fetch MySQL content, error ?

[thaidomizil]

Hello

 

i want to fetch the content from my MySQL db for one table and show it on a page, i'm using this code:

<?php include("inc/connect.php"); ?>
<?php
$result = $_REQUEST['result'];
?>

<?php
$query = mysql_query("select * from listtest order by id asc");
while($result = mysql_fetch_array($query))
{
?>
  <tr>
    <td><?php echo $result['id']; ?></td>
    <td><?php echo $result['code']; ?></td>
    <td align="center"><a href="edit.php?id=<?php echo $result['id']; ?>">View</a></td>

    <td align="center"> </td>
  </tr>
<?php } ?>

 

it's throwing me this error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in 32qgtf.php on line 30

 

What am i doing wrong ?

 

/Edit: line 29/30 is the mysql query / fetch.

[mikesta707]

It seems like your query is failing. Try changing your query line to

 

$query = mysql_query("select * from listtest order by id asc") or die(mysql_error());

 

and post if there is an error (and what the error is)

 

Note: or die() is never good to use in production. its ok for debugging purposes only.

[thaidomizil]

I changed the query line, the result is the same, no more specific error is given.

[mikesta707]

is there anything in your database?

[cunoodle2]

If you have phpmyadmin log into there, select your table and then in the mySql box copy and paste the query in.. "select * from listtest order by id asc"  What happens?  Any results?

 

What if you View>Source of the page.  Is anything showing up then?  It could be you have an issue with your html

[PFMaBiSmAd]

I'm going to guess that in your actual code, you are overwriting the result resource that is in $query, inside of the loop. Is that the actual complete code that is inside the while(){} loop?