[SOLVED] Problem With Showing Records From Database

[amin1982]

Hey Guys,

 

Need some help here!

 

I'm building a site at the moment which is database driven. I have a menu which lists about 20 or so clubs in london.

 

When each of these clubs are clicked the user should be taken to a page which shows content about that club. However this works when I tested it with two venues on the database but when I added more the links kept showing the same data on each page even if the URL would change.

 

if you go to:

 

http://www.omaraguestlist.co.uk/test/database/londonclub3.php

 

You will see a list of 4 venues. If you click on "24 London" everything is ok... as is when you click on funkybuddha (these were the initial two venues in the database) however when I click on Amika or Paper I simply get the same info thats for funkybuddha!!!

 

The URL changes as you would expect so you are taken to a new page!

 

This is driving me nuts! The database has been populated correctly and I cant see whats wrong with the code.

 

<?php 
require_once ('includes/buffer.php');
require_once ('includes/config.inc.php'); 
require_once ('includes/mysql_connect.php'); // Connect to the database.
?>



<?php
$query = "SELECT * FROM `londonclubs2` WHERE active='1' ORDER BY londonclub";
$result = mysql_query($query) or die('Error, query failed');
while ($list = mysql_fetch_array($result)) {
echo "<a href ='londonclub3.php?londonclub={$list['londonclub']}' class='Text' >{$list['clubName']}</a> <br>";
}
?>
<?php

$problem = FALSE; 

if (isset($_GET['londonclub'])) {

$londonclub= (int) $_GET['londonclub'];

$sql = "SELECT * FROM londonclubs2 WHERE londonclub = $londonclub";

$result = mysql_query ($sql, $dbc);

if (mysql_num_rows($result)) { // Good to go!

$row = mysql_fetch_array ($result, MYSQLI_ASSOC);

echo "{$row['image001']}";

echo "{$row['image002']}";

echo "{$row['image003']}";

echo "{$row['image004']}";

echo "{$row['heading']}";

echo "{$row['description']}";

echo "{$row['reviewform']}";

}
}
?>

 

Could there be something missing or incorrect with the code? Or is the database set up incorrectlty?

 

Please help...

[marcus]

1. Why do you have (int) in your $londonclub variable

2. Add an or die statemeont on your second result [before the //good to go! comment]

3. Why waste space, just use mysql_fetch_assoc($result)

[harristweed]

$londonclub= (int) $_GET['londonclub'];

 

will produce a '0'

 

try:

$londonclub= trim($_GET['londonclub']);

[Crew-Portal]

<table width="575" border="0">
  <tr bgcolor="#999999">
    <td><strong>Colum Number 1:<strong></td>
    <td><strong>Colum Number 2:</strong></td>
    <td><strong>Colum Number 3:</strong></td>
    <td><strong>Colum Number 4:</strong></td>
    <td><strong>Colum Number 5:</strong></td>
  </tr>
<?php
$query="select * from database";  // query string stored in a variable
$rt=mysql_query($query);          // query executed 
echo mysql_error();                    // if any error is there that will be printed to the screen
while($nt=mysql_fetch_array($rt)){
echo "
  <tr bgcolor=#990000>
    <td><strong><span class=white>$nt[colum1]</span></strong></td>
    <td><strong><span class=white>$nt[colum2]</span></strong></td>
    <td><strong><span class=white>$nt[colum3]</span></strong></td>
    <td><strong><span class=white>$nt[colum4]</span></strong></td>
    <td><strong><span class=white>$nt[colum5]</span></strong></td>
  </tr>
";
}
?>
</table>

 

^^^ Is the easiest way to do it!

[amin1982]

hey guys

 

 

just wanted to say a big thank you to you all!

 

i took abit of advice from all of you and its seemed to have worked!

 

thanks