Display mysql data in formatted table - php

Question

I am retrieving data from a database and displaying in a table. I would like the table to be in a 4 x 3 layout. The data retrieves just fine, it's the layout that isn't working. Here is my code:
<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');
$db = new mysqli('localhost', 'root', '', 'ezwayautos');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
$sql = "SELECT * FROM vehicles ORDER BY RAND() LIMIT 12";
if(!$result = $db->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}
echo "<table>";
while($row = $result->fetch_assoc())
{
echo "<tr>";
echo "<td height='160' valign='top' class='featured'>";
echo "<div class='Image-Thumbnail'>";
echo "<a href=''>";
echo "<img src='".$row['image']."' width='160' height='54'>";
echo "</a>";
echo "</div> <a href=''>" .$row['vehicle_name']. "</a>";
echo "</td>";
echo "</tr>";
}
echo "</table>";
?>
What is supposed to go in each cell is a picture of the vehicle with the name underneath the picture.
Here is an example from another website as to how I would like it to look:
http://www.denisonmotors.com
I am not stealing any information from their site I am just trying to get my data to display in the format that they have on their site.
I am not concerned about the empty href tag as they will be filled in later.
After doing some research, I think that I have to use 2 for loops to create the table stating how many columns and row that I want.

Answer 1

Is it as simple as moving the last 3 lines that close the outer table outside the while loop?
<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');
$db = new mysqli('localhost', 'root', '', 'ezwayautos');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
$sql = "SELECT * FROM vehicles ORDER BY RAND() LIMIT 12";
if(!$result = $db->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}
echo "<table>";
echo "<tr>";
echo "<td>";
while($row = $result->fetch_assoc())
{
echo "<table border='0' cellpadding='0' cellspacing='0'>";
echo "<tr>";
echo "<td height='160' valign='top' class='featured'>";
echo "<div class='Image-Thumbnail'>";
echo "<a href='inventory/view/7995179/2005-Volvo-XC90-4dr-2.html'>";
echo "<img src='".$row['image']."' width='160' height='54' alt='".$row['vehicle_name']."'>";
echo "</a>";
echo "</div> <a href='inventory/view/7995179/2005-Volvo-XC90-4dr-2.html'>" .$row['vehicle_name']. "</a>";
echo "</td>";
echo "</tr>";
echo "</table>";
}
echo "</td>";
echo "</tr>";
echo "</table>";
?>

Answer 2

The table generation is not correct. I think you don't mean to create a table inside another table, so I think this is the way you should do:
echo "<table>";
while($row = $result->fetch_assoc())
{
echo "<tr>";
echo "<td height='160' valign='top' class='featured'>";
echo "<div class='Image-Thumbnail'>";
echo "<a href='inventory/view/7995179/2005-Volvo-XC90-4dr-2.html'>";
echo "<img src='".$row['image']."' width='160' height='54' alt='".$row['vehicle_name']."'>";
echo "</a>";
echo "</div> <a href='inventory/view/7995179/2005-Volvo-XC90-4dr-2.html'>" .$row['vehicle_name']. "</a>";
echo "</td>";
echo "</tr>";
}
echo "</table>";
You have to create the table once, and then a row for each record.