhile; ?> The error experienced is: Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in /home/cello10/public_html/display.php on line 10 Thanks for your help
  5. Hi guys, I've a little problem i have a fileld in my database that i want to check. If is null to display some text else display another text. I've manage to do something like this but doesn't work. The code is below: <?php $result = @$mysqli->query("SELECT * FROM depanarecuora_clients WHERE paid_status = '$paid_status'"); while ($result = mysqli_fetch_assoc($result)) if (is_null($paid_status)) { echo "some text"; } ?>
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