php in tables...

gat

echo "<TABLE CLASS='table' BORDER='1' CELLPADDING='6' CELLSPACING='0' BORDERCOLOUR='#000000' >";
echo "<TR CLASS='tablebackground'>";
echo "<TD COLSPAN='5'>";
echo "<P CLASS='tabletext'>";
echo "<a href=\"display.php?file=".$result['pic_path']."\">
<img src=\"".$result['thum_path']."\"></a><br>";;
echo "</P>";
echo "</TD>";
echo "</TR>";

This code puts each thumbnail in its own table in one cell.
and i can understand why.

How can i make each thum appear in a different cell of the same table.

do i have to create an array, or is their an easier way?

jpmoriarty

$sql = "select * from table";
$result = mysql_query($sql);
$num_results = mysql_num_rows($result);
echo "<table>\n";
for($i=0;$i<$num_results;$i++)
{
  $row = mysql_fetch_array($result);
  if(($i%3)==0)
    echo "<tr>\n";
  echo "<td>";
  echo "<a href=\"display.php?file=".$row['pic_path']."\"><img src=\"".$row['thum_path']."\"></a>";
  echo "</td>\n";
  if($i%3)==2)
    echo "</tr>\n";
}
echo "</table>\n";

that will give you a 3 by x table - if you want more than 3 across, then change the ($i%3) lines to ($i%however many cols you want) and change the ($i%3)==2 to ($i%however many cols you want)==however many cols you want minus one.

gat

sorry, please bear with my ignorance. I swaped to noobs after my last error....duh..

Anyway, the code looks good to me but i get

parse error, unexpected T_IS_EQUAL in c:\program files\apache group\apache\htdocs\mysql\displaythum_intable2.php on line 26

and line 26 = ' if($i%3)==2) '

thanks.

jpmoriarty

it should be if(($i%3)==2)

gat

im sorry, i noticed that shortly after posting that message.
I should have a good look really before i post.

but that then creates this error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\mysql\displaythum_intable2.php on line 16

line 16 =

$num_results = mysql_num_rows($result);

jpmoriarty

i think that means it's getting no results.

please god dont tell me you're using the SQL statement that i gave you above - you need to replace table with the actual name of your table...

BuzzLY

Perhaps, jp, you should be more clear:

$sql = "select * from [i]replace_this_variable_with_your_own_table_name [/i]";