The Database Forum

**pushedtomb** · 04-17-2005, 02:16 PM

Can anyone c something wrong with this coding? As im getting Resource id#6 when i load it on my site, http://www.pushedtomb.co.uk/news.php , ne ideas?

PHP Code:


			
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>..::Pushedtomb.co.uk::..</title>
<link href="css/css.css" rel="stylesheet" type="text/css">
<script language="JavaScript" type="text/JavaScript">
<!--
function MM_reloadPage(init) {  //reloads the window if Nav4 resized
  if (init==true) with (navigator) {if ((appName=="Netscape")&&(parseInt(appVersion)==4)) {
    document.MM_pgW=innerWidth; document.MM_pgH=innerHeight; onresize=MM_reloadPage; }}
  else if (innerWidth!=document.MM_pgW || innerHeight!=document.MM_pgH) location.reload();
}
MM_reloadPage(true);
//-->
</script>
<style type="text/css">
<!--
.style4 {color: #D0E0F0}
-->
</style>
</head>
<body>
<div align="center">
<table width="100%" border="1" bordercolor="#FFFFFF">
  <tr>
    <th align="center" valign="top" class="border" scope="col"><marquee>
      <img src="images/3dtext_2555.gif" width="512" height="47">
      </marquee></th>
  </tr>
</table>
<div align="left">
<table width="15%" height="75%" border="1" bordercolor="#FFFFFF">
</table>
<?php //navigation
include('nav.php');
?>
<td width="83%" align="center" valign="top" class="bgblend" scope="col" aname="site"><p>
    <?php include('Connections/news.php');
 @mysql_select_db('links') or die('Unable to find database');
 $sql = mysql_query('SELECT * '
        . ' FROM `news` '
        . ' WHERE 1 '
        . ' ORDER BY `id` ASC LIMIT 0, 30');
  echo $sql;?>
  </p></td>
</tr>
</table>
</body>
</html>
<p>
  <?php //footer
include('footer.php');
?>

**leavethisplace** · 04-17-2005, 03:08 PM

Quote:

sql = mysql_query('SELECT * '
. ' FROM `news` '
. ' WHERE 1 '
. ' ORDER BY `id` ASC LIMIT 0, 30');
echo $sql;?>

Pretty sure it's this bit, try doing the query a bit like this, it's taken from my actual news publishing program

PHP Code:


			
$query = "SELECT id, date, topic, content, archive, createdby

          FROM News WHERE id = '$ID'";

    $result = mysql_query($query)

        or die("Query failed: " . mysql_error());



    while ( $row = mysql_fetch_array($result) ) {

        

                                   argument or presented variables

        

    }

You can't echo a sql query because it return it in an array, so use the while loop to go through each result (or "row") and extract the information from each cell using $row['cell_name'];

Hope that makes sense and helps

**pushedtomb** · 04-17-2005, 03:35 PM

im still getting an error.

PHP Code:


			
<?php include('Connections/news.php');

    @mysql_select_db('links') or die('Unable to find database');

 $query = "SELECT id, name, url, FROM News WHERE id = '$ID'"; 

    $result = mysql_query($query) 

        or die("Query failed: " . mysql_error()); 



    while ( $row = mysql_fetch_array($result) )?>

check that page agen

**leavethisplace** · 04-17-2005, 03:39 PM

$query = "SELECT id, name, url, FROM news WHERE id = '$ID'";

Hang on you've changed your code now, do you want to display all news or a specific item of news? if all news, if because you had FROM News, SQL is CaSe SenSiTive! So make sure your news table actually has a capital N or not.

If you want to get all the news, use this:

$query = "SELECT id, name, url, FROM News WHERE 1 ORDER BY id";

**pushedtomb** · 04-17-2005, 03:51 PM

ok.. start agen.. Table name i want to access is news. with colums id name url. name contains Music, Rugby, Football, Lotto, then under url music.php, rugby.php, football.php, lotto.php. and i want to display them as links on news.php

**leavethisplace** · 04-17-2005, 03:54 PM

ok cool, well i sent u a PM, but check this out:

$query = "SELECT id, name, url, FROM News WHERE id = '$ID'"; is wrong, take out the comma before FROM and after url. Thats a SQL error you're getting.

The while loop will allow you to construct a generic template to place all the information you need, using $row['name']; and $row['url']

That any more help?

**pushedtomb** · 04-17-2005, 03:55 PM

yup cheers.. will pm u if i have more trouble

**leavethisplace** · 04-17-2005, 03:56 PM

Cool beans, my sigs gone all funny.