this should be easy.....table problem.

[gat]

This bit of code displays thumbnails, pulled from mysql.
the problem is not that, it is the formatting of the table.

Is their code that will dynamically make the table the right size for the number of images in the DB.

this code works, but the table is always 1 coloumn, with a number of rows.
If i had 30 images in the DB, i would like the table to be 6 coloums by 5 rows.

many thanks for any help

---

$sql = "select * from images";
$result = mysql_query($sql);
$num_results = mysql_num_rows($result);
echo "<images>\n";
for($i=0;$i<$num_results;$i++)
{
$row = mysql_fetch_array($result);
if(($i%6)==0)
echo "<table border='1' cellpadding='0' cellspacing='6' bordercolor='#000000'>";
echo "<tr>\n";
echo "<td>";
echo "<a href=\"display.php?file=".$row['pic_path']."\"><img src=\"".$row['thum_path']."\"></a>";
echo "</td>\n";
if(($i%5)==2)
echo "</tr>\n";
}
echo "</table>\n";
?>

---

[ir4z0r]

echo "<table border='1' cellpadding='0' cellspacing='6' bordercolor='#000000'>";

for($i=0;$i<$num_results;$i++) 
{ 
  $row = mysql_fetch_array($result); 
  if(($i%6)==0)
  {
     echo "<tr>\n"; 
  }
  echo "<td>"; 
  echo "<a href=\"display.php?file=".$row['pic_path']."\"><img src=\"".$row['thum_path']."\"></a>"; 
  echo "</td>\n";

  if (($i%6)==5)
  {
    echo "</tr>\n"; 
  }
}
  echo "</table>\n";

try that one

[gat]

many thanks.... i just want to ask one more thing.
in the db, i also hold "pic_area" and "pic_descrip".

i would like these displayed on the display.php page.

the code for this page (that displays the thumnails parent image)
is as follows.

________________________________---
<?php
echo "<img src=\"".$GET['file']."\" border='3' >";
echo ("Details, $pic_descrip");
?>

_________________________-
this gets the parent image fine.
but it does not get the pic_descrip.

i would also like "pic_area" to be displayed.

do i need to make changes to the previous page???

The current error i get is.

"Notice: Undefined variable: pic_descrip in c:\program files\apache group\apache\htdocs\mysql\display.php on line 13"

line 13 = echo ("Details, $pic_descrip");

im sure this is a simple one too.

many thanks

[jpmoriarty]

well to get to pic descrip you'll have to query your database to return results based on the picture thats being displayed.

[ir4z0r]

append to your created link the infos you need e.g.

echo "<a href=\"display.php?file=".$row['pic_path']."&pic_descrip=" . $row['pic_descrip'] . "\"><img src=\"".$row['thum_path']."\"></a>";

details.php:
echo "Details: " . $_GET['pic_descrip'];

[gat]

all working, great. thanks guys.

one other security question.
bearing in mind that i have user name and password to connect to mysql.

when i do view source, the connection code is not shown, so i presume the log on detail will not be shown?

is this true?

or are there some security issues that i may face.?

thanks

[ir4z0r]

the "view source" only displays the output created by your script
mainly caused by echo().
So nobody should be able to view your pw's and stuff

in general be careful with variables you retrieve via URL ($_GET),
for example if you choose to use sth like open.php?file=bla
check bla for / because one might manually edit the url and do

open.php?file=../../passwdfile