[SOLVED] Php table help please

[hane]

Attached is an eg. of how I want to display info that i get from my mysql database. If you can just point me in the right direction I would really appreciated it.

 

[attachment deleted by admin]

[ILMV]

What? You need to be a thousand times more specific...

 

Are you returning a list of image URLs you want displaying in that format?

 

 

Also, this is in the wrong forum...

 

Edit: Sorry It is int he correct forum

[gevans]

What do you have so far, it's easier to work with your existing code than to write it out in full, plus you'll learn more that way

[hane]

I've got it here is the code

 

<?php
//set the number of columns
$columns = 3;
$query = "SELECT * FROM africanskyjewellery_items";
$result = mysql_query($query);

//we add this line because we need to know the number of rows
$num_rows = mysql_num_rows($result);
echo "<table summary='' border='0' width='100%'>\n";

//changed this to a for loop so we can use the number of rows
for($i = 0; $i < $num_rows; $i++) {
    $row = mysql_fetch_array($result);
    if($i % $columns == 0) {
        //if there is no remainder, we want to start a new row
        echo "<tr>\n";
    }
    echo "<td colspan='2'>"; 
	echo "<img src='images/" . $row['africanskyjewellery_items_ref'] . ".jpg' width='150'></a><br />";
    echo "<b>Ref:</b>". $row['africanskyjewellery_items_ref'] ."<br />";
	echo "<b>Price:</b>R" . $row['africanskyjewellery_items_price'] . "<br />"; 
	echo "</td>\n";
    if(($i % $columns) == ($columns - 1) || ($i + 1) == $num_rows) {
        //if there is a remainder of 1, end the row
        //or if there is nothing left in our result set, end the row
        echo "</tr>\n";
    }
}
echo "</table>\n";
?>