[williamh26]

I have this code that populates a dynamic table data

PHP Code:

mysql_connect("localhost", "taverna", "taverna") or die(mysql_error());
mysql_select_db("taverna") or die(mysql_error());


// SQL query to interact with info from our database
$sql = mysql_query("SELECT gal_id, pic_name FROM galeria ORDER BY gal_id DESC LIMIT 100");
$i = 0;
// Establish the output variable
$dyn_table = '<table border="1" cellpadding="0">';
while($row = mysql_fetch_array($sql)){

    $gal_id = $row["gal_id"];
    $pic_name = $row["pic_name"];
    //<img src='fiestaspic/$row[pic_name]'
    if ($i % 3 == 0) { // if $i is divisible by our target number (in this case "3")
        $dyn_table .= '<tr><td>' .$pic_name.'</td>';
    } else {
        $dyn_table .= '<td>' . $pic_name . '</td>';
    }
    $i++;
}
$dyn_table .= '</tr></table>';
?> 

PHP Code:

<?php echo $dyn_table; ?>

However i want to populate a image,,, but for some reason onlygive the name of the picture not the image.... can you guys help me please thank you

[nafirici]

add the path to the image and put the name in like so:

PHP Code:

$dyn_table .= "<td><img src='path/to/image/" . $pic_name . "' /></td>";