|
PHP - Php Code Help - Database Content Not Showing
Hi, I am hoping I have posted my current problem in the right part of this forum. As it relates to the php code, I presume this is the right place, although it also relates to a mysql database I have too.
Ok, here is my problem.... I have two websites, one is a non members site and one is a paid members area site. I had a directory style script made for the non members site and want to use the same script within the members site. I paid for the script to be made and set up, which cost me quite a bit, hence the reason I am copying it all over to my other site too. The script displays thumbnials from the database, 4 across and 3 down (total of 12 thumbs per page), each linking to a URL and each thumb has a title. On the left of the webpage is categories, which when clicked, shows thumbs relating to that specific category, or all etc. The CMS forms allow us to add categories etc. Ok, now my problem is this.. I backed up the original database and created a new database on the members site where I also want this functionality. I copied all the scripts over then chenged the content of the database. On the members site I only want it to display three categories, All, Photos and Videos. So this is quite simple compared to the first site. All seemed to display fine, until I moved all the scripts into the members folder (which is accessed via ccbill). The page displays ok, but it just states "No updates found" as in, no content in the database is found. The base url in the configuration script is set to the top domain, but I have renamed the folders where the thumbs are saved to members/thumbs, but still it isnt working. I also want to change the folder name from its original name of www.domain.com/sites to the new domain.com/updates but nothing works when I try to change it. If anyone can help me out with this, I would be extremely grateful, as the workings of this script are all there, and it seems pointless having to spend hundreds on a new script to be made when the functionality is already there, but just not working right. Thanks so much! Similar Tutorialshttp://www.compareandchoose.com.au/video_games?shopingfor=US&shopingfor=US http://www.compareandchoose.com.au/video_games?shopingfor=AU&shopingfor=US http://www.compareandchoose.com.au/video_games?shopingfor=UK&shopingfor=US http://www.compareandchoose.com.au/video_games?shopingfor=US&shopingfor=UK http://www.compareandchoose.com.au/video_games?shopingfor=AU&shopingfor=UK http://www.compareandchoose.com.au/video_games?shopingfor=UK&shopingfor=UK http://www.compareandchoose.com.au/video_games?shopingfor=US&shopingfor=AU http://www.compareandchoose.com.au/video_games?shopingfor=AU&shopingfor=AU http://www.compareandchoose.com.au/video_games?shopingfor=UK&shopingfor=AU when it should only show http://www.compareandchoose.com.au/video_games?shopingfor=US http://www.compareandchoose.com.au/video_games?shopingfor=AU http://www.compareandchoose.com.au/video_games?shopingfor=UK why its appending all combinations I am not sure Please find my 3 page code files it usesHi Professionals I have been doing a site check with MOZ tools and it is showing duplicate page content for all my shopping pages Basically I have a site where you can compare products for either the UK,US or AU But it is showing the above results in MOZ I have tried everything to figure this out, but because it was not written by me I cannot pin point it Please find attached the 3 pages it utilises Attached Files check_form.php 433bytes 5 downloads shopping_video_games.php 3.17KB 6 downloads video_games.php 3.2KB 5 downloads Hello - I'm opening my website up to visitors for free, and trying to bypass a login screen to go straight into the data content that was appearing after a user logged in. I have an index.php file that included the following code at the beginning: <?php session_start(); include("database.php"); include("login.php"); include("/vservers/skyranks/db_connect.php"); ?> <?header("Cache-control: private"); ?> <html> I deleted the "include("login.php"); line, and was successful at bypassing the username and login screen. However, the page that is supposed to display the data content is incomplete. In fact, it only displays my company's logo. Any ideas as to why the data content is not showing up? Thank you for any help with this, as my php is quite novice at this point. Regards - Joe I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
Hi all, I have table, named: tb_event (id, f_date, f_title, f_image, f_status) How to manage, while if 'f_status' from all records NOT EQUAL to 'Published' then show ABCDE Note: on tb_event should be only the latest record (1 record only) that will be shown/published I have code: <? $qry = mysql_query( "SELECT * FROM tb_event WHERE f_status = 'Published'") or die(mysql_error()); while($res = mysql_fetch_array($qry)) { if ($res[f_status] = "Published") { echo ' <img src="images/uploads/'.$res[f_image].'" border="0" /> '; } else { echo ' ABCDE '; } } ?> The result is 'ABCDE' not shown while f_image (all records) is OFF Please help me.. thanks Hi guys, I was just wondering if anyone could help me. I've got a My_SQL database containing articles, a summary for the article and a date. I have a basic CMS system set-up, but I want to create a script that when users sign up to a mail list it forwards the summary and dates of the articles database. If that makes sense? But I only want it to forward the most recent 5 rows. I'm pretty new to PHP and I've been mostly following tutorials thus far, but this is quite specific. Thanks in advance! Hello, im trying to make a dropdown show up for every single row gotten from the database table. Instead it shows only one for the first row. Here is the code: <!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"> <head> <title>Example Form</title> <link rel="stylesheet" type="text/css" href="dd.css" /> <script type="text/javascript" src="jquery-1.3.2.min.js"></script> <script type="text/javascript" src="jquery.dd.js"></script> </head> <body> <?php mysql_connect("localhost", "", "")or die("cannot connect"); mysql_select_db("test")or die("cannot select DB"); $tbl_name="test_mysql"; $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql); $count=mysql_num_rows($result); ?> <form name="form1" method="post" action=""> <tr> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td align="center"><strong>Row</strong></td> <td align="center"><strong>Month Date</strong></td> <td align="center"><strong>Message</strong></td> <td align="center"><strong>Title</strong></td> <td align="center"><strong>Icon</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td align="center"><?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?></td> <td align="center"><input name="month[]" MAXLENGTH="3" size="3" type="text" id="month" value="<?php echo $rows['month']; ?>"> <input name="date[]" MAXLENGTH="2" size="2" type="text" id="date" value="<?php echo $rows['date']; ?>"> </td> <td align="center"><input name="message[]" size="50" type="text" id="message" value="<?php echo $rows['message']; ?>"></td> <td align="center"><input name="title[]" size="50" type="text" id="title" value="<?php echo $rows['title']; ?>"></td> <td align="center"> <select name="test[]" style="display:none; width:200px" class="mydds"> <option value="icon_phone.gif" title="icon/icon_phone.gif">Phone</option> <option value="icon_sales.gif" title="icon/icon_sales.gif">Graph</option> </select> </tr> <?php } ?> <tr> <td colspan="4" align="center"><br><input type="submit" name="Submit" value="Submit"></td> </tr> </table> </td> </tr> </form> <script language="javascript" type="text/javascript"> function showvalue(arg) { alert(arg); //arg.visible(false); } $(document).ready(function() { try { oHandler = $(".mydds").msDropDown().data("dd"); oHandler.visible(true); //alert($.msDropDown.version); //$.msDropDown.create("body select"); $("#ver").html($.msDropDown.version); } catch(e) { alert("Error: "+e.message); } }) </script> <hr> <?php if (isset($_POST['Submit'])) { for($i=0;$i<$count;$i++){ $month = $_POST['month']; $date = $_POST['date']; $message = $_POST['message']; $title = $_POST['title']; $monthday = $month[$i]."<br>".$date[$i]; $sql1="UPDATE $tbl_name SET monthday='$monthday', month='$month[$i]', date='$date[$i]', message='" . mysql_real_escape_string($message[$i]) . "', title='" . mysql_real_escape_string($title[$i]) . "' WHERE id='$id[$i]'"; //$sql1="UPDATE $tbl_name SET monthday='$monthday', month='$month[$i]', date='$date[$i]', message='$message[$i]', title='$title[$i]' WHERE id='$id[$i]'"; $result1 = mysql_query($sql1); } header("location:update2.php"); } ?> </body> </html> <BR> I have no idea if it has to do with the PHP, HTML, or JAVASCRIPT. So if someone can tell me what is going on I would be happy. Here is how it looks: Youl see it shows the fields but not the dropdown... Please help and thank You! Pls anyone tell me to connect database hello team, can you please look at this code snippet and tell me why it is showing the following data this way? <?php # jeff Exp $ include("app.php"); $page->set("title", "Business & Commerical"); $what = $page->getvar("what"); $type = $page->getvar("type"); $dt = new xDataTable('width="500"'); $form = new Form(); if($what == "submit") $html->set("readonly", true); $dt->header($_types{$type} . " Quote"); $form->get_contact_form(true); When I run it, it shows like this: set("title", "Business & Commerical"); $what = $page->getvar("what"); $type = $page->getvar("type"); $dt = new xDataTable('width="500"'); $form = new Form(); if($what == "submit") $html->set("readonly", true); $dt->header($_types{$type} . " Quote"); $form->get_contact_form(true); switch ($type) { case "comp": $dt->left("Business Type:"); $dt->cell($html->select("business_type", $_business_type, 1, "", $page->getvar("business_type"), "(select)")); $dt->left("Federal Tax ID:"); $dt->cell($html->textfield("tax_id", $page->getvar("tax_id"), 12)); $dt->left("If Individual, Owner SSN#:"); $dt->cell($html->textfield("ssn", $page->getvar("ssn"), 12)); $dt->cell("Spouse SSN#:"); $dt->cell($html->textfield("spouse_ssn", $page->getvar("spouse_ssn"), 12)); just a small sample. What am I doing wrong? Thanks alot in advance Advance thank you. Can you help please. The error..... Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in C:\wamp\www\test_dabase.php on line 24 code. Code: [Select] <?php //database connection. $DB = mysql_connect("localhost","root") or die(mysql_error()); if($DB){ //database name. $DB_NAME="mysql"; //select database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name"); // if connection. }if($CON){ //show tables. $mysql_show="SHOW TABLES"; //select show and show. $mysql_select2="mysql_query(".$mysql_show.") or die(mysql_error())"; } //if allowed to show. if($mysql_select2){ //while it and while($data=mysql_fetch_assoc($mysql_select2)){ //show it. echo $data; } } ?> so ik you cna get a search box to search items or text in a database, but what i want to know is how to manually insert text and images into it example image (imagine its an image) text name - text price how would i do this? I have created an input field on a website for people to subscribe by their email address. The email address is stored in a database. I am using PHPMyAdmin. The email address is successfully working, but I want to prevent duplicate email address to be stored, however, I am having an error. Here are my codes: HTML codes: Code: [Select] <form action="index.php" method="post"> <input type="text" size="25" placeholder="Your email address..." name="enter"/> <input class="submit" type="submit" value="Subscribe" name="subscribe"/> <br/> PHP with Query codes: Code: [Select] <?php if ( $_SERVER['REQUEST_METHOD'] == "POST" ) { $ee = htmlentities($_POST['enter']); if (!preg_match('/^[^0-9][a-zA-Z0-9_]+([.][a-zA-Z0-9_]+)*[@][a-zA-Z0-9_]+([.][a-zA-Z0-9_]+)*[.][a-zA-Z]{2,4}$/',$ee) || empty($ee)){ echo '<p class="fail">Failed...Try again!</p>'; } else { @mysql_connect ('localhost', 'root', '') or die ('A problem has occurred, refresh the page and try again!'); @mysql_select_db ('links') or die ('A problem has occurred, refresh the page and try again!'); $duplicate = "SELECT * FROM `email` WHERE `emailaadress` = '{$ee}'"; $query = "INSERT INTO email (id, emailaddress) VALUES('NULL', '.$ee')"; $result = mysql_query($duplicate); if ( mysql_num_rows ( $result ) > 1) { /* Username already exists */ echo 'Username already exists'; } else { mysql_query($query); echo '<p class="success">Successfully subscribed!</p>'; } } } ?> Error I am having: Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\softwareportfolio\index.php on line 68 Can someone help me? Thank Lets say you have a database which is packed with content. How can this be used for SEO purposes? Does Google follow all the links on your site and read the echo and indexing that information. Or does the content in a database need to converted into HTML for Google to be able to read it? I tried using Google Custom Search however it did not display anything in my database. If Google crawled my site and ranked it and then carried out the search using Google Custom Search would it then display using Google Custom Search? Hi Guys, I have the following code that I'm getting from my database. This is how it appears in the DB. Code: [Select] <div id="reply_flash"> <table width="100%" border="0" cellpadding="0" cellspacing="0"> <tr> <td style="font-family: Arial; font-size: 14px; line-height: 19px; text-align: left; color: rgb(102, 102, 102);"> <br> <div style="font-family: Georgia; font-size: 19px; line-height: 20px; color: rgb(88, 43, 0); font-weight: normal; text-align: left;">This is a heading</div> <p>This is some text in this area.</p> <a href="#"><strong>Read more</strong></a> </td> </tr> </table> </div> When the user goes to edit the page, I'm retrieving the above from the database: http://www.mysite.com/newsletter/edit/56 Code: [Select] // Controller to handle the request public function edit() { $id = $this->uri->segment(3); $data['get_newsletter'] = $this->Newsletter_model->get_newsletter($id); $this->load->view('header'); $this->load->view('newsletter/edit', $data); $this->load->view('footer'); } // View to display the request foreach($get_newsletter as $row) { $code = $row->code; $newsletter_id = $row->id; } echo $code; Now, when I echo the code below. It appears in the html as Code: [Select] <div id="reply_flash"> </div> // CLOSING THE DIV AND PUTTING THE TABLE OUTSIDE <table width="100%" border="0" cellpadding="0" cellspacing="0"> <tr> <td style="font-family: Arial; font-size: 14px; line-height: 19px; text-align: left; color: rgb(102, 102, 102);"> <br> <div style="font-family: Georgia; font-size: 19px; line-height: 20px; color: rgb(88, 43, 0); font-weight: normal; text-align: left;">This is a heading</div> <p>This is some text in this area.</p> <a href="#"><strong>Read more</strong></a> </td> </tr> </table> </div> As you can see, when it's being retrieved and being displayed on the webpage, the div is being closed and the table appears outside it. Have no isdea what is going on. This is the code that I am using to make a page that is kind of like an administration area for a CMS system. When the page is loading all I get is a blank page. When I remove all of the back code before the HTML code is sent. The page shows up fine. I have looked over and over the code trying to see if I missed a colon or something and I see nothing. I hope that this explains everything well enough and I would appreciate any help. Code: [Select] <?php $today = date("l, F j, Y"); $host =''; $username = ''; $dbname =''; $dbpassword = ''; $table = ''; $table2 = ''; $connection = mysql_connect($host, $username, $dbpassword) or die (mysql_error()); $db = mysql_select_db ($dbname, $connection) or die (mysql_error()); $sql = "SELECT * FROM $table WHERE username = '$_POST[username]' AND password = password('$_POST[password]')"; $result = mysql_query($sql, $connection) or die(mysql_error()); $num = mysql_num_rows($result); while ($row = mysql_fetch_array($result)) { $f_name = $row['f_name']; $l_name = $row['l_name']; $username = $row['username']; } if ($num != 0) { $_SESSION[auth] = "yes"; $msg = "Welcome $f_name "; } //else { //header("Location: index.php"); //exit; //} $get_count = "SELECT count(id) FROM $table2"; $get_count_res = mysql_query($get_count, $connection) or die (mysql_error()); $count = mysql_result($get_count_res, 0, "count(id)") or die (mysql_error()); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link rel="stylesheet" type="text/css" href="css/reset.css" media="screen" /> <link rel="stylesheet" type="text/css" href="css/admin.css" media="screen" /> <title>Administration Area</title> </head> <body> <div id="wrapper"> <?php include('includes/header.php'); ?> <div id="content"> <div class="message"><?php echo "$msg"; ?> <a href="logout.php">Logout</a><span style="float:right; "><?php echo "$today"; ?></span></div> <div class="admin"> <span class="top">Administration</span> <ul> <li><A href="show_add.php">Add a Ad</A></li> <li><a href="pick_modify.php">Modify a Add</a></li> <li>Delete Ad</li> <li><a href="user/index.php">Add a User</a></li> <li><a href="user/pick_user.php">Delete a User</a></li> <li><a href="#">View Database Records</a></li> <li><a href="php_info.php">PHP info</a></li> <li><a href="make_backup.php">Backup Database</a> </li> <li><a href="https://p3nlmysqladm001.secureserver.net/nl50/529/index.php?lang=en-utf-8&token=db052d2ddafce3f6ba841cfe0a6c224a">View Database</a> </li> </ul> </div> <div class="misc"><span class="top">miscellaneous</span>Classified Ads online <?php echo "$count"; ?></div> </div> <!--End #content--> <?php include('includes/footer.php'); ?> </div> <!-- End #wrapper --> </body> </html> This is the script i am using i two areas , but the goal is not the same :( Code: <div class='allDeal'> <div class='date'><?php echo app()->locale->dateFormatter->formatDateTime( utime($data->start,false),'medium',false); ?></div> <div class='content'> <div class='name'><?php echo CHtml::link($data->name,url('/deal/view',array('url'=>$data->url))); ?></div> <div class='clearfix prepend-top'> <div class='column data'> <div class='stats txt-center'> <span class='num'><?php echo $data->stats && $data->stats->bought?$data->stats->bought:'0'; ?></span><br /> <?php echo $this->t('Coupons Bought'); ?> </div> <div class='info prepend-top'> <div class='row price'> <label><?php echo $this->t('Price').':'; ?></label> <?php echo m('payment')->format($data->price); ?> </div> <div class='row'> <label><?php echo $this->t('Value').':'; ?></label> <?php echo m('payment')->format($data->value); ?> </div> <div class='row'> <label><?php echo $this->t('Savings').':'; ?></label> <?php echo m('payment')->format($data->value-$data->price); ?> </div> </div> </div> <div class='column last image'><?php if($data->imageBin) echo CHtml::link( CHtml::image(app()->storage->bin($data->imageBin)->getFileUrl('original')), url('/deal/view',array('url'=>$data->url)) ); ?></div> </div> </div> </div><?php if($index && ($index+1)%2==0) { ?><div class='clearfix'></div><?php } Only change: div class='allDeal' to div class='recentDeal' This is the good: http://kupon.woano.lv/deal/recent and the same code in other place looks ugly :( http://kupon.woano.lv/deal/all What i am doing wrong I am having a big problem in MySQL and a part in my PHP coding as well. I created a registration form, where the user will have to fill his names, email address, choose a username and so on. I do not want duplicate email address and username in my database, thus, if a user choose either an email address or a username which has already taken, he will be notified. To prevent this duplication, I have set both the email address and username fields as UNIQUE KEY in MySQL. My problems a Even by setting both the username and email address fields as UNIQUE KEY, it is not working as I can register using the same email address or username. How to solve this? I have coded also to prevent this problem of duplicate so that the user will be notified to choose another email or username, but I am having a warning. My PHP codes: Code: [Select] <?php if ($_SERVER['REQUEST_METHOD'] == 'POST') { if (isset($_POST['fname']) && isset($_POST['lname'])&& isset($_POST['emailr']) && isset($_POST['user']) && isset($_POST['pass'])) { //Assignng variables $firstname = mysql_real_escape_string($_POST['fname']); $lastname = mysql_real_escape_string($_POST['lname']); $email = mysql_real_escape_string($_POST['emailr']); $uname = mysql_real_escape_string($_POST['user']); $pwd = mysql_real_escape_string($_POST['pass']); $pmd= md5($pwd); //Database $connect = mysql_connect('localhost', 'root', '') or die ('Connection Failed'); mysql_select_db('registration', $connect) or die ('Connection Failed'); //Registration codes if (empty($firstname) || empty($lastname) || empty($email) || empty($uname) || empty($pmd)) { echo '<p class="error">All fields are required to fill!</p>'; return false; } elseif (strlen($firstname) && (strlen($lastname) < '2')) { echo '<p class="error">Invalid first name or last name!</p>'; return false; } elseif (filter_var($firstname, FILTER_VALIDATE_INT) || (filter_var($lastname, FILTER_VALIDATE_INT))) { echo '<p class="error">First name or last name cannot be integers!</p>'; return false; } elseif (!filter_var($email, FILTER_VALIDATE_EMAIL)) { echo '<p class="error">Email address not valid!</p>'; return false; } elseif (strlen($uname) && (strlen($pmd) < '6' )) { echo '<p class="error">Username or password must be minimum 6 characters!</p>'; return false; } else { $query = "INSERT INTO login (id, firstname, lastname, emailaddress, username, password) VALUES('', '$firstname', '$lastname', '$email', '$uname', '$pmd')"; mysql_query($query, $connect); if (mysql_num_rows(mysql_query("SELECT * FROM login WHERE emailaddress = '$email' username = '$uname'"))) { echo '<p class="fail">This email or username is already taken!</p>'; } } } } ?> The warning message I am getting: Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\miniimagehosting\register.php on line 44 Hi , I have one question .. Can I split showing of content of dynamic list in 2 parts , when I echo list in code .. Code: [Select] <?php // Run a select query to get my letest 8 items // Connect to the MySQL database include "../connect_to_mysql.php"; $dynamicList = ""; $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 8"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $product_name = $row["product_name"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $dynamicList .= '<table width="100%" border="2" cellspacing="2" cellpadding="2"> <tr> <td width="17%" valign="top"><a href="product.php?id=' . $id . '"><img style="border:#666 1px solid;" src="inventory_images/' . $id . '.jpg" alt="' . $product_name . '" width="77" height="102" border="2" /></a></td> <td width="83%" valign="top">' . $product_name . '<br /> $' . $price . '<br /> <a href="product.php?id=' . $id . '">View Product Details</a></td> </tr> </table>'; } } else { $dynamicList = "We have no products listed in our store yet"; } mysql_close(); ?> Code: [Select] <p><?php echo $dynamicList; ?><br /> </p> It works ok, and putting my files, everything works, but when I put 8 pictures with price and other details, it just show one image with details and another image below with details, and the third image below and so on .. Can I split dynamic list to show 4 images with details on the left side and 4 on the right side? Thank you in advance for help , if is possible This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=330251.0
<?php
require_once('init.php');
$results = $db->prepare('SELECT file_location, file_type, file_size, id as media_id WHERE id = ? LIMIT 1;');
$results->execute(array($_GET['id']));
while ($row = $results->fetch()) {
header('Content-Type: ' . $row['file_type']);
header('Content-Length: ' . $row['file_size']);
$media_ins = $db->prepare('UPDATE media SET total_clicks = total_clicks + 1 WHERE id = ?');
$media_ins->execute(array($row['id']));
readfile($row['file_location']);
}
For some reason, putting the header('Content-Type....') causes the UPDATE statement to trigger twice increasing the total_clicks by 2. Commenting out the content-type line causes a single update to occur.
init.php contains nothing more than session_start and the database connection. When i comment-out the content-type line the page is blank with no errors or messages.
I'm not sure why this is occurring.
I've been working on developing a CMS blog and now I'm trying to create a slideshow wit Bootstrap Carousel on the homepage to present the dynamic content (images + text) using the data from table 'posts'. I tested this code, and it only presents one post. I mean, It's not possible to go to the next slide. I want to show all the posts on the slides. *The DB connection is already on the includes. The connection was written on a small file called DB.php Home.php
<header>
<div id="carouselExampleIndicators" class="carousel slide" data-ride="carousel">
<ol class="carousel-indicators">
<li data-target="#carouselExampleIndicators" data-slide-to="0" class="active"></li>
<li data-target="#carouselExampleIndicators" data-slide-to="1"></li>
<li data-target="#carouselExampleIndicators" data-slide-to="2"></li>
</ol>
<div class="carousel-inner" role="listbox">
<?php
// The default SQL query
$sql = "SELECT * FROM posts ORDER BY id desc";
$stmt = $ConnectingDB->query($sql);
while ($DataRows = $stmt->fetch()) {
$Id = $DataRows["id"];
$PostTitle = $DataRows["title"];
$Image = $DataRows["image"];
$PostText = $DataRows["post"];
?>
<!-- Slide -->
<div class="carousel-item active" style="background-image: url('uploads/<?php echo $Image; ?>')">
<div class="carousel-caption">
<div class="card-body black">
<h3 class="large-mistral-white"><?php echo $PostTitle; ?></h3>
<p class="small-times-white"><?php echo $PostText; ?></p>
</div>
</div>
</div>
<?php } ?>
</div>
<a class="carousel-control-prev" href="#carouselExampleIndicators" role="button" data-slide="prev">
<span class="carousel-control-prev-icon" aria-hidden="true"></span>
<span class="sr-only">Previous</span>
</a>
<a class="carousel-control-next" href="#carouselExampleIndicators" role="button" data-slide="next">
<span class="carousel-control-next-icon" aria-hidden="true"></span>
<span class="sr-only">Next</span>
</a>
</div>
</header>
DB.php <?php $DSN='mysql:host = localhost; dbname=everybody_blog'; $ConnectingDB = new PDO($DSN,'root',''); ?>
|
|||||||