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PHP - Table Wont Display Properly
I have a table that needs to display data and its formatting should allow a scroll bar after a bit of length. In this case, the below code seems to allow it to continue well past the footer of the page... Did I miss something obvious?
<div class="col-md-12">
<div class="card card-plain">
<div class="header">
<h4 class="title">Current Vendors</h4>
<p class="category">Vendors listed as active within VendorBase.</p>
</div>
<?php
$con=mysqli_connect("localhost","root","test","vendors");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM vendor_data");
echo " <div class='content table-responsive table-full-width'>
<table class='table table-hover'>
<thead>
<th>Name</th>
<th>Type</th>
<th>Company</th>
<th>Email</th>
<th>SOC2 Report</th>
<th>Status</th>
</thead>
<tbody>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['type'] . "</td>";
echo "<td>" . $row['company'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['soc'] . "</td>";
echo "<td>" . $row['status'] . "</td>";
echo "</tr>";
}
echo "</table>";
echo "</div>";
mysqli_close($con);
?>
</table>
</div>
</div>
</div>
Similar TutorialsHI im having a problem displaying an image in an iframe or img tag on my php page if i set the source to google it displays fine in the iframe! but if i set the source to a page on my server it doesnt display. It displays in both fire fox and chrome!???? Any help would be greatly appreciated Thanks Sean Hi i am trying to display image using 'CAT_BIO' => (isset($_GET['c'])) ? '<br />' '<img src="images/' . $row['image'] . ' " width="150" height="200" />' '', but i get the error Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in i think a comma or something missing can anyone spot where i am going wrong?? the other way i tried is <img src=\"images/{$row['image']}\" width=\"150\" height=\"200\" />" this one give no error bt doesnt display the anything it might be cz wrong path how do i make to go out one folder the get inside images folder?? not sure if its the solution help ok I am getting an error over something simple, can anyone help
Also how do i delete previous threads that I dont wont to continue?
Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Post.user_id' in 'on clause' SQL Query: SELECT `Post`.`id`, `Post`.`title`, `Post`.`body`, `Post`.`created`, `Post`.`modified`, `User`.`two`, `User`.`three`, `User`.`id` FROM `jagguy`.`posts` AS `Post` LEFT JOIN `jagguy`.`users` AS `User` ON (`Post`.`user_id` = `User`.`id`) WHERE 1 = 1 class User extends AppModel { public $hasMany = 'Post'; } class Post extends AppModel { /*var $name='User';*/ public $belongsTo = array('User'); } controller //// class PostsController extends AppController { public function home2() { $everything= $this->Post->find('all', array('contain' => array('User'))); } view <?php foreach ($everything as $item): echo '<tr><td>'. $item['Post']['id']. '</td>'; echo '<td>'. $item['Post']['title'].'</td>'; echo '<td>'. $item['Post']['body'].'</td>'; Hi Guys, I have a php mail script that calculates a branch to send it to dependant on the postcode which works fine as it goes to the right email and when I echo the script to show branch name it comes up with the correct branch name. However when you open the email I cannot get it to display the 'branch name'. It will display all the other bits of information that has been 'posted' from the form. But wont display the branch name form an sql database table. The code below proves that it finds the branch name. while($row = mysql_fetch_array( $result )) { $ter = $row['Ter']; $branchemail = $row['BranchEmail']; $branchname = $row['BranchName']; } This is the PHP Mail in HTML part below <td bgcolor=#FFFFFF bordercolor=#666666 valign=top>' . ( $row['BranchName']) . '</td> But when the email comes through the branch name is always blank. However on this page I can get it to echo the branch name and it displays the branch name and also emails to the branch email address. Any ideas why it doesn't work. Cheers, S Hey guys so my code below is not working, it will get the app information for the user but its only display 3 results like every time a new user "installs" an new app like it stops showing the last result and starts displaying the new result. Like for example right now it should be showing 7 results but only displaying 3 is there something wrong with my query. (PS: this is just developement testing purposes so thats why my code is sorta sloppy) Thanks!
$default_apps = mysql_query("SELECT * FROM apps WHERE `default`='1'") or die(mysql_error());
$user_apps = mysql_query("SELECT * FROM user_apps WHERE `user_id`='$user_id'") or die(mysql_error());
while($row = mysql_fetch_array($default_apps))
{
$url = $row['download_url'];
$name = $row['name'];
echo $row['name'];
echo "<a href='$url'>$name</a><br />";
}
while($raw = mysql_fetch_array($user_apps)){
$app_id = $raw['app_id'];
}
$select_user_apps = mysql_query("SELECT * FROM apps WHERE `app_id`='$app_id' ");
while($rop = mysql_fetch_array($select_user_apps))
{
$name = $rop['name'];
$url = $rop['download_url'];
echo $name;
echo $url;
}
Hi, Im trying to get an image to display from a reference in the database but am having no joy. At the minute Im just playing about with PHP to get familiar with it so you can ignore most of the code. The line im interested in is: <td><?php echo "<img src=\"C:\wamp\www\fermpix\Pics\'{$row["Name"]}'\">";?></td> When I view the page in my browser I see the attached. Can anybody see what im doing wrong? Heres the full page: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <title>Browse Upload Files</title> </head> <body bgcolor="white"> <?php error_reporting(E_ALL); include 'db.inc'; $query = "SELECT ID, Name, Path, Date, Description FROM Pics"; if (!($connection = @ mysql_pconnect($hostName, $username, $password))) showerror(); if (!mysql_select_db("fermpics", $connection)) showerror(); if (!($result = @ mysql_query ($query, $connection))) showerror(); ?> <h1>Image database</h1> <h3>Click <a href="insert.php">here</a> to upload an image.</h3> <?php //require 'disclaimer'; if ($row = @ mysql_fetch_array($result)) { ?> <table> <col span="1" align="right"> <tr> <th>File Name</th> <th>Date</th> <th>Image</th> </tr> <?php do { ?> <tr> <td><?php echo "{$row["Name"]}";?></td> <td><?php echo "{$row["Date"]}";?></td> <td><?php echo "<img src=\"C:\wamp\www\fermpix\Pics\'{$row["Name"]}'\">";?></td> </tr> <?php } while ($row = @ mysql_fetch_array($result)); ?> </table> <?php } // if mysql_fetch_array() else echo "<h3>There are no images to display</h3>\n"; ?> </body> </html> I would appreciate any help. Cheers Paul Hey, Im having a problem where it wont show the members name only TBA, i was wondering if anyone could spot whats wrong Code - http://pastebin.com/8mEGcTqd page - http://tw.tghq.org/rosters/hto_roster_blistig.php Hi all,
I need some help with my code as I've a trouble with sending the emails with images attachments. I'm using Pear Mail library to send the emails so when I send the emails with attachments to gmail, I am unable to see the attached images in my gmail inbox as there is no images show on the bottom of the subject unless when I open on my email so I can see the images attachments. And I am unable to see the images on yahoo when i sent the emails, so I sent a test email on my webmail with the images as attachments and I can see the images on gmail and yahoo with no problem. I think there is a problem with my PHP script that need to be resolve.
//attachments
if (is_array($email_attachments)) {
foreach ($email_attachments as $attachments) {
$filename = str_replace('uploads/', '', $attachments);
$attachment = '';
$file_path = '';
$type = '';
if (strpos($filename, '.png') !== false) {
$type .= 'image/png';
}
// ADD attachment(s)
$attachment1 .= "--$boundary1\r\n";
$attachment1 .= "Content-Type: $type; name=\"$filename\"\r\n";
$attachment1 .= "Content-Transfer-Encoding: base64\r\n";
$attachment1 .= "Content-Disposition: attachment; filename=\"$filename\"\r\n";
$attachment1 .= "\r\n\r\n";
$attachment1 .= $attachment;
$attachment1 .= "\r\n\r\n";
$body .= $attachment1;
$mime->addAttachment($file_path, $type);
}
}
Here is the full code:
<?php
require_once('Mail.php');
require_once('Mail/mime.php');
require_once('Mail/IMAPv2.php');
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
//Connect to the database
include('config.php');
// Connect to the server:
$username = 'myusername';
$password = 'mypassword';
$smtp_hostname = "smtp.mydomain.com";
$port = "587";
$attached_files = array();
$inline_images = array();
global $inline_images;
if (isset($_POST['send_to']))
{
$from = "Mark <name@mydomain.com>";
$to_email = $_POST['send_to'];
$subject = $_POST['email_subject'];
$message ='<div dir="ltr">' . $_POST['email_body'] . '</div>';
$email_attachments = $_POST['email_attachment'];
$total_emails = count($to_email);
$email_number = $_POST['email_number'];
$sent_message = $message;
$attachment1 = '';
$inline_id = 0.1;
$success = '';
$base64 = [];
$src = [];
//Check if the images is base64
if (strpos($message, 'data:image/') !== false)
{
// read all image tags into an array
preg_match_all('/<img.*?src="data:image\/.*;.*,(.*)".*?>/i', $message, $match, PREG_PATTERN_ORDER);
$base64 = array_pop($match);
if (is_array($src))
{
foreach($base64 as $images)
{
$type = end(explode('/', (explode(';', $images))[0]));
$filename = md5(time().uniqid()). '.' . $type;
$base64_string = str_replace('data:image/png;base64,', '', $images);
$base64_string = str_replace(' ', '+', $base64_string);
$decoded = base64_decode($base64_string);
$fp = fopen("uploads/". $filename, "w+");
fwrite($fp, $decoded);
fclose($fp);
}
}
}
//check if the inline images is in the array
if (strpos($message, '<img src=') !== false)
{
// read all image tags into an array
preg_match_all('@src="([^"]+)"@' , $message, $match);
$src = array_pop($match);
if (is_array($src))
{
foreach ($src as $key => $value)
{
$parsed = parse_url($src[$key]);
$filename = basename($parsed['path']);
$content = file_get_contents($src[$key]);
if (!file_exists($filename))
{
$fp = fopen("uploads/". $filename, "w+");
fwrite($fp, $content);
fclose($fp);
}
}
}
}
$boundary1 = '###'.md5(microtime()).'###';
$boundary2 = '###'.md5(microtime().rand(99,999)).'###';
foreach ($to_email as $to)
{
$name = '';
$email = '';
if (strpos($to, ' <') !== false) {
$name_str = explode(' <', $to);
$email_str = explode(' <', $to);
$name = $name_str[0];
$email = str_replace('>', '', $email_str);
$email = $email[1];
}
$messageID = sprintf("<%s.%s@%s>",
base_convert(microtime(), 10, 36),
base_convert(bin2hex(openssl_random_pseudo_bytes(8)), 16, 36),
'mydomain.com');
$message_id = getMessageid(isset($message_id));
$now = new DateTime();
$email_id = $now->getTimestamp();
$sent_date = date('Y-m-d H:i:s');
$sent_mailbox1 = $link->prepare("SELECT * FROM sent WHERE message_id = ?");
$sent_mailbox1->execute([$message_id]);
$emailID = '';
if (!$name == '' && !$email == '') {
if ($name == $email) {
$to = $email;
}
}
if ($sent_mailbox1->rowCount() == 0)
{
$sent_mailbox1 = $link->prepare("INSERT INTO sent (from_email, to_email, message_id) VALUES (?,?,?)");
if ($sent_mailbox1->execute([$from, $to, $message_id]))
{
$emailID = $link->lastInsertId();
}
}
//check if the images is base64
if (is_array($base64))
{
$sent_message .= 'http://mydomain.com/u/?id='.$email_id.'&attid='.$inline_id.'&msgid='.$message_id.'&view=attachment&display=view';
$inline_id += 0.1;
}
else if (!$base64 == '') {
$sent_message .= 'http://mydomain.com/u/?id='.$email_id.'&attid='.$inline_id.'&msgid='.$message_id.'&view=attachment&display=view';
$inline_id += 0.1;
}
//check if the inline images is in the array
if (is_array($src))
{
$sent_message .= 'http://mydomain.com/u/?id='.$email_id.'&attid='.$inline_id.'&msgid='.$message_id.'&view=attachment&display=view';
$inline_id += 0.1;
}
else if (!$src == '') {
$sent_message .= 'http://mydomain.com/u/?id='.$email_id.'&attid='.$inline_id.'&msgid='.$message_id.'&view=attachment&display=view';
$inline_id += 0.1;
}
$message .= '<img src="http://mydomain.com/track/Images/signature.gif?id='.$emailID.'&etc='.time(). '" ' . 'style="width:0;max-height:0;overflow:hidden" alt="">';
$headers = array ('From' => $from,
'To' => $to,
'Subject' => $subject,
'Reply-To' => $from,
//'Content-Type' => 'Content-Type: text/plain; charset="UTF-8"',
'Content-Type' => 'Content-Type: multipart/mixed; boundary="=_d909e7abc497193ad3b6636530382391"',
'MIME-Version' => '1.0',
'Received' => 'from mail.mydomain.com',
'Date' => date("r"),
'Message-ID' => '<'.sha1(microtime(true)).'@mydomain.com>');
$crlf = "\r\n";
$mime = new Mail_mime(array('eol' => $crlf));
//$mime = new Mail_mime("\r\n");
$html = $message;
$text = strip_tags($html);
$body = $html;
$mime->setTXTBody($text);
$mime->setHTMLBody($html);
//check if the img tags have url called display=view
if (strpos($message, 'display=view') !== false) {
$pattern = '@src="([^"]+)"@';
$message = preg_replace_callback($pattern,"setImageLinks", $message);
foreach ($inline_images as $inline_image) {
$file_path = $inline_image;
$typeInt = imagetype($file_path);
//code goes here to find the imagetype case
switch ($typeInt) {
case IMG_GIF:
$imageType = 'image/gif';
break;
case IMG_JPG:
$imageType = 'image/jpg';
break;
case IMG_JPEG:
$imageType = 'image/jpeg';
break;
case IMG_PNG:
$imageType = 'image/png';
break;
case IMG_WBMP:
$imageType = 'image/wbmp';
break;
case IMG_XPM:
$imageType = 'image/xpm';
break;
default:
$imageType = 'unknown';
}
$mime->addHTMLImage($file_path, $imageType);
}
}
//attachments
if (is_array($email_attachments)) {
foreach ($email_attachments as $attachments) {
$filename = str_replace('uploads/', '', $attachments);
$attachment = '';
$file_path = '';
$type = '';
if (strpos($filename, '.png') !== false) {
$type .= 'image/png';
}
// ADD attachment(s)
$attachment1 .= "--$boundary1\r\n";
$attachment1 .= "Content-Type: $type; name=\"$filename\"\r\n";
$attachment1 .= "Content-Transfer-Encoding: base64\r\n";
$attachment1 .= "Content-Disposition: attachment; filename=\"$filename\"\r\n";
$attachment1 .= "\r\n\r\n";
$attachment1 .= $attachment;
$attachment1 .= "\r\n\r\n";
$body .= $attachment1;
$mime->addAttachment($file_path, $type);
}
}
// always call these methods in this order
$mime_params = array(
'text_encoding' => '7bit',
'text_charset' => '"UTF-8"',
'html_charset' => '"UTF-8"',
'head_charset' => '"UTF-8"'
);
//$body = $mime->get(array('text_encoding' => '8bit','html_encoding' => '8bit'));
$body = $mime->get($mime_params);
$headers = $mime->headers($headers);
$smtp_params = array ('host' => $smtp_hostname,
'port' => $port,
'auth' => true, // Note 1
'username' => $username,
'password' => $password);
$smtp = Mail::factory('smtp', $smtp_params);
$mail = $smtp->send($to, $headers, $body);
if (PEAR::isError($mail))
{
echo("<p>" . $mail->getMessage() . "</p>");
}
else
{
echo("<p>Email has been sent!</p>");
$response = array("success"=>$success);
echo json_encode($response);
}
}
}
Do you know why the images wont show up on yahoo and gmail when I send the images as attachment?
I have checked on the image path and i have put the correct image path so it should work fine.
Any advice would be much appreciated. Thanks in advance.
Edited June 5, 2020 by mark107 This is the query which should likes salestrack productid with printers productid, but even though there are data no result is showing and I really dont know why. Please help Code: [Select] SELECT salestrack.orderid AS orderid, salestrack.salesman AS salesman, printers.rrp AS price, salestrack.name AS name, salestrack.phone AS phone, salestrack.email AS email, printers.name AS printername, salestrack.orderdate AS orderdate, salestrack.status AS STATUS FROM printers, salestrack WHERE salestrack.productid = printers.productid AND salestrack.status = 'OPEN' ORDER BY salestrack.orderdate hi there, the code below is suppose to display something like the attachement "code2" but instead it displays something like attachement "code1" please assist in find what is wrong with my echo lines.
<?php
$counter = 2;
$sqlq="select * from state WHERE status = 0 ";
$categorysqlq = mysql_query($sqlq);
$varq = mysql_num_rows($categorysqlq);
while($catfetchq = mysql_fetch_array($categorysqlq))
{
$cnty = $catfetchq[0];
$sqllq="select * from vehicle WHERE country = '$cnty' ";
$categorysqllq = mysql_query($sqllq);
$numsql = "select * from branchaddr WHERE state = '$cnty' ";
$numquery = mysql_query($numsql);
$varqa = mysql_num_rows($numquery);
$cntyfetchq= mysql_fetch_array($numquery);
if($varq != 0){
if($counter == 2){
echo "<tr><td><a href=\"transport2.php?id=".$cntyfetchq['state'].">".$catfetchq[1]."(<span style=\"color:red\">".$varqa."</span>)</a></td>";
$counter--;
}
else{
echo "<td><a href=\"transport2.php?id=".$cntyfetchq['state'].">".$catfetchq[1]."(<span style=\"color:red\">".$varqa."</span>)</a></td></tr>";
$counter = 2;
}
}
}
?>
Attached Files
code1.png 12.65KB
0 downloads
code2.png 25.34KB
0 downloadsWell I have a script file that was originally written like this: Code: [Select] $query = "UPDATE ".$prefix."users SET nickname='".$nickname."' WHERE username='".$loggedinname."'"; mysql_query($query); $query = "UPDATE ".$prefix."users SET gender='".$gender."' WHERE username='".$loggedinname."'"; mysql_query($query); $query = "UPDATE ".$prefix."users SET color='".$color."' WHERE username='".$loggedinname."'"; mysql_query($query); $query = "UPDATE ".$prefix."users SET profile='".$profile."' WHERE username='".$loggedinname."'"; mysql_query($query); $query = "UPDATE ".$prefix."users SET favpet='".$favpet."' WHERE username='".$loggedinname."'"; mysql_query($query); $query = "UPDATE ".$prefix."users SET about='".$about."' WHERE username='".$loggedinname."'"; mysql_query($query); I tried to simplify it by rewriting the following codes below, but unfortunately it did not work. It wouldnt give any errors, but the columns aint updated at all: Code: [Select] mysql_query("UPDATE ".$prefix."users SET nickname='".$nickname."' , gender='".$gender."' , color='".$color."' , profile='".$profile."' , favpet='".$favpet."' , about='".$about."' , WHERE username='".$loggedinname."'"); Did I make any mistake rewriting the codes? Or is it actually impossible to update six columns using only one mysql_query? Please help... Hello to all, I have problem figuring out how to properly display data fetched from MySQL database in a HTML table. In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table. The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'
Can someone help me on this, I guess the problem lies in the while loop? <thead> <tr> <th data-column-id="id" data-type="numeric">ID</th> <th data-column-id="email">Subscriber's Email</th> <th data-column-id="validity">Validity</th> <th data-column-id="valid_from">Valid From</th> <th data-column-id="valid_to">Valid To</th> </tr> </thead> Here is part of the PHP code:
<?php
while($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
echo '
<tr>
<td>'.$row["id"].'</td>
';
while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) {
echo '
<td>'.$row1["email"].'</td>
';
}
if($row["validity"] == 1) {
echo '<td>'.$row["validity"].' month</td>';
}else{
echo '<td>'.$row["validity"].' months</td>';
}
echo ' <td>'.$row["valid_from"].'</td>
<td>'.$row["valid_to"].'</td>
</tr>';
}
?>
Thank you. Hi guys,
I've got the following command im trying to push out how it would look if i ran it in ssh
<?php
if (!function_exists("ssh2_connect")) die("function ssh2_connect doesn't exist");
if(!($con = ssh2_connect("hostname", 22))){
echo "fail: unable to establish connection\n";
} else {
if(!ssh2_auth_password($con, "username", "password")) {
echo "fail: unable to authenticate\n";
} else {
echo "okay: logged in...\n <br>";
if (!($stream = ssh2_exec($con, "showspace" ))) {
echo "fail: unable to execute command\n";
} else {
stream_set_blocking($stream, true);
$data = "";
while ($buf = fread($stream,4096)) {
echo $data .= $buf;
}
fclose($stream);
}
}
}
?>
This displays as:
---Estimated(MB)---- ---Estimated(MB)---- RawFree UsableFree ---Estimated(MB)---- RawFree UsableFree 135770112 67885056
in putty it displays as:
TestRepo cli% showspaceHey everyone, I have a problem here.Now as u see my data is being displayed one after another in vertical manner.But what do i want to do is the entire table being displayed in same page continuously one after another in horizontal manner.how would i do that? (below a screen shot is given how my table looks like.) <code> <?php for($i=0;$i<=25;$i++) { if($i%5==0) { ?> <table> <tr> <th scope="row">U_Id :</th> <td><?php echo 'uid'; ?></td> </tr> <tr> <th scope="row">Name :</th> <td><?php echo "name"; ?></td> </tr> <tr> <th scope="row">Teamname :</th> <td><?php echo "teamname"; ?></td> </tr> <tr> <th scope="row">Coins :</th> <td><?php echo 'coins';?></td> </tr> <tr> <th scope="row">Cash:</th> <td><?php echo 'cash'; ?></td> </tr> </table> <br/> <?php //echo "$i<br/>"; } } ?> </code> i have a stupid simple problem here, but ive never done this exactly this way before and im having a tough time working it out. was looking for any suggestions. my script is working fine: $table_name = "plan"; $sql = "SELECT id, plan_name FROM $table_name ORDER BY plan_name"; $query = mysql_query($sql); if(mysql_num_rows($query) > 0) { echo "<table>"; while($row = mysql_fetch_array($query)) { echo "<tr><td>" . $row['plan_name'] ."</tr></td>"; } echo "</table>"; ... but i have another table company that are related to plan_name i have the company_id field in my plan table for the relationship. and that id is related to the company id field of course. all i am trying to do is display the company_name next to $row['plan_name'], so i know which plans are related to which company. i think i need to create two seperate queries, but is there a way to include everyone in one query? is there a better way? here is my sql: plan table: `id` int(11) NOT NULL AUTO_INCREMENT, `company_id` int(11) NOT NULL, `plan_name` varchar(255) NOT NULL, ... company table `id` int(11) NOT NULL AUTO_INCREMENT, `plan_id` int(11) NOT NULL, `company_name` varchar(255) NOT NULL,... it also seemed to be overkill to have two while loops running.. i am just thinking out loud on the best approach to this. This may be in here already and I'm sorry for not being able to find it but.. I just want to display some information that I get from the database in a HTML table only using 2 columns. Do I just set i=1 and run a if statement to see what column I am on? like Code: [Select] $i =0 While ($row = msyql_fetch_array($result)) { if ($i==0) { // start a new row echo "<tr>"; $i = $i++; }else{ //columns echo "<td>info</td>"; $i=$i++; } if ($i < 2) { //end the row reset $i echo "</tr>"; $i =0; } } Or something like that, am I heading in right direction? Thanks Stephen Hi, i cant seem to get something working, should be simple but its not working for me. I just need to only display a table if a variable in my table = a certain value. The column in the table is called 'option1_available' and if its value is set to 'Y' i want it to display a table. Appreciate any help, Thanks Hello, I want to make a list with 2 columns on one of my pages with link to categories on my page. But they appear in only one column (on below the other) and i want them to be like: Pictures Videos Pictures2 Videos2 These are my codes <li> <a href="user_album_add.php">Pictures</a> <li> <a href="user_album2_add.php">Pictures2</a> <li> <a href="user_video_add.php">Video</a> <li> <a href="user_video2_add.php">Video2</a> What should i do? Thank you! Hi I have the following code: Code: [Select] $result = mysql_query("SELECT * FROM xbox_games order by gameid"); $row = mysql_fetch_assoc($result); $id = $row["gameid"]; $title = $row["gametitle"]; $cover = $row["cover"]; <?php echo $title;?> Which displays only the first result from my database. How can i change this to display all the results either as a list, or in a table? Thanks I have the code below displaying images and image names. I want these to display in a table 2 rows high by the needed number of columns to show all the images in the directory. I have no idea what to do. What I am getting now is a single column with each image in its own row. <?php $path = "./uploaded/"; $dir_handle = @opendir($path) or die("Unable to open folder"); while (false != ($file = readdir($dir_handle))) { if($file == "index.php") continue; if($file == ".") continue; if($file == "..") continue; //show in a table 2 rows by required/needed number of columns echo'<div>'; echo '<table border="1">'; echo "<img src='$path/$file' alt='$file'>"."<img src='$file' alt='$file'>"; echo'</table>'; echo '<div>'; } ?> |
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