need help with code.

The code bellow trows me an error on line 11... can anyone help me as to why is do that ?

<?php
$con = mysql_connect("localhost","asd","das");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("asd", $con);
$result = mysql_query('SELECT * FROM `images` WHERE `tags` LIKE '%' . $_GET['tag'] . '%');
while($row = mysql_fetch_array($result))
  {
  echo '<a href='http://mydomain.com/out.php/'.$row['name'].'</a>';
  echo "<br/>";
  }
?>			

PHP:

 

$result = mysql_query("SELECT * FROM `images` WHERE `tags` LIKE '%" . mysql_real_escape_string($_GET['tag']) . "%'");

PHP:

 

thanks nico!

after putting that in i get Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/izlik/public_html/tags.php on line 14

and using crimson Editor i already se one on line 14

<?php
$con = mysql_connect("localhost","asd","asd");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("das", $con);
$result = mysql_query("SELECT * FROM `images` WHERE `tags` LIKE '%" . mysql_real_escape_string($_GET['tag']) . "%'");
while($row = mysql_fetch_array($result))
  {
  echo '<a href='http://mydomain.com/out.php/'.$row['name'].'</a>';
  echo "<br/>";
  }
?>

PHP:

 

  echo '<a href="http://mydomain.com/out.php/">' .$row['name'].'</a>';

PHP:

 

now this one poped up up!

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/izlik/public_html/tags.php on line 11

<?php
$con = mysql_connect("localhost","asd","asd");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("asd", $con);
$result = $result = mysql_query("SELECT * FROM `images` WHERE `tags` LIKE '%" . mysql_real_escape_string($_GET['tag']) . "%'");
while($row = mysql_fetch_array($result))
  {
  echo '<a href="http://mydomain/out.php/">' .$row['name'].'</a>';
  echo "<br/>";
  }
?>			
	

PHP:

 

test your query to see if it return something ..

 

<?php

$con = mysql_connect("localhost","asd","asd") OR die('Could not connect: ' . mysql_error());
mysql_select_db("asd", $con);

$result = mysql_query("
	SELECT *
	FROM `images`
	WHERE `tags`
	LIKE '%" . mysql_real_escape_string($_GET['tag']) . "%'
")
OR die(mysql_error());

while($row = mysql_fetch_array($result))
{
	echo '<a href="http://mydomain/out.php/">' .$row['name'].'</a>';
	echo "<br/>\n";
}

?>

PHP:

 

thanks a lot nicko, your just awsome! adding rep for you and thanks a lot!

however i have one last question, is it someway to make this easy on the database to load? like let's say that there is a huge ammount of data to load and many people do it at the same time.

 

$filename = preg_replace('~\.[a-z0-9]+$~i', '.html', $row['name']);
echo '<a href="http://mydomain/out.php/' . $filename .'">' .$row['name'].'</a>';

PHP:

 

thanks again nico

i know i promised that was the last question, but i have another problem i just cant get my hand around "need to study more php"

as you se in the code bellow i but a limit to 30 images. now i know "limit" has more parameters so that if there is more then 30 images it could make a "next page" avaliable to view the rest of the images. do you think you could help me with this?

i know i asked a lot already but if you could help me with this i would love you forever!

<?php

$con = mysql_connect("localhost","asd","asd") OR die('Could not connect: ' . mysql_error());
mysql_select_db("das", $con);

$result = mysql_query("
    SELECT *
    FROM `images`
    WHERE `tags`
    LIKE '%" . mysql_real_escape_string($_GET['tag']) . "%'
    ORDER BY views DESC LIMIT 50
")
OR die(mysql_error());

while($row = mysql_fetch_array($result))
{
    echo '<a href="http://mydomain.net/show.php/' .$row['id'].'_' .$row['name'].'"><img src="http://www.mydomain.net/out.php/t' .$row['id'].'_' .$row['name'].'"></a>';
    echo "<br/>\n";
}

?>		

PHP: