Problems with images and tables.

[Anon]

Hello all.

I am relatively new with php and mysql but my problem is this:

I have an image gallery where I have stored images according to their matching unique id-numbers. This is simple and works just fine but could someone please tell me who can I bring the images to the webpage and format them into tables.

I want to use a 3 column table where I'd have one picture per column and list would go on at least to 50.

It's difficult to explain but my idea is this:

<table width="450" cellspacing="1" cellpadding="1">
<tr>
<td>---first image-----</td>
<td>---secong image----</td>
<td>---third image----</td>
</tr>
<tr>
<td>---fourth image---</td>
<td>---fifth image----</td>
<td>---sixth iamge----</td>
</tr>
...and so on
</table>

Could someone help me please!

[rulian]

this should be relativily simple, all you gotta do is set up a for loop to perform the task , and a switch to tell you where in the table are you. for example you run a query saying you want it to get the url of the images in your db.

$query=mysql_query("select imageurl from images");
$queryamount=mysql_num_rows($query);
echo "
<table width='100%' border=0>
";
for ($i=0; $i<$howmany; $i++){
if (!isset($what)) $what='one';
switch($what){
case "one";
$do="<tr><td>";
$what='two';
break;
case "two":
$do="</td><td>";
$what="three";
break;
case "three":
$do="</td><td>";
$do2="</tr>";
$what="one";
break;
}

$image=mysql_result($query, $i, 'imageurl');
echo $do;
echo "<a href='$imageurl'><img src='$imageurl' border=0></a>";
if (isset($do2)) echo $do2;
}
echo "</table>";

I dont know if this code will work, but this should give you an idea